Pengertian Limit Fungsi Trigonometri
Soal dan Pembahasan Limit Fungsi Trigonometri. Limit fungsi trigonometri adalah limit fungsi yang melibatkan fungsi trigonometri seperti fungsi sinus, cosinus, tangen, dan lain-lain. Penyelesaian soal-soal limit selalu melakukan uji substitusi secara langsung. Jika nilai fungsi $f(x)$ untuk $x$ mendekati atau menuju suatu nilai tertentu, baik dari kiri maupun dari kanan memiliki nilai tertentu dan tidak $\dfrac{0}{0},$ maka uji substitusi disebut sukses. Dengan demikian nilai dari limitnya telah ditemukan, yaitu $\displaystyle \lim_{x \to a} = c$. Tetapi jika hasilnya adalah $\dfrac{0}{0},$ maka harus dilakukan upaya untuk menemukan hasil yang sebenarnya. Berikut ini adalah rumus-rumus penting yang dapat membantu mencari solusi untuk masalah-masalah limit fungsi trigonometri.Rumus-rumus Limit Fungsi Trigonometri
$1.\ \displaystyle \lim_{x \to 0}\dfrac{sin\ x}{x} = 1$
$2.\ \displaystyle \lim_{x \to 0}\dfrac{x}{sin\ x} = 1$
$3.\ \displaystyle \lim_{x \to 0}\dfrac{tan\ x}{x} = 1$
$4.\ \displaystyle \lim_{x \to 0}\dfrac{x}{tan\ x} = 1$
$5.\ \displaystyle \lim_{x \to 0}\dfrac{tan\ x}{sin\ x} = 1$
$6.\ \displaystyle \lim_{x \to 0}\dfrac{sin\ x}{tan\ x} = 1$
$7.\ \displaystyle \lim_{x \to 0}\dfrac{sin\ mx}{nx} = \dfrac{m}{n}$
$8.\ \displaystyle \lim_{x \to 0}\dfrac{mx}{sin\ nx} = \dfrac{m}{n}$
$9.\ \displaystyle \lim_{x \to 0}\dfrac{tan\ mx}{nx} = \dfrac{m}{n}$
$10.\ \displaystyle \lim_{x \to 0}\dfrac{mx}{tan\ nx} = \dfrac{m}{n}$
$11.\ \displaystyle \lim_{x \to 0}\dfrac{tan\ mx}{sin\ nx} = \dfrac{m}{n}$
$12.\ \displaystyle \lim_{x \to 0}\dfrac{sin\ mx}{tan\ nx} = \dfrac{m}{n}$
$13.\ \displaystyle \lim_{x \to 0}\dfrac{sin\ m(x - a)}{n(x - a)} = \dfrac{m}{n}$
$14.\ \displaystyle \lim_{x \to 0}\dfrac{m(x - a)}{sin\ n(x - a)} = \dfrac{m}{n}$
$15.\ \displaystyle \lim_{x \to 0}\dfrac{tan\ m(x - a)}{n(x - a)} = \dfrac{m}{n}$
$16.\ \displaystyle \lim_{x \to 0}\dfrac{m(x - a)}{tan\ n(x - a)} = \dfrac{m}{n}$
$17.\ \displaystyle \lim_{x \to 0}\dfrac{tan\ m(x - a)}{sin\ n(x - a)} = \dfrac{m}{n}$
$18.\ \displaystyle \lim_{x \to 0}\dfrac{sin\ m(x - a)}{tan\ n(x - a)} = \dfrac{m}{n}$
Rumus-rumus Pendukung
$1.\ sin^2\ ax + cos^2\ ax = 1$
$2.\ sin^2\ ax = 1 - cos^2\ ax$
$3.\ cos^2\ ax = 1 - sin^2\ x$
$4.\ cos\ 2x = cos^2\ x - sin^2\ x$
$5.\ 1 - cos\ 2x = 2sin^2\ x$
$6.\ cos\ 2x + 1 = 2cos^2\ x$
$7.\ sin\ 2x = 2sinxcosx$
$8.\ tan\ x = \dfrac{sin\ x}{cos\ x}$
$9.\ cot\ x = \dfrac{cos\ x}{sin\ x}$
Supaya lebih jelas, simak soal dan pembahasan limit fungsi trigonometri berikut.$2.\ \displaystyle \lim_{x \to 0}\dfrac{x}{sin\ x} = 1$
$3.\ \displaystyle \lim_{x \to 0}\dfrac{tan\ x}{x} = 1$
$4.\ \displaystyle \lim_{x \to 0}\dfrac{x}{tan\ x} = 1$
$5.\ \displaystyle \lim_{x \to 0}\dfrac{tan\ x}{sin\ x} = 1$
$6.\ \displaystyle \lim_{x \to 0}\dfrac{sin\ x}{tan\ x} = 1$
$7.\ \displaystyle \lim_{x \to 0}\dfrac{sin\ mx}{nx} = \dfrac{m}{n}$
$8.\ \displaystyle \lim_{x \to 0}\dfrac{mx}{sin\ nx} = \dfrac{m}{n}$
$9.\ \displaystyle \lim_{x \to 0}\dfrac{tan\ mx}{nx} = \dfrac{m}{n}$
$10.\ \displaystyle \lim_{x \to 0}\dfrac{mx}{tan\ nx} = \dfrac{m}{n}$
$11.\ \displaystyle \lim_{x \to 0}\dfrac{tan\ mx}{sin\ nx} = \dfrac{m}{n}$
$12.\ \displaystyle \lim_{x \to 0}\dfrac{sin\ mx}{tan\ nx} = \dfrac{m}{n}$
$13.\ \displaystyle \lim_{x \to 0}\dfrac{sin\ m(x - a)}{n(x - a)} = \dfrac{m}{n}$
$14.\ \displaystyle \lim_{x \to 0}\dfrac{m(x - a)}{sin\ n(x - a)} = \dfrac{m}{n}$
$15.\ \displaystyle \lim_{x \to 0}\dfrac{tan\ m(x - a)}{n(x - a)} = \dfrac{m}{n}$
$16.\ \displaystyle \lim_{x \to 0}\dfrac{m(x - a)}{tan\ n(x - a)} = \dfrac{m}{n}$
$17.\ \displaystyle \lim_{x \to 0}\dfrac{tan\ m(x - a)}{sin\ n(x - a)} = \dfrac{m}{n}$
$18.\ \displaystyle \lim_{x \to 0}\dfrac{sin\ m(x - a)}{tan\ n(x - a)} = \dfrac{m}{n}$
Rumus-rumus Pendukung
$1.\ sin^2\ ax + cos^2\ ax = 1$
$2.\ sin^2\ ax = 1 - cos^2\ ax$
$3.\ cos^2\ ax = 1 - sin^2\ x$
$4.\ cos\ 2x = cos^2\ x - sin^2\ x$
$5.\ 1 - cos\ 2x = 2sin^2\ x$
$6.\ cos\ 2x + 1 = 2cos^2\ x$
$7.\ sin\ 2x = 2sinxcosx$
$8.\ tan\ x = \dfrac{sin\ x}{cos\ x}$
$9.\ cot\ x = \dfrac{cos\ x}{sin\ x}$
Contoh Soal dan Pembahasan Limit Fungsi Trigonometri
$1.\ \displaystyle \lim_{x \to 0}\dfrac{1 - cos\ 4x}{2x\ sin\ 4x} =\ .\ .\ .\ .$
$A.\ 1$
$B.\ \dfrac12$
$C.\ 0$
$D.\ -\dfrac12$
$E.\ -1$
[Soal dan Pembahasan Limit Fungsi Trigonometri UN 2016 MtkIPA]
$A.\ 1$
$B.\ \dfrac12$
$C.\ 0$
$D.\ -\dfrac12$
$E.\ -1$
[Soal dan Pembahasan Limit Fungsi Trigonometri UN 2016 MtkIPA]
$cos\ 4x = cos(2x + 2x)$
$cos\ 4x = cos^2\ 2x - sin^2\ 2x$
$cos\ 4x = 1 - sin^2\ 2x - sin^2\ 2x$
$cos\ 4x = 1 - 2sin^2\ 2x$
$2sin^2\ 2x = 1 - cos\ 4x$ . . . . *
$sin\ 4x = sin(2x + 2x)$
$sin\ 4x = 2sin\ 2xcos\ 2x$ . . . . **
$\displaystyle \lim_{x \to 0}\dfrac{1 - cos\ 4x}{2x\ sin\ 4x} = \displaystyle \lim_{x \to 0}\dfrac{2sin^2\ 2x}{2x.2sin\ 2xcos\ 2x}$
$= \displaystyle \lim_{x \to 0}\dfrac{sin\ 2x}{2x.cos\ 2x}$
$= \displaystyle \lim_{x \to 0}\dfrac{sin\ 2x}{2x}.\displaystyle \lim_{x \to 0}\dfrac{1}{cos\ 2x}$
$= 1.1$
$= 1$
$Jawab:\ A.$
$cos\ 4x = cos^2\ 2x - sin^2\ 2x$
$cos\ 4x = 1 - sin^2\ 2x - sin^2\ 2x$
$cos\ 4x = 1 - 2sin^2\ 2x$
$2sin^2\ 2x = 1 - cos\ 4x$ . . . . *
$sin\ 4x = sin(2x + 2x)$
$sin\ 4x = 2sin\ 2xcos\ 2x$ . . . . **
$\displaystyle \lim_{x \to 0}\dfrac{1 - cos\ 4x}{2x\ sin\ 4x} = \displaystyle \lim_{x \to 0}\dfrac{2sin^2\ 2x}{2x.2sin\ 2xcos\ 2x}$
$= \displaystyle \lim_{x \to 0}\dfrac{sin\ 2x}{2x.cos\ 2x}$
$= \displaystyle \lim_{x \to 0}\dfrac{sin\ 2x}{2x}.\displaystyle \lim_{x \to 0}\dfrac{1}{cos\ 2x}$
$= 1.1$
$= 1$
$Jawab:\ A.$
$2.\ Nilai\ \displaystyle \lim_{x \to 0}\dfrac{xtan\ 3x}{1 - cos^2\ x}\ adalah$ . . . .
$A.\ 0$
$B.\ \dfrac14$
$C.\ \dfrac24$
$D.\ \dfrac34$
$E.\ 1$
[Soal dan Pembahasan Limit Fungsi Trigonometri UN 2015 MtkIPA]
$A.\ 0$
$B.\ \dfrac14$
$C.\ \dfrac24$
$D.\ \dfrac34$
$E.\ 1$
[Soal dan Pembahasan Limit Fungsi Trigonometri UN 2015 MtkIPA]
$1 - cos^2\ x = sin^2\ x$ . . . . *
$\displaystyle \lim_{x \to 0}\dfrac{xtan\ 3x}{1 - cos^2\ x} = \displaystyle \lim_{x \to 0}\dfrac{xtan\ 3x}{sin^2\ x}$
$= \displaystyle \lim_{x \to 0}\dfrac{x}{sin\ x}.\displaystyle \lim_{x \to 0}\dfrac{tan\ 3x}{sin\ x}$
$= 1.3$
$= 3$
$Jawab:\ -$
$\displaystyle \lim_{x \to 0}\dfrac{xtan\ 3x}{1 - cos^2\ x} = \displaystyle \lim_{x \to 0}\dfrac{xtan\ 3x}{sin^2\ x}$
$= \displaystyle \lim_{x \to 0}\dfrac{x}{sin\ x}.\displaystyle \lim_{x \to 0}\dfrac{tan\ 3x}{sin\ x}$
$= 1.3$
$= 3$
$Jawab:\ -$
$3.\ Nilai\ \displaystyle \lim_{x \to 0}\dfrac{4xcos\ x}{sin\ x + sin\ 3x} =$ . . . .
$A.\ 4$
$B.\ 3$
$C.\ \dfrac43$
$D.\ 1$
$E.\ \dfrac34$
[Soal dan Pembahasan Limit Fungsi Trigonometri UN 2014 MtkIPA]
$A.\ 4$
$B.\ 3$
$C.\ \dfrac43$
$D.\ 1$
$E.\ \dfrac34$
[Soal dan Pembahasan Limit Fungsi Trigonometri UN 2014 MtkIPA]
$sin\ A + sin\ B = 2sin\dfrac12(A + B)cos\dfrac12(A - B)$
$sin\ x + sin\ 3x = 2sin\dfrac12(x + 3x)cos\dfrac12(x - 3x)$
$= 2sin\ 2xcos\ (-x)$
$= 2sin\ 2xcos\ x → ingat:\ cos\ (-x) = cos\ x$
$\displaystyle \lim_{x \to 0}\dfrac{4xcos\ x}{sin\ x + sin\ 3x} = \displaystyle \lim_{x \to 0}\dfrac{4xcos\ x}{2sin\ 2xcos\ x}$
$=\displaystyle \lim_{x \to 0}\dfrac{4x}{2sin\ 2x} $
$= \dfrac12\displaystyle \lim_{x \to 0}\dfrac{4x}{sin\ 2x}$
$= \dfrac12.2$
$= 1$
$Jawab:\ D.$
$sin\ x + sin\ 3x = 2sin\dfrac12(x + 3x)cos\dfrac12(x - 3x)$
$= 2sin\ 2xcos\ (-x)$
$= 2sin\ 2xcos\ x → ingat:\ cos\ (-x) = cos\ x$
$\displaystyle \lim_{x \to 0}\dfrac{4xcos\ x}{sin\ x + sin\ 3x} = \displaystyle \lim_{x \to 0}\dfrac{4xcos\ x}{2sin\ 2xcos\ x}$
$=\displaystyle \lim_{x \to 0}\dfrac{4x}{2sin\ 2x} $
$= \dfrac12\displaystyle \lim_{x \to 0}\dfrac{4x}{sin\ 2x}$
$= \dfrac12.2$
$= 1$
$Jawab:\ D.$
$4.\ Nilai\ \displaystyle \lim_{x \to 1}\dfrac{sin^2(x - 1)}{x^2 - 2x + 1} =$ . . . .
$A.\ 0$
$B.\ 1$
$C.\ 2$
$D.\ 4$
$E.\ \infty$
[Soal dan Pembahasan Limit Fungsi Trigonometri UN 2013 MtkIPA]
$A.\ 0$
$B.\ 1$
$C.\ 2$
$D.\ 4$
$E.\ \infty$
[Soal dan Pembahasan Limit Fungsi Trigonometri UN 2013 MtkIPA]
$\displaystyle \lim_{x \to 1}\dfrac{sin^2(x - 1)}{x^2 - 2x + 1} = \displaystyle \lim_{x \to 1}\dfrac{sin^2(x - 1)}{(x - 1)^2}$
$= \displaystyle \lim_{(x - 1) \to 0}\left(\dfrac{sin\ (x - 1)}{(x - 1)}\right)^2$
$Misalkan\ x - 1 = p$
$= \displaystyle \lim_{p \to 0}\left(\dfrac{sin\ p}{p}\right)^2$
$= \left(\displaystyle \lim_{p \to 0}\dfrac{sin\ p}{p}\right)^2$
$= 1^2$
$= 1$
$Jawab:\ B.$
$= \displaystyle \lim_{(x - 1) \to 0}\left(\dfrac{sin\ (x - 1)}{(x - 1)}\right)^2$
$Misalkan\ x - 1 = p$
$= \displaystyle \lim_{p \to 0}\left(\dfrac{sin\ p}{p}\right)^2$
$= \left(\displaystyle \lim_{p \to 0}\dfrac{sin\ p}{p}\right)^2$
$= 1^2$
$= 1$
$Jawab:\ B.$
$5.\ Nilai\ \displaystyle \lim_{x \to 0}\dfrac{cos\ 4x - 1}{x tan\ 2x} =$ . . . .
$A.\ 4$
$B.\ 2$
$C.\ -1$
$D.\ -2$
$E.\ -4$
[Soal dan Pembahasan Limit Fungsi Trigonometri UN 2012 MtkIPA]
$A.\ 4$
$B.\ 2$
$C.\ -1$
$D.\ -2$
$E.\ -4$
[Soal dan Pembahasan Limit Fungsi Trigonometri UN 2012 MtkIPA]
$2sin^2\ 2x = 1 - cos\ 4x$ (lihat pembahasan soal nomor 1.)
$-2sin^2\ 2x = cos\ 4x - 1$ . . . . *
$\displaystyle \lim_{x \to 0}\dfrac{cos\ 4x - 1}{x tan\ 2x} = \displaystyle \lim_{x \to 0}\dfrac{-2sin^2\ 2x}{x tan\ 2x}$
$= -2.\displaystyle \lim_{x \to 0}\dfrac{sin\ 2x}{x}.\displaystyle \lim_{x \to 0}\dfrac{sin\ 2x}{tan\ 2x}$
$= -2.2.1$
$= -4$
$Jawab:\ E.$
$-2sin^2\ 2x = cos\ 4x - 1$ . . . . *
$\displaystyle \lim_{x \to 0}\dfrac{cos\ 4x - 1}{x tan\ 2x} = \displaystyle \lim_{x \to 0}\dfrac{-2sin^2\ 2x}{x tan\ 2x}$
$= -2.\displaystyle \lim_{x \to 0}\dfrac{sin\ 2x}{x}.\displaystyle \lim_{x \to 0}\dfrac{sin\ 2x}{tan\ 2x}$
$= -2.2.1$
$= -4$
$Jawab:\ E.$
$6.\ Nilai\ \displaystyle \lim_{x \to 0}\dfrac{1 - cos\ 2x}{1 - cos\ 4x} =$ . . . .
$A.\ -\dfrac12$
$B.\ -\dfrac14$
$C.\ 0$
$D.\ \dfrac{1}{16}$
$E.\ \dfrac14$
[Soal dan Pembahasan Limit Fungsi Trigonometri UN 2011 MtkIPA]
$A.\ -\dfrac12$
$B.\ -\dfrac14$
$C.\ 0$
$D.\ \dfrac{1}{16}$
$E.\ \dfrac14$
[Soal dan Pembahasan Limit Fungsi Trigonometri UN 2011 MtkIPA]
$1 - cos\ 4x = 2sin^2\ 2x$ . . . . * Lihat pembahasan soal no 1.
$1 - cos\ 2x = 2sin^2\ x$ . . . . *
$\displaystyle \lim_{x \to 0}\dfrac{1 - cos\ 2x}{1 - cos\ 4x} = \displaystyle \lim_{x \to 0}\dfrac{2sin^2\ x}{2sin^2\ 2x}$
$= \displaystyle \lim_{x \to 0}\left(\dfrac{sin\ x}{sin\ 2x}\right)^2$
$= \left(\displaystyle \lim_{x \to 0}\dfrac{sin\ x}{sin\ 2x}\right)^2$
$= \left(\dfrac12 \right)^2$
$= \dfrac14$
$Jawab:\ E.$
$1 - cos\ 2x = 2sin^2\ x$ . . . . *
$\displaystyle \lim_{x \to 0}\dfrac{1 - cos\ 2x}{1 - cos\ 4x} = \displaystyle \lim_{x \to 0}\dfrac{2sin^2\ x}{2sin^2\ 2x}$
$= \displaystyle \lim_{x \to 0}\left(\dfrac{sin\ x}{sin\ 2x}\right)^2$
$= \left(\displaystyle \lim_{x \to 0}\dfrac{sin\ x}{sin\ 2x}\right)^2$
$= \left(\dfrac12 \right)^2$
$= \dfrac14$
$Jawab:\ E.$
$7.\ Nilai\ \displaystyle \lim_{x \to 0}\left(\dfrac{sin\ x + sin\ 5x}{6x} \right) =$ . . . .
$A.\ 2$
$B.\ 1$
$C.\ \dfrac12$
$D.\ \dfrac13$
$E.\ -1$
[Soal dan Pembahasan Limit Fungsi Trigonometri UN 2010 MtkIPA]
$A.\ 2$
$B.\ 1$
$C.\ \dfrac12$
$D.\ \dfrac13$
$E.\ -1$
[Soal dan Pembahasan Limit Fungsi Trigonometri UN 2010 MtkIPA]
cara I:
$sin\ x + sin\ 5x = 2sin\dfrac12(x + 5x)cos\dfrac12(x - 5x)$
$sin\ x + sin\ 5x = 2sin\ 3xcos\ (-2x)$
$sin\ x + sin\ 5x = 2sin\ 3xcos\ 2x$
$\displaystyle \lim_{x \to 0}\left(\dfrac{sin\ x + sin\ 5x}{6x} \right) = \displaystyle \lim_{x \to 0}\left(\dfrac{2sin\ 3xcox\ 2x}{6x} \right)$
$= 2.\displaystyle \lim_{x \to 0}\left(\dfrac{sin\ 3x}{6x} \right).\displaystyle \lim_{x \to 0}cos\ 2x$
$= 2.\dfrac12.1$
$= 1$
cara II:
$Bagi\ pembilang\ dan\ penyebut\ dengan\ x$
$\displaystyle \lim_{x \to 0}\left(\dfrac{sin\ x + sin\ 5x}{6x} \right) = \displaystyle \lim_{x \to 0}\left(\dfrac{\dfrac{sin\ x}{x} + \dfrac{sin\ 5x}{x}}{\dfrac{6x}{x}} \right)$
$= \left(\dfrac{\displaystyle \lim_{x \to 0}\dfrac{sin\ x}{x} + \displaystyle \lim_{x \to 0}\dfrac{sin\ 5x}{x}}{\displaystyle \lim_{x \to 0}\dfrac{6x}{x}} \right)$
$= \dfrac{1 + 5}{6}$
$= 1$
$Jawab:\ B.$
$sin\ x + sin\ 5x = 2sin\dfrac12(x + 5x)cos\dfrac12(x - 5x)$
$sin\ x + sin\ 5x = 2sin\ 3xcos\ (-2x)$
$sin\ x + sin\ 5x = 2sin\ 3xcos\ 2x$
$\displaystyle \lim_{x \to 0}\left(\dfrac{sin\ x + sin\ 5x}{6x} \right) = \displaystyle \lim_{x \to 0}\left(\dfrac{2sin\ 3xcox\ 2x}{6x} \right)$
$= 2.\displaystyle \lim_{x \to 0}\left(\dfrac{sin\ 3x}{6x} \right).\displaystyle \lim_{x \to 0}cos\ 2x$
$= 2.\dfrac12.1$
$= 1$
cara II:
$Bagi\ pembilang\ dan\ penyebut\ dengan\ x$
$\displaystyle \lim_{x \to 0}\left(\dfrac{sin\ x + sin\ 5x}{6x} \right) = \displaystyle \lim_{x \to 0}\left(\dfrac{\dfrac{sin\ x}{x} + \dfrac{sin\ 5x}{x}}{\dfrac{6x}{x}} \right)$
$= \left(\dfrac{\displaystyle \lim_{x \to 0}\dfrac{sin\ x}{x} + \displaystyle \lim_{x \to 0}\dfrac{sin\ 5x}{x}}{\displaystyle \lim_{x \to 0}\dfrac{6x}{x}} \right)$
$= \dfrac{1 + 5}{6}$
$= 1$
$Jawab:\ B.$
$8.$ Nilai dari $\displaystyle \lim_{x \to 3}\dfrac{x^2 + 6x + 9}{2 - 2cos(2x + 6)}$ adalah . . . .
$A.\ 3$
$B.\ 1$
$C.\ \dfrac12$
$D.\ \dfrac13$
$E.\ \dfrac14$
[Soal dan Pembahasan Limit Fungsi Trigonometri UN 2009 MtkIPA]
$A.\ 3$
$B.\ 1$
$C.\ \dfrac12$
$D.\ \dfrac13$
$E.\ \dfrac14$
[Soal dan Pembahasan Limit Fungsi Trigonometri UN 2009 MtkIPA]
$\displaystyle \lim_{x \to -3}\dfrac{x^2 + 6x + 9}{2 - 2cos(2x + 6)}$
$= \displaystyle \lim_{x \to -3}\dfrac{(x + 3)^2}{2(1 - cos2(x + 3))}$
$Analog:$
$1 - cos\ 2x = 2sin^2\ x$
$1 - cos\ 2(x + 3) = 2sin^2\ (x + 3)$
$= \displaystyle \lim_{(x + 3) \to 0}\dfrac{(x + 3)^2}{2.2sin^2\ (x + 3)}$
$= \dfrac14.\displaystyle \lim_{(x + 3) \to 0}\left(\dfrac{(x + 3)}{sin\ (x + 3)}\right)^2$
$= \dfrac14.\left(\displaystyle \lim_{(x + 3) \to 0}\dfrac{(x + 3)}{sin\ (x + 3)}\right)^2$
$Misalkan\ x + 3 = p$
$= \dfrac14.\left(\displaystyle \lim_{p \to 0}\dfrac{p}{sin\ p}\right)^2$
$= \dfrac14.1^2$
$= \dfrac14$
$Jawab:\ E.$
$= \displaystyle \lim_{x \to -3}\dfrac{(x + 3)^2}{2(1 - cos2(x + 3))}$
$Analog:$
$1 - cos\ 2x = 2sin^2\ x$
$1 - cos\ 2(x + 3) = 2sin^2\ (x + 3)$
$= \displaystyle \lim_{(x + 3) \to 0}\dfrac{(x + 3)^2}{2.2sin^2\ (x + 3)}$
$= \dfrac14.\displaystyle \lim_{(x + 3) \to 0}\left(\dfrac{(x + 3)}{sin\ (x + 3)}\right)^2$
$= \dfrac14.\left(\displaystyle \lim_{(x + 3) \to 0}\dfrac{(x + 3)}{sin\ (x + 3)}\right)^2$
$Misalkan\ x + 3 = p$
$= \dfrac14.\left(\displaystyle \lim_{p \to 0}\dfrac{p}{sin\ p}\right)^2$
$= \dfrac14.1^2$
$= \dfrac14$
$Jawab:\ E.$
$9.\ Nilai\ \displaystyle \lim_{x \to 0}\dfrac{2x sin\ 3x}{1 - cos\ 6x} =$ . . . .
$A.\ -1$
$B.\ -\dfrac13$
$C.\ 0$
$D.\ \dfrac13$
$E.\ 1$
[Soal dan Pembahasan Limit Fungsi Trigonometri UN 2007 MtkIPA]
$A.\ -1$
$B.\ -\dfrac13$
$C.\ 0$
$D.\ \dfrac13$
$E.\ 1$
[Soal dan Pembahasan Limit Fungsi Trigonometri UN 2007 MtkIPA]
$1 - cos\ 6x = 2sin^2\ 3x$ . . . . *
$\displaystyle \lim_{x \to 0}\dfrac{2x sin\ 3x}{1 - cos\ 6x} = \displaystyle \lim_{x \to 0}\dfrac{2x sin\ 3x}{2sin^2\ 3x}$
$= \displaystyle \lim_{x \to 0}\dfrac{x}{sin\ 3x}.\displaystyle \lim_{x \to 0}\dfrac{sin\ 3x}{sin\ 3x}$
$= \dfrac13.1$
$= \dfrac13$
$Jawab:\ D.$
$\displaystyle \lim_{x \to 0}\dfrac{2x sin\ 3x}{1 - cos\ 6x} = \displaystyle \lim_{x \to 0}\dfrac{2x sin\ 3x}{2sin^2\ 3x}$
$= \displaystyle \lim_{x \to 0}\dfrac{x}{sin\ 3x}.\displaystyle \lim_{x \to 0}\dfrac{sin\ 3x}{sin\ 3x}$
$= \dfrac13.1$
$= \dfrac13$
$Jawab:\ D.$
$10.$ Nilai dari $\displaystyle \lim_{x \to 0}\dfrac{tan\ 2xcos\ 8x - tan\ 2x}{16x^3}$ adalah . . . .
$A.\ -4$
$B.\ -6$
$C.\ -8$
$D.\ -16$
$E.\ -32$
[Soal dan Pembahasan Limit Fungsi Trigonometri UN 2005 MtkIPA]
$A.\ -4$
$B.\ -6$
$C.\ -8$
$D.\ -16$
$E.\ -32$
[Soal dan Pembahasan Limit Fungsi Trigonometri UN 2005 MtkIPA]
$\displaystyle \lim_{x \to 0}\dfrac{tan\ 2xcos\ 8x - tan\ 2x}{16x^3}$
$= \displaystyle \lim_{x \to 0}\dfrac{tan\ 2x(cos\ 8x - 1)}{16x^3}$
$1 - cos\ 8x = 2sin^2\ 4x$
$cos\ 8x - 1 = -2sin^2\ 4x$
$=\displaystyle \lim_{x \to 0}\dfrac{tan\ 2x.(-2sin^2\ 4x)}{x.16x^2}$
$= -2.\displaystyle \lim_{x \to 0}\dfrac{tan\ 2x}{x}.\displaystyle \lim_{x \to 0}\dfrac{sin^2\ 4x}{16x^2}$
$= -2.\displaystyle \lim_{x \to 0}\dfrac{tan\ 2x}{x}.\displaystyle \lim_{x \to 0}\left(\dfrac{sin\ 4x}{4x}\right)^2$
$= -2.\displaystyle \lim_{x \to 0}\dfrac{tan\ 2x}{x}.\left(\displaystyle \lim_{x \to 0}\dfrac{sin\ 4x}{4x}\right)^2$
$= -2.2.1^2$
$= -4$
$Jawab:\ A.$
$= \displaystyle \lim_{x \to 0}\dfrac{tan\ 2x(cos\ 8x - 1)}{16x^3}$
$1 - cos\ 8x = 2sin^2\ 4x$
$cos\ 8x - 1 = -2sin^2\ 4x$
$=\displaystyle \lim_{x \to 0}\dfrac{tan\ 2x.(-2sin^2\ 4x)}{x.16x^2}$
$= -2.\displaystyle \lim_{x \to 0}\dfrac{tan\ 2x}{x}.\displaystyle \lim_{x \to 0}\dfrac{sin^2\ 4x}{16x^2}$
$= -2.\displaystyle \lim_{x \to 0}\dfrac{tan\ 2x}{x}.\displaystyle \lim_{x \to 0}\left(\dfrac{sin\ 4x}{4x}\right)^2$
$= -2.\displaystyle \lim_{x \to 0}\dfrac{tan\ 2x}{x}.\left(\displaystyle \lim_{x \to 0}\dfrac{sin\ 4x}{4x}\right)^2$
$= -2.2.1^2$
$= -4$
$Jawab:\ A.$
$11.\ Nilai\ \displaystyle \lim_{x \to -2}\dfrac{(x + 6)sin(x + 2)}{x^2 - 3x - 10} =$ . . . .
$A.\ -\dfrac43$
$B.\ -\dfrac47$
$C.\ -\dfrac25$
$D.\ 0$
$E.\ 1$
[Soal dan Pembahasan Limit Fungsi Trigonometri UN 2004 MtkIPA]
$A.\ -\dfrac43$
$B.\ -\dfrac47$
$C.\ -\dfrac25$
$D.\ 0$
$E.\ 1$
[Soal dan Pembahasan Limit Fungsi Trigonometri UN 2004 MtkIPA]
$\displaystyle \lim_{x \to -2}\dfrac{(x + 6)sin(x + 2)}{x^2 - 3x - 10}$
$= \displaystyle \lim_{x \to -2}\dfrac{(x + 6)sin(x + 2)}{(x - 5)(x + 2)}$
$= \displaystyle \lim_{x \to -2}\dfrac{(x + 6)}{(x - 5)}.\displaystyle \lim_{(x + 2) \to 0}\dfrac{sin(x + 2)}{(x + 2)}$
$= \dfrac{-2 + 6}{-2 - 5}.1$
$= -\dfrac47$
$Jawab:\ B.$
$= \displaystyle \lim_{x \to -2}\dfrac{(x + 6)sin(x + 2)}{(x - 5)(x + 2)}$
$= \displaystyle \lim_{x \to -2}\dfrac{(x + 6)}{(x - 5)}.\displaystyle \lim_{(x + 2) \to 0}\dfrac{sin(x + 2)}{(x + 2)}$
$= \dfrac{-2 + 6}{-2 - 5}.1$
$= -\dfrac47$
$Jawab:\ B.$
$12.\ \displaystyle \lim_{x \to 2}\dfrac{sin(2x - 4)}{2 - \sqrt{6 - x}} =$ . . . .
$A.\ -8$
$B.\ -2$
$C.\ 0$
$D.\ 2$
$E.\ 8$
[Soal dan Pembahasan Limit Fungsi Trigonometri SBMPTN 2018 MtkIPA]
$A.\ -8$
$B.\ -2$
$C.\ 0$
$D.\ 2$
$E.\ 8$
[Soal dan Pembahasan Limit Fungsi Trigonometri SBMPTN 2018 MtkIPA]
$\displaystyle \lim_{x \to 2}\dfrac{sin(2x - 4)}{2 - \sqrt{6 - x}}$
$= \displaystyle \lim_{x \to 2}\dfrac{sin2(x - 2)}{2 - \sqrt{6 - x}}.\dfrac{2 + \sqrt{6 - x}}{2 + \sqrt{6 - x}}$
$= \displaystyle \lim_{x \to 2}\dfrac{sin2(x - 2)}{(x - 2)}.(2 + \sqrt{6 - x})$
$= \displaystyle \lim_{(x - 2) \to 0}\dfrac{sin2(x - 2)}{(x - 2)}.\displaystyle \lim_{x \to 2}(2 + \sqrt{6 - x})$
$= 2.(2 + \sqrt{6 - 2})$
$= 2.4$
$= 8$
$Jawab:\ E.$
$= \displaystyle \lim_{x \to 2}\dfrac{sin2(x - 2)}{2 - \sqrt{6 - x}}.\dfrac{2 + \sqrt{6 - x}}{2 + \sqrt{6 - x}}$
$= \displaystyle \lim_{x \to 2}\dfrac{sin2(x - 2)}{(x - 2)}.(2 + \sqrt{6 - x})$
$= \displaystyle \lim_{(x - 2) \to 0}\dfrac{sin2(x - 2)}{(x - 2)}.\displaystyle \lim_{x \to 2}(2 + \sqrt{6 - x})$
$= 2.(2 + \sqrt{6 - 2})$
$= 2.4$
$= 8$
$Jawab:\ E.$
$13.$ Nilai dari $\displaystyle \lim_{x \to \dfrac{\pi}{2}}\dfrac{xcot^2\ x}{1 - sin\ x}$ adalah . . . .
$A.\ \dfrac12$
$B.\ 1$
$C.\ \dfrac{\pi}{2}$
$D.\ 2$
$E.\ \pi$
[Soal dan Pembahasan Limit Fungsi Trigonometri SBMPTN 2017 MtkIPA]
$A.\ \dfrac12$
$B.\ 1$
$C.\ \dfrac{\pi}{2}$
$D.\ 2$
$E.\ \pi$
[Soal dan Pembahasan Limit Fungsi Trigonometri SBMPTN 2017 MtkIPA]
$\displaystyle \lim_{x \to \dfrac{\pi}{2}}\dfrac{xcot^2\ x}{1 - sin\ x} = \displaystyle \lim_{x \to \dfrac{\pi}{2}}\dfrac{xcos^2\ x}{sin^2\ x(1 - sin\ x)}$
$= \displaystyle \lim_{x \to \dfrac{\pi}{2}}\dfrac{x(1 - sin^2\ x)}{sin^2\ x(1 - sin\ x)}$
$= \displaystyle \lim_{x \to \dfrac{\pi}{2}}\dfrac{x(1 - sin\ x)(1 + sin\ x)}{sin^2\ x(1 - sin\ x)}$
$= \displaystyle \lim_{x \to \dfrac{\pi}{2}}\dfrac{x(1 + sin\ x)}{sin^2\ x}$
$= \dfrac{\dfrac{\pi}{2}(1 + sin\ \dfrac{\pi}{2})}{sin^2\ \dfrac{\pi}{2}}$
$= \dfrac{\dfrac{\pi}{2}(1 + 1)}{1}$
$= \pi$
$Jawab:\ E.$
$= \displaystyle \lim_{x \to \dfrac{\pi}{2}}\dfrac{x(1 - sin^2\ x)}{sin^2\ x(1 - sin\ x)}$
$= \displaystyle \lim_{x \to \dfrac{\pi}{2}}\dfrac{x(1 - sin\ x)(1 + sin\ x)}{sin^2\ x(1 - sin\ x)}$
$= \displaystyle \lim_{x \to \dfrac{\pi}{2}}\dfrac{x(1 + sin\ x)}{sin^2\ x}$
$= \dfrac{\dfrac{\pi}{2}(1 + sin\ \dfrac{\pi}{2})}{sin^2\ \dfrac{\pi}{2}}$
$= \dfrac{\dfrac{\pi}{2}(1 + 1)}{1}$
$= \pi$
$Jawab:\ E.$
$14.\ \displaystyle \lim_{x \to \infty}\dfrac{2x^2 tan\left(\dfrac{1}{x}\right) - xsin\left(\dfrac{1}{x} \right) + \dfrac{1}{x}}{xcos\left(\dfrac{2}{x}\right)} =$ . . . .
$A.\ 2$
$B.\ 1$
$C.\ 0$
$D.\ -1$
$E.\ -2$
[Soal dan Pembahasan Limit Fungsi Trigonometri SBMPTN 2017 MtkIPA]
$A.\ 2$
$B.\ 1$
$C.\ 0$
$D.\ -1$
$E.\ -2$
[Soal dan Pembahasan Limit Fungsi Trigonometri SBMPTN 2017 MtkIPA]
$\displaystyle \lim_{x \to \infty}\dfrac{2x^2 tan\left(\dfrac{1}{x}\right) - xsin\left(\dfrac{1}{x} \right) + \dfrac{1}{x}}{xcos\left(\dfrac{2}{x}\right)}$
$= \displaystyle \lim_{x \to \infty}\dfrac{2x tan\left(\dfrac{1}{x}\right) - sin\left(\dfrac{1}{x} \right) + \dfrac{1}{x^2}}{cos\left(\dfrac{2}{x}\right)}$
$= \displaystyle \lim_{\dfrac{1}{x} \to 0}\dfrac{2tan\left(\dfrac{1}{x}\right) - sin\left(\dfrac{1}{x} \right) + \dfrac{1}{x^2}}{\dfrac{1}{x}.cos2\left(\dfrac{1}{x}\right)}$
$Misalkan\ \dfrac{1}{x} = p$
$= \displaystyle \lim_{p \to 0}\dfrac{2tan\ p - sin\ p + p^2}{p.cos\ 2p}$
$= \displaystyle \lim_{p \to 0}\left(\dfrac{2tan\ p}{p} - \dfrac{sin\ p}{p} + p\right).\displaystyle \lim_{p \to 0}\dfrac{1}{cos\ 2p}$
$= (2 - 1 + 0).\dfrac{1}{1}$
$= 1$
$Jawab:\ B.$
$= \displaystyle \lim_{x \to \infty}\dfrac{2x tan\left(\dfrac{1}{x}\right) - sin\left(\dfrac{1}{x} \right) + \dfrac{1}{x^2}}{cos\left(\dfrac{2}{x}\right)}$
$= \displaystyle \lim_{\dfrac{1}{x} \to 0}\dfrac{2tan\left(\dfrac{1}{x}\right) - sin\left(\dfrac{1}{x} \right) + \dfrac{1}{x^2}}{\dfrac{1}{x}.cos2\left(\dfrac{1}{x}\right)}$
$Misalkan\ \dfrac{1}{x} = p$
$= \displaystyle \lim_{p \to 0}\dfrac{2tan\ p - sin\ p + p^2}{p.cos\ 2p}$
$= \displaystyle \lim_{p \to 0}\left(\dfrac{2tan\ p}{p} - \dfrac{sin\ p}{p} + p\right).\displaystyle \lim_{p \to 0}\dfrac{1}{cos\ 2p}$
$= (2 - 1 + 0).\dfrac{1}{1}$
$= 1$
$Jawab:\ B.$
$15.\ \displaystyle \lim_{x \to 0}\dfrac{x(\sqrt{x + 1} - 1)}{1 - cos\ x} =$ . . . .
$A.\ 2$
$B.\ 1$
$C.\ \dfrac12$
$D.\ -\dfrac12$
$E.\ -1$
[Soal dan Pembahasan Limit Fungsi Trigonometri SBMPTN 2016 MtkIPA]
$A.\ 2$
$B.\ 1$
$C.\ \dfrac12$
$D.\ -\dfrac12$
$E.\ -1$
[Soal dan Pembahasan Limit Fungsi Trigonometri SBMPTN 2016 MtkIPA]
$\displaystyle \lim_{x \to 0}\dfrac{x(\sqrt{x + 1} - 1)}{1 - cos\ x}$
$= \displaystyle \lim_{x \to 0}\dfrac{x(\sqrt{x + 1} - 1)}{1 - cos\ x}.\dfrac{(\sqrt{x + 1} + 1)}{(\sqrt{x + 1} + 1)}$
$1 - cos\ x = 2sin^2\ \dfrac12x$
$= \displaystyle \lim_{x \to 0}\dfrac{x^2}{2sin^2\ \dfrac12x}.\dfrac{1}{(\sqrt{x + 1} + 1)}$
$= \dfrac12.\displaystyle \lim_{x \to 0}\dfrac{x^2}{sin^2\ \dfrac12x}.\displaystyle \lim_{x \to 0}\dfrac{1}{(\sqrt{x + 1} + 1)}$
$= \dfrac12.\displaystyle \lim_{x \to 0}\left(\dfrac{x}{sin\ \dfrac12x}\right)^2.\displaystyle \lim_{x \to 0}\dfrac{1}{(\sqrt{x + 1} + 1)}$
$= \dfrac12.\left(\displaystyle \lim_{x \to 0}\dfrac{x}{sin\ \dfrac12x}\right)^2.\displaystyle \lim_{x \to 0}\dfrac{1}{(\sqrt{x + 1} + 1)}$
$= \dfrac12.\displaystyle \left( 2\right)^2.\dfrac{1}{(\sqrt{0 + 1} + 1)}$
$= \dfrac12.4.\dfrac12$
$= 1$
$Jawab:\ B.$
$= \displaystyle \lim_{x \to 0}\dfrac{x(\sqrt{x + 1} - 1)}{1 - cos\ x}.\dfrac{(\sqrt{x + 1} + 1)}{(\sqrt{x + 1} + 1)}$
$1 - cos\ x = 2sin^2\ \dfrac12x$
$= \displaystyle \lim_{x \to 0}\dfrac{x^2}{2sin^2\ \dfrac12x}.\dfrac{1}{(\sqrt{x + 1} + 1)}$
$= \dfrac12.\displaystyle \lim_{x \to 0}\dfrac{x^2}{sin^2\ \dfrac12x}.\displaystyle \lim_{x \to 0}\dfrac{1}{(\sqrt{x + 1} + 1)}$
$= \dfrac12.\displaystyle \lim_{x \to 0}\left(\dfrac{x}{sin\ \dfrac12x}\right)^2.\displaystyle \lim_{x \to 0}\dfrac{1}{(\sqrt{x + 1} + 1)}$
$= \dfrac12.\left(\displaystyle \lim_{x \to 0}\dfrac{x}{sin\ \dfrac12x}\right)^2.\displaystyle \lim_{x \to 0}\dfrac{1}{(\sqrt{x + 1} + 1)}$
$= \dfrac12.\displaystyle \left( 2\right)^2.\dfrac{1}{(\sqrt{0 + 1} + 1)}$
$= \dfrac12.4.\dfrac12$
$= 1$
$Jawab:\ B.$
$16.\ \displaystyle \lim_{x \to a}\dfrac{sin(x - a)tan\sqrt[3]{x - a}} {2\sqrt[3]{x - a} - \sqrt[3]{(x - a)^2}} =$ . . . .
$A.\ -a$
$B.\ -\dfrac{1}{a}$
$C.\ 0$
$D.\ \dfrac{1}{a}$
$E.\ a$
[Soal dan Pembahasan Limit Fungsi Trigonometri SIMAK UI 2017 MtkIPA]
$A.\ -a$
$B.\ -\dfrac{1}{a}$
$C.\ 0$
$D.\ \dfrac{1}{a}$
$E.\ a$
[Soal dan Pembahasan Limit Fungsi Trigonometri SIMAK UI 2017 MtkIPA]
$\displaystyle \lim_{x \to a}\dfrac{sin(x - a)tan\sqrt[3]{x - a}} {2\sqrt[3]{x - a} - \sqrt[3]{(x - a)^2}}$
$= \displaystyle \lim_{x \to a}\dfrac{sin(x - a)tan\sqrt[3]{x - a}} {\sqrt[3]{x - a}(1 - \sqrt[3]{(x - a)})}$
$= \displaystyle \lim_{x \to a}\dfrac{sin(x - a)}{(1 - \sqrt[3]{(x - a)}} .\displaystyle \lim_{(x - a) \to 0}\dfrac{ tan\sqrt[3]{x - a}}{\sqrt[3]{x - a}}$
$= \dfrac{sin(a - a)}{(1 - \sqrt[3]{(a - a)}}.1$
$= \dfrac{sin\ 0}{(1 - \sqrt[3]{0})}$
$= \dfrac{0}{1}$
$= 0$
$Jawab:\ C.$
$= \displaystyle \lim_{x \to a}\dfrac{sin(x - a)tan\sqrt[3]{x - a}} {\sqrt[3]{x - a}(1 - \sqrt[3]{(x - a)})}$
$= \displaystyle \lim_{x \to a}\dfrac{sin(x - a)}{(1 - \sqrt[3]{(x - a)}} .\displaystyle \lim_{(x - a) \to 0}\dfrac{ tan\sqrt[3]{x - a}}{\sqrt[3]{x - a}}$
$= \dfrac{sin(a - a)}{(1 - \sqrt[3]{(a - a)}}.1$
$= \dfrac{sin\ 0}{(1 - \sqrt[3]{0})}$
$= \dfrac{0}{1}$
$= 0$
$Jawab:\ C.$
$17.\ \displaystyle \lim_{x \to 0}\dfrac{sin\ xcos\ x}{\sqrt{\pi + 2sin\ x} - \sqrt{\pi}} =$ . . . .
$A.\ -2\sqrt{\pi}$
$B.\ -\sqrt{\pi}$
$C.\ 0$
$D.\ \sqrt{\pi}$
$E.\ 2\sqrt{\pi}$
[Soal dan Pembahasan Limit Fungsi Trigonometri SBMPTN 2018 MtkIPA]
$A.\ -2\sqrt{\pi}$
$B.\ -\sqrt{\pi}$
$C.\ 0$
$D.\ \sqrt{\pi}$
$E.\ 2\sqrt{\pi}$
[Soal dan Pembahasan Limit Fungsi Trigonometri SBMPTN 2018 MtkIPA]
$\displaystyle \lim_{x \to 0}\dfrac{sin\ xcos\ x}{\sqrt{\pi + 2sin\ x} - \sqrt{\pi}}$
$= \displaystyle \lim_{x \to 0}\dfrac{sin\ xcos\ x}{\sqrt{\pi + 2sin\ x} - \sqrt{\pi}}.\dfrac{(\sqrt{\pi + 2sin\ x} + \sqrt{\pi})}{(\sqrt{\pi + 2sin\ x} + \sqrt{\pi})}$
$= \displaystyle \lim_{x \to 0}\dfrac{sin\ xcos\ x}{2sin\ x}.(\sqrt{\pi + 2sin\ x} + \sqrt{\pi})$
$= \dfrac12.\displaystyle \lim_{x \to 0}(cos\ x).(\sqrt{\pi + 2sin\ x} + \sqrt{\pi})$
$= \dfrac12.(cos\ 0).(\sqrt{\pi + 2sin\ 0} + \sqrt{\pi})$
$= \dfrac12.1.(\sqrt{\pi} + \sqrt{\pi})$
$= \dfrac12\sqrt{\pi}$
$= \sqrt{\pi}$
$Jawab:\ D.$
$= \displaystyle \lim_{x \to 0}\dfrac{sin\ xcos\ x}{\sqrt{\pi + 2sin\ x} - \sqrt{\pi}}.\dfrac{(\sqrt{\pi + 2sin\ x} + \sqrt{\pi})}{(\sqrt{\pi + 2sin\ x} + \sqrt{\pi})}$
$= \displaystyle \lim_{x \to 0}\dfrac{sin\ xcos\ x}{2sin\ x}.(\sqrt{\pi + 2sin\ x} + \sqrt{\pi})$
$= \dfrac12.\displaystyle \lim_{x \to 0}(cos\ x).(\sqrt{\pi + 2sin\ x} + \sqrt{\pi})$
$= \dfrac12.(cos\ 0).(\sqrt{\pi + 2sin\ 0} + \sqrt{\pi})$
$= \dfrac12.1.(\sqrt{\pi} + \sqrt{\pi})$
$= \dfrac12\sqrt{\pi}$
$= \sqrt{\pi}$
$Jawab:\ D.$
$18.\ \displaystyle \lim_{x \to \pi}\dfrac{sin\ x}{\sqrt{\pi + tan\ x} - \sqrt{\pi - tan\ x}} =$ . . . .
$A.\ -2\sqrt{\pi}$
$B.\ -\sqrt{\pi}$
$C.\ 0$
$D.\ \sqrt{\pi}$
$E.\ 2\sqrt{\pi}$
[Soal dan Pembahasan Limit Fungsi Trigonometri SBMPTN 2018 MtkIPA]
$A.\ -2\sqrt{\pi}$
$B.\ -\sqrt{\pi}$
$C.\ 0$
$D.\ \sqrt{\pi}$
$E.\ 2\sqrt{\pi}$
[Soal dan Pembahasan Limit Fungsi Trigonometri SBMPTN 2018 MtkIPA]
$\displaystyle \lim_{x \to \pi}\dfrac{sin\ x}{\sqrt{\pi + tan\ x} - \sqrt{\pi - tan\ x}}$
$= \displaystyle \lim_{x \to \pi}\dfrac{sin\ x}{\sqrt{\pi + tan\ x} - \sqrt{\pi - tan\ x}}.\dfrac{\sqrt{\pi + tan\ x} + \sqrt{\pi - tan\ x}}{\sqrt{\pi + tan\ x} + \sqrt{\pi - tan\ x}}$
$= \displaystyle \lim_{x \to \pi}\dfrac{sin\ x}{2tan\ x}.(\sqrt{\pi + tan\ x} + \sqrt{\pi - tan\ x})$
$= \dfrac12.\displaystyle \lim_{x \to \pi}\dfrac{sin\ x}{tan\ x}.(\sqrt{\pi + tan\ x} + \sqrt{\pi - tan\ x})$
$= \dfrac12.\displaystyle \lim_{x \to \pi}\dfrac{sin\ x}{\dfrac{sin\ x}{cos\ x}}.(\sqrt{\pi + tan\ x} + \sqrt{\pi - tan\ x})$
$= \dfrac12.\displaystyle \lim_{x \to \pi}cos\ x.(\sqrt{\pi + tan\ x} + \sqrt{\pi - tan\ x})$
$= \dfrac12.cos\ \pi.(\sqrt{\pi + tan\ \pi} + \sqrt{\pi - tan\ \pi})$
$= \dfrac12.-1.(\sqrt{\pi + 0} + \sqrt{\pi - 0})$
$= -\dfrac12.2\sqrt{\pi}$
$= -\sqrt{\pi}$
$Jawab:\ B.$
$= \displaystyle \lim_{x \to \pi}\dfrac{sin\ x}{\sqrt{\pi + tan\ x} - \sqrt{\pi - tan\ x}}.\dfrac{\sqrt{\pi + tan\ x} + \sqrt{\pi - tan\ x}}{\sqrt{\pi + tan\ x} + \sqrt{\pi - tan\ x}}$
$= \displaystyle \lim_{x \to \pi}\dfrac{sin\ x}{2tan\ x}.(\sqrt{\pi + tan\ x} + \sqrt{\pi - tan\ x})$
$= \dfrac12.\displaystyle \lim_{x \to \pi}\dfrac{sin\ x}{tan\ x}.(\sqrt{\pi + tan\ x} + \sqrt{\pi - tan\ x})$
$= \dfrac12.\displaystyle \lim_{x \to \pi}\dfrac{sin\ x}{\dfrac{sin\ x}{cos\ x}}.(\sqrt{\pi + tan\ x} + \sqrt{\pi - tan\ x})$
$= \dfrac12.\displaystyle \lim_{x \to \pi}cos\ x.(\sqrt{\pi + tan\ x} + \sqrt{\pi - tan\ x})$
$= \dfrac12.cos\ \pi.(\sqrt{\pi + tan\ \pi} + \sqrt{\pi - tan\ \pi})$
$= \dfrac12.-1.(\sqrt{\pi + 0} + \sqrt{\pi - 0})$
$= -\dfrac12.2\sqrt{\pi}$
$= -\sqrt{\pi}$
$Jawab:\ B.$
$19.\ \displaystyle \lim_{x \to 0}\dfrac{4x + 3xcos\ 2x}{sin\ xcos\ x} =$ . . . .
$A.\ 8$
$B.\ 7$
$C.\ 6$
$D.\ 5$
$E.\ 2$
[Soal dan Pembahasan Limit Fungsi Trigonometri SBMPTN 2017 MtkIPA]
$A.\ 8$
$B.\ 7$
$C.\ 6$
$D.\ 5$
$E.\ 2$
[Soal dan Pembahasan Limit Fungsi Trigonometri SBMPTN 2017 MtkIPA]
$Bagi\ pembilang\ dan\ penyebut\ dengan\ x\ !$
$\displaystyle \lim_{x \to 0}\dfrac{4x + 3xcos\ 2x}{sin\ xcos\ x} = \displaystyle \lim_{x \to 0}\dfrac{\dfrac{4x}{x} + \dfrac{3xcos\ 2x}{x}}{\dfrac{sin\ xcos\ x}{x}}$
$= \dfrac{\displaystyle \lim_{x \to 0}4 + \displaystyle \lim_{x \to 0}3cos\ 2x}{\displaystyle \lim_{x \to 0}\dfrac{sin\ x}{x}.\displaystyle \lim_{x \to 0}cos\ x}$
$= \dfrac{4 + 3.cos\ 2.0}{1.cos\ 0}$
$= \dfrac{4 + 3.1}{1.1}$
$= 7$
$Jawab:\ B.$
$\displaystyle \lim_{x \to 0}\dfrac{4x + 3xcos\ 2x}{sin\ xcos\ x} = \displaystyle \lim_{x \to 0}\dfrac{\dfrac{4x}{x} + \dfrac{3xcos\ 2x}{x}}{\dfrac{sin\ xcos\ x}{x}}$
$= \dfrac{\displaystyle \lim_{x \to 0}4 + \displaystyle \lim_{x \to 0}3cos\ 2x}{\displaystyle \lim_{x \to 0}\dfrac{sin\ x}{x}.\displaystyle \lim_{x \to 0}cos\ x}$
$= \dfrac{4 + 3.cos\ 2.0}{1.cos\ 0}$
$= \dfrac{4 + 3.1}{1.1}$
$= 7$
$Jawab:\ B.$
$20.\ \displaystyle \lim_{x \to \infty}\left(sec\ \dfrac{1}{\sqrt{x}} - 1 \right) =$ . . . .
$A.\ 1$
$B.\ \dfrac12$
$C.\ 0$
$D.\ -\dfrac12$
$E.\ -1$
[Soal dan Pembahasan Limit Fungsi Trigonometri SBMPTN 2017 Matematika IPA]
$A.\ 1$
$B.\ \dfrac12$
$C.\ 0$
$D.\ -\dfrac12$
$E.\ -1$
[Soal dan Pembahasan Limit Fungsi Trigonometri SBMPTN 2017 Matematika IPA]
$\displaystyle \lim_{x \to \infty}\left(sec\ \dfrac{1}{\sqrt{x}} - 1 \right) = \displaystyle \lim_{\dfrac{1}{x} \to 0}\left(sec\ \sqrt{\dfrac{1}{x}} - 1 \right)$
$Misalkan\ \dfrac{1}{x} = p$
$= \displaystyle \lim_{p \to 0}\left(sec\ \sqrt{p} - 1 \right)$
$= sec\ \sqrt{0} - 1$
$= sec\ 0 - 1$
$= 1 - 1$
$= 0$
$Jawab:\ C.$
$Misalkan\ \dfrac{1}{x} = p$
$= \displaystyle \lim_{p \to 0}\left(sec\ \sqrt{p} - 1 \right)$
$= sec\ \sqrt{0} - 1$
$= sec\ 0 - 1$
$= 1 - 1$
$= 0$
$Jawab:\ C.$
$21.\ Nilai\ dari\ \displaystyle \lim_{x \to 2}\dfrac{\sqrt{1 - cos(x - 2)}}{\sqrt{x^2 - 2x}}\ adalah$ . . . .
$A.\ 0$
$B.\ \dfrac12$
$C.\ \dfrac34$
$D.\ 1$
$E.\ \infty$
[Soal dan Pembahasan Limit Fungsi Trigonometri SBMPTN 2016 MtkIPA]
$A.\ 0$
$B.\ \dfrac12$
$C.\ \dfrac34$
$D.\ 1$
$E.\ \infty$
[Soal dan Pembahasan Limit Fungsi Trigonometri SBMPTN 2016 MtkIPA]
$\displaystyle \lim_{x \to 2}\dfrac{\sqrt{1 - cos(x - 2)}}{\sqrt{x^2 - 2x}} = \displaystyle \lim_{x \to 2}\sqrt{\dfrac{1 - cos(x - 2)}{x^2 - 2x}}$
$= \sqrt{\displaystyle \lim_{x \to 2}\dfrac{1 - cos(x - 2)}{x^2 - 2x}}$
$= \sqrt{\displaystyle \lim_{x \to 2}\dfrac{2sin^2\dfrac12(x - 2)}{x(x - 2)}}$
$= \sqrt{\displaystyle \lim_{(x - 2) \to 0}\dfrac{2sin\dfrac12(x - 2)}{(x - 2)}.\displaystyle \lim_{x \to 2}\dfrac{sin\dfrac12(x - 2)}{x}}$
$= \sqrt{2.\dfrac12.\dfrac{sin\dfrac12(2 - 2)}{2}}$
$= \sqrt{1.\dfrac{sin\ 0}{2}}$
$= \sqrt{\dfrac{0}{2}}$
$= \sqrt{0}$
$= 0$
$Jawab:\ A.$
$= \sqrt{\displaystyle \lim_{x \to 2}\dfrac{1 - cos(x - 2)}{x^2 - 2x}}$
$= \sqrt{\displaystyle \lim_{x \to 2}\dfrac{2sin^2\dfrac12(x - 2)}{x(x - 2)}}$
$= \sqrt{\displaystyle \lim_{(x - 2) \to 0}\dfrac{2sin\dfrac12(x - 2)}{(x - 2)}.\displaystyle \lim_{x \to 2}\dfrac{sin\dfrac12(x - 2)}{x}}$
$= \sqrt{2.\dfrac12.\dfrac{sin\dfrac12(2 - 2)}{2}}$
$= \sqrt{1.\dfrac{sin\ 0}{2}}$
$= \sqrt{\dfrac{0}{2}}$
$= \sqrt{0}$
$= 0$
$Jawab:\ A.$
$22.\ \displaystyle \lim_{x \to 0}\dfrac{\sqrt{2x^2 + 1} - 1}{\sqrt{3sin^5\ x + x^4}} =$ . . . .
$A.\ 0$
$B.\ \dfrac{\sqrt{2}}{\sqrt{3}}$
$C.\ \dfrac{\sqrt{3}}{\sqrt{4}}$
$D.\ \dfrac12$
$E.\ 1$
[Soal dan Pembahasan Limit Fungsi Trigonometri SBMPTN 2016 MtkIPA]
$A.\ 0$
$B.\ \dfrac{\sqrt{2}}{\sqrt{3}}$
$C.\ \dfrac{\sqrt{3}}{\sqrt{4}}$
$D.\ \dfrac12$
$E.\ 1$
[Soal dan Pembahasan Limit Fungsi Trigonometri SBMPTN 2016 MtkIPA]
$\displaystyle \lim_{x \to 0}\dfrac{\sqrt{2x^2 + 1} - 1}{\sqrt{3sin^5\ x + x^4}}$
$= \displaystyle \lim_{x \to 0}\dfrac{\sqrt{2x^2 + 1} - 1}{\sqrt{3sin^5\ x + x^4}}.\dfrac{(\sqrt{2x^2 + 1} + 1)}{(\sqrt{2x^2 + 1} + 1)}$
$= \displaystyle \lim_{x \to 0}\dfrac{2x^2}{\sqrt{3sin^5\ x + x^4}}.\dfrac{1}{(\sqrt{2x^2 + 1} + 1)}$
$= \displaystyle \lim_{x \to 0}\dfrac{2x^2}{\sqrt{3sin^5\ x + x^4}}.\dfrac{1}{(\sqrt{2x^2 + 1} + 1)}\ \times\ \dfrac{\dfrac{1}{x^2}}{\dfrac{1}{x^2}}$
$= \displaystyle \lim_{x \to 0}\dfrac{2}{\sqrt{\dfrac{3sin^5\ x + x^4}{x^4}}}.\dfrac{1}{(\sqrt{2x^2 + 1} + 1)}$
$= \displaystyle \lim_{x \to 0}\dfrac{2}{\sqrt{3\dfrac{sin^4\ x}{x^4}.sin\ x + 1}}.\dfrac{1}{(\sqrt{2x^2 + 1} + 1)}$
$= \displaystyle \lim_{x \to 0}\dfrac{2}{\sqrt{3.sin\ x + 1}}.\dfrac{1}{(\sqrt{2x^2 + 1} + 1)}$
$= \dfrac{2}{\sqrt{3.sin\ 0 + 1}}.\dfrac{1}{(\sqrt{2.0^2 + 1} + 1)}$
$= 2.\dfrac12$
$= 1$
$Jawab:\ E.$
$= \displaystyle \lim_{x \to 0}\dfrac{\sqrt{2x^2 + 1} - 1}{\sqrt{3sin^5\ x + x^4}}.\dfrac{(\sqrt{2x^2 + 1} + 1)}{(\sqrt{2x^2 + 1} + 1)}$
$= \displaystyle \lim_{x \to 0}\dfrac{2x^2}{\sqrt{3sin^5\ x + x^4}}.\dfrac{1}{(\sqrt{2x^2 + 1} + 1)}$
$= \displaystyle \lim_{x \to 0}\dfrac{2x^2}{\sqrt{3sin^5\ x + x^4}}.\dfrac{1}{(\sqrt{2x^2 + 1} + 1)}\ \times\ \dfrac{\dfrac{1}{x^2}}{\dfrac{1}{x^2}}$
$= \displaystyle \lim_{x \to 0}\dfrac{2}{\sqrt{\dfrac{3sin^5\ x + x^4}{x^4}}}.\dfrac{1}{(\sqrt{2x^2 + 1} + 1)}$
$= \displaystyle \lim_{x \to 0}\dfrac{2}{\sqrt{3\dfrac{sin^4\ x}{x^4}.sin\ x + 1}}.\dfrac{1}{(\sqrt{2x^2 + 1} + 1)}$
$= \displaystyle \lim_{x \to 0}\dfrac{2}{\sqrt{3.sin\ x + 1}}.\dfrac{1}{(\sqrt{2x^2 + 1} + 1)}$
$= \dfrac{2}{\sqrt{3.sin\ 0 + 1}}.\dfrac{1}{(\sqrt{2.0^2 + 1} + 1)}$
$= 2.\dfrac12$
$= 1$
$Jawab:\ E.$
$23.\ \displaystyle \lim_{x \to 0}\sqrt{\dfrac{xtan\ x}{sin^2\ x - cos\ 2x + 1}} =$ . . . .
$A.\ 3$
$B.\ \sqrt{3}$
$C.\ \dfrac{\sqrt{3}}{3}$
$D.\ \dfrac13$
$E.\ \dfrac{\sqrt{3}}{2}$
[Soal dan Pembahasan Limit Fungsi Trigonometri SBMPTN 2013 MtkIPA]
$A.\ 3$
$B.\ \sqrt{3}$
$C.\ \dfrac{\sqrt{3}}{3}$
$D.\ \dfrac13$
$E.\ \dfrac{\sqrt{3}}{2}$
[Soal dan Pembahasan Limit Fungsi Trigonometri SBMPTN 2013 MtkIPA]
$\displaystyle \lim_{x \to 0}\sqrt{\dfrac{xtan\ x}{sin^2\ x - cos\ 2x + 1}}$
$= \displaystyle \lim_{x \to 0}\sqrt{\dfrac{xtan\ x}{sin^2\ x - (1 - 2sin^2\ x) + 1}}$
$= \displaystyle \lim_{x \to 0}\sqrt{\dfrac{xtan\ x}{3sin^2\ x}}$
$= \sqrt{\displaystyle \lim_{x \to 0}\dfrac{xtan\ x}{3sin^2\ x}}$
$= \sqrt{\dfrac13.\displaystyle \lim_{x \to 0}\dfrac{x}{sin\ x}.\displaystyle \lim_{x \to 0}\dfrac{tan\ x}{sin\ x}}$
$= \sqrt{\dfrac13.1.1}$
$= \sqrt{\dfrac13}$
$= \dfrac{\sqrt{3}}{3}$
$Jawab:\ C.$
$= \displaystyle \lim_{x \to 0}\sqrt{\dfrac{xtan\ x}{sin^2\ x - (1 - 2sin^2\ x) + 1}}$
$= \displaystyle \lim_{x \to 0}\sqrt{\dfrac{xtan\ x}{3sin^2\ x}}$
$= \sqrt{\displaystyle \lim_{x \to 0}\dfrac{xtan\ x}{3sin^2\ x}}$
$= \sqrt{\dfrac13.\displaystyle \lim_{x \to 0}\dfrac{x}{sin\ x}.\displaystyle \lim_{x \to 0}\dfrac{tan\ x}{sin\ x}}$
$= \sqrt{\dfrac13.1.1}$
$= \sqrt{\dfrac13}$
$= \dfrac{\sqrt{3}}{3}$
$Jawab:\ C.$
$24.\ \displaystyle \lim_{x \to 0}\dfrac{sin^2\ x - cos\ x + 1}{xtan\ x} =$ . . . .
$A.\ \dfrac32$
$B.\ \dfrac12$
$C.\ -\dfrac12$
$D.\ -1$
$E.\ -2$
[Soal dan Pembahasan Limit Fungsi Trigonometri SBMPTN 2013 MtkIPA]
$A.\ \dfrac32$
$B.\ \dfrac12$
$C.\ -\dfrac12$
$D.\ -1$
$E.\ -2$
[Soal dan Pembahasan Limit Fungsi Trigonometri SBMPTN 2013 MtkIPA]
$\displaystyle \lim_{x \to 0}\dfrac{sin^2\ x - cos\ x + 1}{xtan\ x}$
$= \displaystyle \lim_{x \to 0}\dfrac{sin^2\ x - (1 - 2sin^2\ \dfrac12x) + 1}{xtan\ x}$
$= \displaystyle \lim_{x \to 0}\dfrac{sin^2\ x + 2sin^2\ \dfrac12x}{xtan\ x}$
$= \displaystyle \lim_{x \to 0}\left(\dfrac{sin^2\ x}{xtan\ x} + \dfrac{2sin^2\ \dfrac12x}{xtan\ x}\right)$
$= \left(\displaystyle \lim_{x \to 0}\dfrac{sin^2\ x}{xtan\ x} + \displaystyle \lim_{x \to 0}\dfrac{2sin^2\ \dfrac12x}{xtan\ x}\right)$
$= \left(\displaystyle \lim_{x \to 0}\dfrac{sin\ x.sin\ x}{x.tan\ x} + \displaystyle \lim_{x \to 0}\dfrac{2sin\ \dfrac12x.sin\ \dfrac12x}{x.tan\ x}\right)$
$= 1.1 + 2.\dfrac12.\dfrac12$
$= 1 + \dfrac12$
$= \dfrac32$
$Jawab:\ A.$
$= \displaystyle \lim_{x \to 0}\dfrac{sin^2\ x - (1 - 2sin^2\ \dfrac12x) + 1}{xtan\ x}$
$= \displaystyle \lim_{x \to 0}\dfrac{sin^2\ x + 2sin^2\ \dfrac12x}{xtan\ x}$
$= \displaystyle \lim_{x \to 0}\left(\dfrac{sin^2\ x}{xtan\ x} + \dfrac{2sin^2\ \dfrac12x}{xtan\ x}\right)$
$= \left(\displaystyle \lim_{x \to 0}\dfrac{sin^2\ x}{xtan\ x} + \displaystyle \lim_{x \to 0}\dfrac{2sin^2\ \dfrac12x}{xtan\ x}\right)$
$= \left(\displaystyle \lim_{x \to 0}\dfrac{sin\ x.sin\ x}{x.tan\ x} + \displaystyle \lim_{x \to 0}\dfrac{2sin\ \dfrac12x.sin\ \dfrac12x}{x.tan\ x}\right)$
$= 1.1 + 2.\dfrac12.\dfrac12$
$= 1 + \dfrac12$
$= \dfrac32$
$Jawab:\ A.$
$25.\ \displaystyle \lim_{x \to 0}\dfrac{1 - cos^2\ 2x}{x^2tan\left(x + \dfrac{\pi}{4}\right)} =$ . . . .
$A.\ -2$
$B.\ 0$
$C.\ \sqrt{2}$
$D.\ \sqrt{3}$
$E.\ 4$
[Soal dan Pembahasan Limit Fungsi Trigonometri SNMPTN 2012 MtkIPA]
$A.\ -2$
$B.\ 0$
$C.\ \sqrt{2}$
$D.\ \sqrt{3}$
$E.\ 4$
[Soal dan Pembahasan Limit Fungsi Trigonometri SNMPTN 2012 MtkIPA]
$sin^2\ 2x + cos^2\ 2x = 1$
$sin^2\ 2x = 1 - cos^2\ 2x$ . . . . *
$\displaystyle \lim_{x \to 0}\dfrac{1 - cos^2\ 2x}{x^2tan\left(x + \dfrac{\pi}{4}\right)} = \displaystyle \lim_{x \to 0}\dfrac{sin^2\ 2x}{x^2tan\left(x + \dfrac{\pi}{4}\right)}$
$= \displaystyle \lim_{x \to 0}\dfrac{sin^2\ 2x}{x^2}.\displaystyle \lim_{x \to 0}\dfrac{1}{tan\left(x + \dfrac{\pi}{4}\right)}$
$= \displaystyle \lim_{x \to 0}\dfrac{sin\ 2x.sin\ 2x}{x.x}.\displaystyle \lim_{x \to 0}\dfrac{1}{tan\left(x + \dfrac{\pi}{4}\right)}$
$= 2.2.\dfrac{1}{tan\left(0 + \dfrac{\pi}{4}\right)}$
$= 2.2.\dfrac{1}{tan\left(0 + \dfrac{\pi}{4}\right)}$
$= 2.2.1$
$= 4$
$Jawab:\ E.$
$sin^2\ 2x = 1 - cos^2\ 2x$ . . . . *
$\displaystyle \lim_{x \to 0}\dfrac{1 - cos^2\ 2x}{x^2tan\left(x + \dfrac{\pi}{4}\right)} = \displaystyle \lim_{x \to 0}\dfrac{sin^2\ 2x}{x^2tan\left(x + \dfrac{\pi}{4}\right)}$
$= \displaystyle \lim_{x \to 0}\dfrac{sin^2\ 2x}{x^2}.\displaystyle \lim_{x \to 0}\dfrac{1}{tan\left(x + \dfrac{\pi}{4}\right)}$
$= \displaystyle \lim_{x \to 0}\dfrac{sin\ 2x.sin\ 2x}{x.x}.\displaystyle \lim_{x \to 0}\dfrac{1}{tan\left(x + \dfrac{\pi}{4}\right)}$
$= 2.2.\dfrac{1}{tan\left(0 + \dfrac{\pi}{4}\right)}$
$= 2.2.\dfrac{1}{tan\left(0 + \dfrac{\pi}{4}\right)}$
$= 2.2.1$
$= 4$
$Jawab:\ E.$
$26.\ \displaystyle \lim_{x \to -y}\dfrac{tan\ x + tan\ y}{\left(\dfrac{x^2 - y^2}{-2y^2}\right)(1 - tan\ xtan\ y)} =$ . . . .
$A.\ -1$
$B.\ 1$
$C.\ 0$
$D.\ y$
$E.\ -y$
[Soal dan Pembahasan Limit Fungsi Trigonometri Utul 2017 MtkIPA]
$A.\ -1$
$B.\ 1$
$C.\ 0$
$D.\ y$
$E.\ -y$
[Soal dan Pembahasan Limit Fungsi Trigonometri Utul 2017 MtkIPA]
$tan(x + y) = \dfrac{tan\ x + tan\ y}{1 - tan\ xtan\ y}$
$\displaystyle \lim_{x \to -y}\dfrac{tan\ x + tan\ y}{\left(\dfrac{x^2 - y^2}{-2y^2}\right)(1 - tan\ xtan\ y)}$
$= \displaystyle \lim_{x \to -y}\dfrac{tan( x + y)}{\left(\dfrac{x^2 - y^2}{-2y^2}\right)}$
$= \displaystyle \lim_{x \to -y}\dfrac{-2y^2tan( x + y)}{x^2 - y^2}$
$= \displaystyle \lim_{x \to -y}\dfrac{-2y^2tan( x + y)}{(x - y)(x + y)}$
$= \displaystyle \lim_{x \to -y}\dfrac{-2y^2}{(x - y)}.\displaystyle \lim_{(x + y) \to 0}\dfrac{tan( x + y)}{(x + y)}$
$= \dfrac{-2y^2}{(-y - y)}.1$
$= \dfrac{-2y^2}{-2y}$
$= y$
$Jawab:\ D.$
$\displaystyle \lim_{x \to -y}\dfrac{tan\ x + tan\ y}{\left(\dfrac{x^2 - y^2}{-2y^2}\right)(1 - tan\ xtan\ y)}$
$= \displaystyle \lim_{x \to -y}\dfrac{tan( x + y)}{\left(\dfrac{x^2 - y^2}{-2y^2}\right)}$
$= \displaystyle \lim_{x \to -y}\dfrac{-2y^2tan( x + y)}{x^2 - y^2}$
$= \displaystyle \lim_{x \to -y}\dfrac{-2y^2tan( x + y)}{(x - y)(x + y)}$
$= \displaystyle \lim_{x \to -y}\dfrac{-2y^2}{(x - y)}.\displaystyle \lim_{(x + y) \to 0}\dfrac{tan( x + y)}{(x + y)}$
$= \dfrac{-2y^2}{(-y - y)}.1$
$= \dfrac{-2y^2}{-2y}$
$= y$
$Jawab:\ D.$
$27.\ \displaystyle \lim_{x \to -4}\dfrac{1 - cos(x + 4)}{x^2 + 8x + 16} =$ . . . .
$A. -2$
$B.\ -\dfrac12$
$C.\ \dfrac13$
$D.\ \dfrac12$
$E.\ 2$
[Soal dan Pembahasan Limit Fungsi Trigonometri Utul 2017 MtkIPA]
$A. -2$
$B.\ -\dfrac12$
$C.\ \dfrac13$
$D.\ \dfrac12$
$E.\ 2$
[Soal dan Pembahasan Limit Fungsi Trigonometri Utul 2017 MtkIPA]
$\displaystyle \lim_{x \to -4}\dfrac{1 - cos(x + 4)}{x^2 + 8x + 16} = \displaystyle \lim_{x \to -4}\dfrac{2sin^2\ \dfrac12(x + 4)}{(x + 4)^2}$
$= 2.\left(\displaystyle \lim_{(x + 4) \to 0}\dfrac{sin\ \dfrac12(x + 4)}{(x + 4)}\right)^2$
$= 2.\left(\dfrac12\right)^2$
$= 2.\dfrac14$
$= \dfrac12$
$Jawab:\ D.$
$= 2.\left(\displaystyle \lim_{(x + 4) \to 0}\dfrac{sin\ \dfrac12(x + 4)}{(x + 4)}\right)^2$
$= 2.\left(\dfrac12\right)^2$
$= 2.\dfrac14$
$= \dfrac12$
$Jawab:\ D.$
$28.\ \displaystyle \lim_{x \to 3}\dfrac{(x + 6)tan(2x - 6)}{x^2 - x - 6} =$ . . . .
$A.\ -\dfrac{18}{5}$
$B.\ -\dfrac95$
$C.\ \dfrac95$
$D.\ \dfrac{18}{5}$
$E.\ \dfrac{27}{5}$
[Soal dan Pembahasan Limit Fungsi Trigonometri Utul 2016 MtkIPA]
$A.\ -\dfrac{18}{5}$
$B.\ -\dfrac95$
$C.\ \dfrac95$
$D.\ \dfrac{18}{5}$
$E.\ \dfrac{27}{5}$
[Soal dan Pembahasan Limit Fungsi Trigonometri Utul 2016 MtkIPA]
$\displaystyle \lim_{x \to 3}\dfrac{(x + 6)tan(2x - 6)}{x^2 - x - 6}$
$= \displaystyle \lim_{x \to 3}\dfrac{(x + 6)tan2(x - 3)}{(x + 2)(x - 3)}$
$= \displaystyle \lim_{x \to 3}\dfrac{(x + 6)}{(x + 2)}.\displaystyle \lim_{(x - 3) \to 0}\dfrac{tan2(x - 3)}{(x - 3)}$
$= \dfrac{(3 + 6)}{(3 + 2)}.2$
$= \dfrac{18}{5}$
$Jawab:\ D.$
$= \displaystyle \lim_{x \to 3}\dfrac{(x + 6)tan2(x - 3)}{(x + 2)(x - 3)}$
$= \displaystyle \lim_{x \to 3}\dfrac{(x + 6)}{(x + 2)}.\displaystyle \lim_{(x - 3) \to 0}\dfrac{tan2(x - 3)}{(x - 3)}$
$= \dfrac{(3 + 6)}{(3 + 2)}.2$
$= \dfrac{18}{5}$
$Jawab:\ D.$
$29.\ \displaystyle \lim_{x \to 0}\dfrac{1 - cos^3\ x}{xtan\ x} =$ . . . .
$A.\ 0$
$B.\ \dfrac12$
$C.\ \dfrac34$
$D.\ \dfrac32$
$E.\ 3$
[Soal dan Pembahasan Limit Fungsi Trigonometri Utul 2013 MtkIPA]
$A.\ 0$
$B.\ \dfrac12$
$C.\ \dfrac34$
$D.\ \dfrac32$
$E.\ 3$
[Soal dan Pembahasan Limit Fungsi Trigonometri Utul 2013 MtkIPA]
$\left(1 - cos\ x\right)^3 = 1 - 3cos\ x + 3cos^2\ x - cos^3\ x$
$\left(1 - cos\ x\right)^3 + 3cos\ x - 3cos^2\ x = 1 - cos^3\ x$
$\left(1 - cos\ x\right)^3 + 3cos\ x(1 - cos\ x) = 1 - cos^3\ x$
$\left(2sin^2\ \dfrac12x \right)^3 + 3cos\ x\left(2sin^2\ \dfrac12x\right) = 1 - cos^3\ x$
$\left(8sin^6\ \dfrac12x \right) + 6cos\ x\left(sin^2\ \dfrac12x\right) = 1 - cos^3\ x$
$\displaystyle \lim_{x \to 0}\dfrac{1 - cos^3\ x}{xtan\ x} = \displaystyle \lim_{x \to 0}\dfrac{\left(8sin^6\ \dfrac12x \right) + 6cos\ x(sin^2\ \dfrac12x)}{xtan\ x}$
$= \displaystyle \lim_{x \to 0}\left(\dfrac{8sin^6\ \dfrac12x}{xtan\ x} \right) + \displaystyle \lim_{x \to 0}6cos\ x\left(\dfrac{2sin^2\ \dfrac12x}{xtan\ x}\right)$
$= 8\displaystyle \lim_{x \to 0}\left(\dfrac{sin^2\ \dfrac12x}{xtan\ x}.sin^4\ \dfrac12x \right) + 6\displaystyle \lim_{x \to 0}cos\ x\left(\dfrac{sin^2\ \dfrac12x}{xtan\ x}\right)$
$= 8.\dfrac14.0 + 6.1.\dfrac14$
$= 0 + \dfrac32$
$= \dfrac32$
$Jawab:\ D.$
$\left(1 - cos\ x\right)^3 + 3cos\ x - 3cos^2\ x = 1 - cos^3\ x$
$\left(1 - cos\ x\right)^3 + 3cos\ x(1 - cos\ x) = 1 - cos^3\ x$
$\left(2sin^2\ \dfrac12x \right)^3 + 3cos\ x\left(2sin^2\ \dfrac12x\right) = 1 - cos^3\ x$
$\left(8sin^6\ \dfrac12x \right) + 6cos\ x\left(sin^2\ \dfrac12x\right) = 1 - cos^3\ x$
$\displaystyle \lim_{x \to 0}\dfrac{1 - cos^3\ x}{xtan\ x} = \displaystyle \lim_{x \to 0}\dfrac{\left(8sin^6\ \dfrac12x \right) + 6cos\ x(sin^2\ \dfrac12x)}{xtan\ x}$
$= \displaystyle \lim_{x \to 0}\left(\dfrac{8sin^6\ \dfrac12x}{xtan\ x} \right) + \displaystyle \lim_{x \to 0}6cos\ x\left(\dfrac{2sin^2\ \dfrac12x}{xtan\ x}\right)$
$= 8\displaystyle \lim_{x \to 0}\left(\dfrac{sin^2\ \dfrac12x}{xtan\ x}.sin^4\ \dfrac12x \right) + 6\displaystyle \lim_{x \to 0}cos\ x\left(\dfrac{sin^2\ \dfrac12x}{xtan\ x}\right)$
$= 8.\dfrac14.0 + 6.1.\dfrac14$
$= 0 + \dfrac32$
$= \dfrac32$
$Jawab:\ D.$
$30.$ Nilai $\displaystyle \lim_{x \to \dfrac{\pi}{4}}sin(\dfrac{\pi}{4} - x)tan(x + \dfrac{\pi}{4})$ adalah . . . .
$A.\ 2$
$B.\ 1$
$C.\ 0$
$D.\ -1$
$E.\ -2$
[Soal dan Pembahasan Limit Fungsi Trigonometri Utul 2010 MtkIPA]
$A.\ 2$
$B.\ 1$
$C.\ 0$
$D.\ -1$
$E.\ -2$
[Soal dan Pembahasan Limit Fungsi Trigonometri Utul 2010 MtkIPA]
$Ingat:\ sudut-sudut\ berelasi\ !$
$sin(\dfrac{\pi}{4} - x) = cos(\dfrac{\pi}{2} - (\dfrac{\pi}{4} - x))$
$sin(\dfrac{\pi}{4} - x) = cos(x + \dfrac{\pi}{4})$ . . . . *
$\displaystyle \lim_{x \to \dfrac{\pi}{4}}sin(\dfrac{\pi}{4} - x)tan(x + \dfrac{\pi}{4})$
$= \displaystyle \lim_{x \to \dfrac{\pi}{4}}cos(x + \dfrac{\pi}{4})\dfrac{sin(x + \dfrac{\pi}{4})}{cos(x + \dfrac{\pi}{4})}$
$= \displaystyle \lim_{x \to \dfrac{\pi}{4}}sin(x + \dfrac{\pi}{4})$
$= sin(\dfrac{\pi}{4} + \dfrac{\pi}{4})$
$= sin(\dfrac{\pi}{2})$
$= 1$
$Jawab:\ B.$
$sin(\dfrac{\pi}{4} - x) = cos(\dfrac{\pi}{2} - (\dfrac{\pi}{4} - x))$
$sin(\dfrac{\pi}{4} - x) = cos(x + \dfrac{\pi}{4})$ . . . . *
$\displaystyle \lim_{x \to \dfrac{\pi}{4}}sin(\dfrac{\pi}{4} - x)tan(x + \dfrac{\pi}{4})$
$= \displaystyle \lim_{x \to \dfrac{\pi}{4}}cos(x + \dfrac{\pi}{4})\dfrac{sin(x + \dfrac{\pi}{4})}{cos(x + \dfrac{\pi}{4})}$
$= \displaystyle \lim_{x \to \dfrac{\pi}{4}}sin(x + \dfrac{\pi}{4})$
$= sin(\dfrac{\pi}{4} + \dfrac{\pi}{4})$
$= sin(\dfrac{\pi}{2})$
$= 1$
$Jawab:\ B.$
$31.\ \displaystyle \lim_{x \to 0}\dfrac{5x - tan\ 5x}{x^3} =$ . . . .
$A.\ \dfrac{125}{3}$
$B.\ \dfrac{115}{3}$
$C.\ \dfrac{125}{6}$
$D.\ \dfrac{-125}{6}$
$E.\ \dfrac{-125}{3}$
[Soal dan Pembahasan Limit Fungsi Trigonometri SIMAK UI 2017 MtkIPA]
$A.\ \dfrac{125}{3}$
$B.\ \dfrac{115}{3}$
$C.\ \dfrac{125}{6}$
$D.\ \dfrac{-125}{6}$
$E.\ \dfrac{-125}{3}$
[Soal dan Pembahasan Limit Fungsi Trigonometri SIMAK UI 2017 MtkIPA]
$Gunakan\ Aturan\ L'Hospital$
$\displaystyle \lim_{x \to 0}\dfrac{5x - tan\ 5x}{x^3} = \displaystyle \lim_{x \to 0}\dfrac{5 - 5sec^2\ 5x}{3x^2}$
$= \displaystyle \lim_{x \to 0}\dfrac{5(1 - sec^2\ 5x)}{3x^2}$
$Ingat:\ 1 - sec^2\ ax = -tan^2\ ax$
$= \displaystyle \lim_{x \to 0}\dfrac{5(-tan^2\ 5x)}{3x^2}$
$= -\dfrac53\displaystyle \lim_{x \to 0}\dfrac{(tan^2\ 5x)}{x^2}$
$= -\dfrac53\displaystyle \lim_{x \to 0}\dfrac{tan\ 5x}{x}.\displaystyle \lim_{x \to 0}\dfrac{tan\ 5x}{x}$
$= -\dfrac53.5.5$
$= -\dfrac{125}{3}$
$Jawab:\ E.$
$\displaystyle \lim_{x \to 0}\dfrac{5x - tan\ 5x}{x^3} = \displaystyle \lim_{x \to 0}\dfrac{5 - 5sec^2\ 5x}{3x^2}$
$= \displaystyle \lim_{x \to 0}\dfrac{5(1 - sec^2\ 5x)}{3x^2}$
$Ingat:\ 1 - sec^2\ ax = -tan^2\ ax$
$= \displaystyle \lim_{x \to 0}\dfrac{5(-tan^2\ 5x)}{3x^2}$
$= -\dfrac53\displaystyle \lim_{x \to 0}\dfrac{(tan^2\ 5x)}{x^2}$
$= -\dfrac53\displaystyle \lim_{x \to 0}\dfrac{tan\ 5x}{x}.\displaystyle \lim_{x \to 0}\dfrac{tan\ 5x}{x}$
$= -\dfrac53.5.5$
$= -\dfrac{125}{3}$
$Jawab:\ E.$
$32.\ \displaystyle \lim_{x \to 0}\dfrac{\displaystyle \int_{0}^{x}\sqrt{1 + cos\ t}\ dt}{x} =$ . . . .
$A.\ 0$
$B.\ 1$
$C.\ \sqrt{2}$
$D.\ \sqrt{3}$
$E.\ \dfrac12\sqrt{2}$
[Soal dan Pembahasan Limit Fungsi Trigonometri SIMAK UI 2016 MtkIPA]
$A.\ 0$
$B.\ 1$
$C.\ \sqrt{2}$
$D.\ \sqrt{3}$
$E.\ \dfrac12\sqrt{2}$
[Soal dan Pembahasan Limit Fungsi Trigonometri SIMAK UI 2016 MtkIPA]
$1 + cos\ t = 2cos^2\ \dfrac12t$
$\displaystyle \int_{0}^{x}\sqrt{2cos^2\ \dfrac12t}\ dt = \displaystyle \sqrt{2}\int_{0}^{x}cos\ \dfrac12t\ dt$
$= \left[\displaystyle 2\sqrt{2}sin\ \dfrac12t\ \right]_{0}^{x}$
$= \displaystyle 2\sqrt{2}sin\ \dfrac12x$
$\displaystyle \lim_{x \to 0}\dfrac{\displaystyle \int_{0}^{x}\sqrt{1 + cos\ t}\ dt}{x} = \displaystyle \lim_{x \to 0}\dfrac{\displaystyle 2\sqrt{2}sin\ \dfrac12x}{x}$
$= 2\sqrt{2}.\dfrac12$
$= \sqrt{2}$
$Jawab:\ C.$
$\displaystyle \int_{0}^{x}\sqrt{2cos^2\ \dfrac12t}\ dt = \displaystyle \sqrt{2}\int_{0}^{x}cos\ \dfrac12t\ dt$
$= \left[\displaystyle 2\sqrt{2}sin\ \dfrac12t\ \right]_{0}^{x}$
$= \displaystyle 2\sqrt{2}sin\ \dfrac12x$
$\displaystyle \lim_{x \to 0}\dfrac{\displaystyle \int_{0}^{x}\sqrt{1 + cos\ t}\ dt}{x} = \displaystyle \lim_{x \to 0}\dfrac{\displaystyle 2\sqrt{2}sin\ \dfrac12x}{x}$
$= 2\sqrt{2}.\dfrac12$
$= \sqrt{2}$
$Jawab:\ C.$
$33.\ Jika\ f(x) = sin\ 2x,\ maka$
$\displaystyle \lim_{h \to 0}\dfrac{f\left(x + \dfrac h2 \right) - 2f(x) + f\left(x - \dfrac h2 \right)}{h^2} =$ . . . .
$A.\ 2sin\ 2x$
$B.\ sin\ 2x$
$C.\ 0$
$D.\ -sin\ 2x$
$E.\ -2sin\ 2x$
[Soal dan Pembahasan Limit Fungsi Trigonometri SIMAK UI 2015 MtkIPA]
$\displaystyle \lim_{h \to 0}\dfrac{f\left(x + \dfrac h2 \right) - 2f(x) + f\left(x - \dfrac h2 \right)}{h^2} =$ . . . .
$A.\ 2sin\ 2x$
$B.\ sin\ 2x$
$C.\ 0$
$D.\ -sin\ 2x$
$E.\ -2sin\ 2x$
[Soal dan Pembahasan Limit Fungsi Trigonometri SIMAK UI 2015 MtkIPA]
$f(x) = sin\ 2x$
$f\left(x + \dfrac h2 \right) = sin(2x + h)$
$= sin\ 2xcos\ h + cos\ 2xsin\ h$
$f\left(x - \dfrac h2 \right) = sin(2x - h)$
$= sin\ 2xcos\ h - cos\ 2xsin\ h$
$f\left(x + \dfrac h2 \right) - 2f(x) + f\left(x - \dfrac h2 \right)$
$= sin\ 2xcos\ h + cos\ 2xsin\ h - 2sin\ 2x $
$\ \ \ + sin\ 2xcos\ h - cos\ 2xsin\ h$
$= 2sin\ 2xcos\ h - 2sin\ 2x$
$= 2sin\ 2x(cos\ h - 1)$
$= 2sin\ 2x(-2sin^2\ \dfrac h2)$
$= -4sin\ 2x(sin^2\ \dfrac h2)$
$\displaystyle \lim_{h \to 0}\dfrac{f\left(x + \dfrac h2 \right) - 2f(x) + f\left(x - \dfrac h2 \right)}{h^2}$
$= \displaystyle \lim_{h \to 0}\dfrac{-4sin\ 2x(sin^2\ \dfrac h2)}{h^2}$
$= -4sin\ 2x.\displaystyle \lim_{h \to 0}\dfrac{(sin^2\ \dfrac h2)}{h^2}$
$= -4sin\ 2x.\dfrac 12. \dfrac 12$
$= -sin\ 2x$
$Jawab:\ D.$
$f\left(x + \dfrac h2 \right) = sin(2x + h)$
$= sin\ 2xcos\ h + cos\ 2xsin\ h$
$f\left(x - \dfrac h2 \right) = sin(2x - h)$
$= sin\ 2xcos\ h - cos\ 2xsin\ h$
$f\left(x + \dfrac h2 \right) - 2f(x) + f\left(x - \dfrac h2 \right)$
$= sin\ 2xcos\ h + cos\ 2xsin\ h - 2sin\ 2x $
$\ \ \ + sin\ 2xcos\ h - cos\ 2xsin\ h$
$= 2sin\ 2xcos\ h - 2sin\ 2x$
$= 2sin\ 2x(cos\ h - 1)$
$= 2sin\ 2x(-2sin^2\ \dfrac h2)$
$= -4sin\ 2x(sin^2\ \dfrac h2)$
$\displaystyle \lim_{h \to 0}\dfrac{f\left(x + \dfrac h2 \right) - 2f(x) + f\left(x - \dfrac h2 \right)}{h^2}$
$= \displaystyle \lim_{h \to 0}\dfrac{-4sin\ 2x(sin^2\ \dfrac h2)}{h^2}$
$= -4sin\ 2x.\displaystyle \lim_{h \to 0}\dfrac{(sin^2\ \dfrac h2)}{h^2}$
$= -4sin\ 2x.\dfrac 12. \dfrac 12$
$= -sin\ 2x$
$Jawab:\ D.$
$34.\ \displaystyle \lim_{x \to 0}\dfrac{\sqrt{1 + tan\ x} - \sqrt{1 + sin\ x}}{x^3} =$ . . . .
$A.\ -1$
$B.\ -\dfrac 14$
$C.\ 0$
$D.\ \dfrac 14$
$E.\ 1$
[Soal dan Pembahasan Limit Fungsi Trigonometri SIMAK UI 2013 MtkIPA]
$A.\ -1$
$B.\ -\dfrac 14$
$C.\ 0$
$D.\ \dfrac 14$
$E.\ 1$
[Soal dan Pembahasan Limit Fungsi Trigonometri SIMAK UI 2013 MtkIPA]
$\displaystyle \lim_{x \to 0}\dfrac{\sqrt{1 + tan\ x} - \sqrt{1 + sin\ x}}{x^3}$
$= \displaystyle \lim_{x \to 0}\dfrac{\sqrt{1 + tan\ x} - \sqrt{1 + sin\ x}}{x^3}.\dfrac{(\sqrt{1 + tan\ x} + \sqrt{1 + sin\ x})}{(\sqrt{1 + tan\ x} + \sqrt{1 + sin\ x})}$
$= \displaystyle \lim_{x \to 0}\dfrac{tan\ x - sin\ x}{x^3}.\dfrac{1}{(\sqrt{1 + tan\ x} + \sqrt{1 + sin\ x})}$
$= \displaystyle \lim_{x \to 0}\dfrac{\dfrac{sin\ x}{cos\ x} - sin\ x}{x^3}.\dfrac{1}{(\sqrt{1 + tan\ x} + \sqrt{1 + sin\ x})}$
$= \displaystyle \lim_{x \to 0}\dfrac{sin\ x - sin\ x.cos\ x}{(cos\ x)x^3}.\dfrac{1}{(\sqrt{1 + tan\ x} + \sqrt{1 + sin\ x})}$
$= \displaystyle \lim_{x \to 0}\dfrac{sin\ x(1 - cos\ x)}{(cos\ x)x^3}.\dfrac{1}{(\sqrt{1 + tan\ x} + \sqrt{1 + sin\ x})}$
$= \displaystyle \lim_{x \to 0}\dfrac{sin\ x(2sin^2\ \dfrac 12x)}{(cos\ x)x^3}.\dfrac{1}{(\sqrt{1 + tan\ x} + \sqrt{1 + sin\ x})}$
$= 2.\displaystyle \lim_{x \to 0}\dfrac{sin\ x.sin\ \dfrac 12x.sin\ \dfrac 12x}{x.x.x}.\dfrac{1}{cos\ x(\sqrt{1 + tan\ x} + \sqrt{1 + sin\ x})}$
$= 2.1.\dfrac12.\dfrac12.\dfrac{1}{cos\ 0(\sqrt{1 + tan\ 0} + \sqrt{1 + sin\ 0})}$
$= \dfrac12.\dfrac{1}{1.(\sqrt{1 + 0} + \sqrt{1 + 0})}$
$= \dfrac12.\dfrac12$
$= \dfrac14$
$Jawab:\ D.$
$= \displaystyle \lim_{x \to 0}\dfrac{\sqrt{1 + tan\ x} - \sqrt{1 + sin\ x}}{x^3}.\dfrac{(\sqrt{1 + tan\ x} + \sqrt{1 + sin\ x})}{(\sqrt{1 + tan\ x} + \sqrt{1 + sin\ x})}$
$= \displaystyle \lim_{x \to 0}\dfrac{tan\ x - sin\ x}{x^3}.\dfrac{1}{(\sqrt{1 + tan\ x} + \sqrt{1 + sin\ x})}$
$= \displaystyle \lim_{x \to 0}\dfrac{\dfrac{sin\ x}{cos\ x} - sin\ x}{x^3}.\dfrac{1}{(\sqrt{1 + tan\ x} + \sqrt{1 + sin\ x})}$
$= \displaystyle \lim_{x \to 0}\dfrac{sin\ x - sin\ x.cos\ x}{(cos\ x)x^3}.\dfrac{1}{(\sqrt{1 + tan\ x} + \sqrt{1 + sin\ x})}$
$= \displaystyle \lim_{x \to 0}\dfrac{sin\ x(1 - cos\ x)}{(cos\ x)x^3}.\dfrac{1}{(\sqrt{1 + tan\ x} + \sqrt{1 + sin\ x})}$
$= \displaystyle \lim_{x \to 0}\dfrac{sin\ x(2sin^2\ \dfrac 12x)}{(cos\ x)x^3}.\dfrac{1}{(\sqrt{1 + tan\ x} + \sqrt{1 + sin\ x})}$
$= 2.\displaystyle \lim_{x \to 0}\dfrac{sin\ x.sin\ \dfrac 12x.sin\ \dfrac 12x}{x.x.x}.\dfrac{1}{cos\ x(\sqrt{1 + tan\ x} + \sqrt{1 + sin\ x})}$
$= 2.1.\dfrac12.\dfrac12.\dfrac{1}{cos\ 0(\sqrt{1 + tan\ 0} + \sqrt{1 + sin\ 0})}$
$= \dfrac12.\dfrac{1}{1.(\sqrt{1 + 0} + \sqrt{1 + 0})}$
$= \dfrac12.\dfrac12$
$= \dfrac14$
$Jawab:\ D.$
$35.\ \displaystyle \lim_{x \to 0}\dfrac{cos\ x\ sin\ x - tan\ x}{x^2sin\ x} =$ . . . .
$A.\ -1$
$B.\ -\dfrac12$
$C.\ 0$
$D.\ \dfrac12$
$E.\ 1$
[Soal dan Pembahasan Limit Fungsi Trigonometri SIMAK UI 2013 MtkIPA]
$A.\ -1$
$B.\ -\dfrac12$
$C.\ 0$
$D.\ \dfrac12$
$E.\ 1$
[Soal dan Pembahasan Limit Fungsi Trigonometri SIMAK UI 2013 MtkIPA]
$\displaystyle \lim_{x \to 0}\dfrac{cos\ x\ sin\ x - tan\ x}{x^2sin\ x}$
$= \displaystyle \lim_{x \to 0}\dfrac{cos^2\ x\ sin\ x - sin\ x}{(cos\ x)x^2sin\ x}$
$= \displaystyle \lim_{x \to 0}\dfrac{sin\ x(cos^2\ x - 1)}{(cos\ x)x^2sin\ x}$
$= \displaystyle \lim_{x \to 0}\dfrac{(cos^2\ x - 1)}{(cos\ x)x^2}$
$= \displaystyle \lim_{x \to 0}\dfrac{-sin^2\ x}{(cos\ x)x^2}$
$= -\displaystyle \lim_{x \to 0}\dfrac{sin^2\ x}{x^2}.\displaystyle \lim_{x \to 0}\dfrac{1}{cos\ x}$
$= -\displaystyle \lim_{x \to 0}\dfrac{sin\ x.sin\ x}{x.x}.\displaystyle \lim_{x \to 0}\dfrac{1}{cos\ x}$
$= -1.1.\dfrac{1}{cos\ 0}$
$= -1$
$Jawab:\ A.$
$= \displaystyle \lim_{x \to 0}\dfrac{cos^2\ x\ sin\ x - sin\ x}{(cos\ x)x^2sin\ x}$
$= \displaystyle \lim_{x \to 0}\dfrac{sin\ x(cos^2\ x - 1)}{(cos\ x)x^2sin\ x}$
$= \displaystyle \lim_{x \to 0}\dfrac{(cos^2\ x - 1)}{(cos\ x)x^2}$
$= \displaystyle \lim_{x \to 0}\dfrac{-sin^2\ x}{(cos\ x)x^2}$
$= -\displaystyle \lim_{x \to 0}\dfrac{sin^2\ x}{x^2}.\displaystyle \lim_{x \to 0}\dfrac{1}{cos\ x}$
$= -\displaystyle \lim_{x \to 0}\dfrac{sin\ x.sin\ x}{x.x}.\displaystyle \lim_{x \to 0}\dfrac{1}{cos\ x}$
$= -1.1.\dfrac{1}{cos\ 0}$
$= -1$
$Jawab:\ A.$
$36.\ \displaystyle \lim_{a \to b}\dfrac{tan\ a - tan\ b}{1 + (1 - \dfrac ab)tan\ a\ tan\ b - \dfrac ab} =$ . . . .
$A.\ \dfrac 1b$
$B.\ b$
$C.\ -b$
$D.\ \dfrac {-1}{b}$
$E.\ 1$
[Soal dan Pembahasan Limit Fungsi Trigonometri SIMAK UI 2011 MtkIPA]
$A.\ \dfrac 1b$
$B.\ b$
$C.\ -b$
$D.\ \dfrac {-1}{b}$
$E.\ 1$
[Soal dan Pembahasan Limit Fungsi Trigonometri SIMAK UI 2011 MtkIPA]
$\displaystyle \lim_{a \to b}\dfrac{tan\ a - tan\ b}{1 + (1 - \dfrac ab)tan\ a\ tan\ b - \dfrac ab}$
$= \displaystyle \lim_{a \to b}\dfrac{tan\ a - tan\ b}{1 + tan\ a\ tan\ b - \dfrac abtan\ a\ tan\ b - \dfrac ab}$
$= \displaystyle \lim_{a \to b}\dfrac{tan\ a - tan\ b}{1 + tan\ a\ tan\ b - \dfrac ab(1 + tan\ a\ tan\ b)}$
$= \displaystyle \lim_{a \to b}\dfrac{tan\ a - tan\ b}{(1 + tan\ a\ tan\ b)(1 - \dfrac ab)}$
$Ingat:\ tan(a - b) = \dfrac{tan\ a - tan\ b}{1 + tan\ a\ tan\ b}$
$= \displaystyle \lim_{a \to b}\dfrac{tan(a - b)}{1 - \dfrac ab}$
$= \displaystyle \lim_{a \to b}\dfrac{tan(a - b)}{\dfrac{b - a}{b}}$
$= b.\displaystyle \lim_{a \to b}\dfrac{tan(a - b)}{-(a - b)}$
$= -b.\displaystyle \lim_{(a - b) \to 0}\dfrac{tan(a - b)}{(a - b)}$
$= -b.1$
$= -b$
$Jawab:\ C.$
$= \displaystyle \lim_{a \to b}\dfrac{tan\ a - tan\ b}{1 + tan\ a\ tan\ b - \dfrac abtan\ a\ tan\ b - \dfrac ab}$
$= \displaystyle \lim_{a \to b}\dfrac{tan\ a - tan\ b}{1 + tan\ a\ tan\ b - \dfrac ab(1 + tan\ a\ tan\ b)}$
$= \displaystyle \lim_{a \to b}\dfrac{tan\ a - tan\ b}{(1 + tan\ a\ tan\ b)(1 - \dfrac ab)}$
$Ingat:\ tan(a - b) = \dfrac{tan\ a - tan\ b}{1 + tan\ a\ tan\ b}$
$= \displaystyle \lim_{a \to b}\dfrac{tan(a - b)}{1 - \dfrac ab}$
$= \displaystyle \lim_{a \to b}\dfrac{tan(a - b)}{\dfrac{b - a}{b}}$
$= b.\displaystyle \lim_{a \to b}\dfrac{tan(a - b)}{-(a - b)}$
$= -b.\displaystyle \lim_{(a - b) \to 0}\dfrac{tan(a - b)}{(a - b)}$
$= -b.1$
$= -b$
$Jawab:\ C.$
$37.\ \displaystyle \lim_{x \to \dfrac {\pi}{4}}\dfrac{\dfrac{1}{\sqrt{2}}sin(\dfrac{\pi}{4} - 2x) + \dfrac{1}{\sqrt{2}}cos(\dfrac{\pi}{4} - 2x) }{4x - \pi} =$ . . . .
$A.\ \dfrac 14$
$B.\ \dfrac 12$
$C.\ 0$
$D.\ -\dfrac 14$
$E.\ -\dfrac 12$
[Soal dan Pembahasan Limit Fungsi Trigonometri Utul 2009 MtkIPA]
$A.\ \dfrac 14$
$B.\ \dfrac 12$
$C.\ 0$
$D.\ -\dfrac 14$
$E.\ -\dfrac 12$
[Soal dan Pembahasan Limit Fungsi Trigonometri Utul 2009 MtkIPA]
$Ingat\ !$
$acos\ P + bsin\ P = kcos(P - \theta)$
$k = \sqrt{a^2 + b^2}$
$tan\ \theta = \dfrac ba$
$\theta = arc\ tan\left(\dfrac ba\right)$
$Soal\ !$
$\displaystyle \lim_{x \to \dfrac {\pi}{4}}\dfrac{\dfrac{1}{\sqrt{2}}sin(\dfrac{\pi}{4} - 2x) + \dfrac{1}{\sqrt{2}}cos(\dfrac{\pi}{4} - 2x) }{4x - \pi}$
$Lihat\ soal\ dan\ perhatikan\ pembilangnya\ !$
$k^2 = \left(\dfrac{1}{\sqrt{2}} \right)^2 + \left(\dfrac{1}{\sqrt{2}} \right)^2$
$K^2 = 1$
$k = 1$
$tan\ \theta = \dfrac{\dfrac{1}{\sqrt{2}}}{\dfrac{1}{\sqrt{2}}} = 1$
$\theta = \dfrac{\pi}{4}$
$= \displaystyle \lim_{x \to \dfrac {\pi}{4}}\dfrac{cos\ \left(\dfrac{\pi}{4} - 2x - \dfrac{\pi}{4} \right) }{4x - \pi}$
$= \displaystyle \lim_{x \to \dfrac {\pi}{4}}\dfrac{cos\ \left(-2x \right) }{4x - \pi}$
$= \displaystyle \lim_{x \to \dfrac {\pi}{4}}\dfrac{cos\ \left(2x \right) }{4x - \pi}$
$Gunakan\ Aturan\ L'Hospital\ !$
$= \displaystyle \lim_{x \to \dfrac {\pi}{4}}\dfrac{-2sin\ \left(2x \right) }{4}$
$= \displaystyle \lim_{x \to \dfrac {\pi}{4}}\dfrac{-2sin\ \left(2.\dfrac{\pi}{4} \right) }{4}$
$= \dfrac{-2sin\ \left(\dfrac{\pi}{2} \right) }{4}$
$= -\dfrac12$
$Jawab:\ E.$
$acos\ P + bsin\ P = kcos(P - \theta)$
$k = \sqrt{a^2 + b^2}$
$tan\ \theta = \dfrac ba$
$\theta = arc\ tan\left(\dfrac ba\right)$
$Soal\ !$
$\displaystyle \lim_{x \to \dfrac {\pi}{4}}\dfrac{\dfrac{1}{\sqrt{2}}sin(\dfrac{\pi}{4} - 2x) + \dfrac{1}{\sqrt{2}}cos(\dfrac{\pi}{4} - 2x) }{4x - \pi}$
$Lihat\ soal\ dan\ perhatikan\ pembilangnya\ !$
$k^2 = \left(\dfrac{1}{\sqrt{2}} \right)^2 + \left(\dfrac{1}{\sqrt{2}} \right)^2$
$K^2 = 1$
$k = 1$
$tan\ \theta = \dfrac{\dfrac{1}{\sqrt{2}}}{\dfrac{1}{\sqrt{2}}} = 1$
$\theta = \dfrac{\pi}{4}$
$= \displaystyle \lim_{x \to \dfrac {\pi}{4}}\dfrac{cos\ \left(\dfrac{\pi}{4} - 2x - \dfrac{\pi}{4} \right) }{4x - \pi}$
$= \displaystyle \lim_{x \to \dfrac {\pi}{4}}\dfrac{cos\ \left(-2x \right) }{4x - \pi}$
$= \displaystyle \lim_{x \to \dfrac {\pi}{4}}\dfrac{cos\ \left(2x \right) }{4x - \pi}$
$Gunakan\ Aturan\ L'Hospital\ !$
$= \displaystyle \lim_{x \to \dfrac {\pi}{4}}\dfrac{-2sin\ \left(2x \right) }{4}$
$= \displaystyle \lim_{x \to \dfrac {\pi}{4}}\dfrac{-2sin\ \left(2.\dfrac{\pi}{4} \right) }{4}$
$= \dfrac{-2sin\ \left(\dfrac{\pi}{2} \right) }{4}$
$= -\dfrac12$
$Jawab:\ E.$
$38.\ \displaystyle \lim_{x \to \dfrac{\pi}{2}} \dfrac{sin\ xtan\ (2x - \pi)}{2\pi - 4x} =$ . . . .
$A.\ -\dfrac12$
$B.\ \dfrac12$
$C.\ \dfrac13\sqrt{3}$
$D. 1$
$E.\ \sqrt{3}$
[Soal dan Pembahasan Limit Fungsi Trigonometri SPMB 2006 MtkIPA]
$A.\ -\dfrac12$
$B.\ \dfrac12$
$C.\ \dfrac13\sqrt{3}$
$D. 1$
$E.\ \sqrt{3}$
[Soal dan Pembahasan Limit Fungsi Trigonometri SPMB 2006 MtkIPA]
$\displaystyle \lim_{x \to \dfrac{\pi}{2}} \dfrac{sin\ xtan\ (2x - \pi)}{2\pi - 4x}$
$= \displaystyle \lim_{x \to \dfrac{\pi}{2}} \dfrac{sin\ xtan\ 2(x - \dfrac{\pi}{2})}{-4(x - \dfrac{\pi}{2})}$
$= \left(-\dfrac 14\right).\left(\displaystyle \lim_{x \to \dfrac{\pi}{2}} sin\ x\right) .\left(\displaystyle \lim_{(x - \dfrac{\pi}{2}) \to 0}\dfrac{tan\ 2(x - \dfrac{\pi}{2})}{(x - \dfrac{\pi}{2})}\right)$
$= -\dfrac 14.sin\ \dfrac {\pi}{2}.2$
$= -\dfrac 14.1.2$
$= -\dfrac 12$
$Jawab:\ A.$
$= \displaystyle \lim_{x \to \dfrac{\pi}{2}} \dfrac{sin\ xtan\ 2(x - \dfrac{\pi}{2})}{-4(x - \dfrac{\pi}{2})}$
$= \left(-\dfrac 14\right).\left(\displaystyle \lim_{x \to \dfrac{\pi}{2}} sin\ x\right) .\left(\displaystyle \lim_{(x - \dfrac{\pi}{2}) \to 0}\dfrac{tan\ 2(x - \dfrac{\pi}{2})}{(x - \dfrac{\pi}{2})}\right)$
$= -\dfrac 14.sin\ \dfrac {\pi}{2}.2$
$= -\dfrac 14.1.2$
$= -\dfrac 12$
$Jawab:\ A.$
$39.\ \displaystyle \lim_{x \to \dfrac{\pi}{4}} \dfrac{\left(x - \dfrac{\pi}{4} \right)tan\left(3x - \dfrac{3\pi}{4} \right)} {2(1 - sin\ 2x)} =$ . . . .
$A.\ 0$
$B.\ -\dfrac32$
$C.\ \dfrac32$
$D.\ -\dfrac34$
$E.\ \dfrac34$
[Soal dan Pembahasan Limit Fungsi Trigonometri Utul 2005 MtkIPA]
$A.\ 0$
$B.\ -\dfrac32$
$C.\ \dfrac32$
$D.\ -\dfrac34$
$E.\ \dfrac34$
[Soal dan Pembahasan Limit Fungsi Trigonometri Utul 2005 MtkIPA]
$\displaystyle \lim_{x \to \dfrac{\pi}{4}} \dfrac{\left(x - \dfrac{\pi}{4} \right)tan\left(3x - \dfrac{3\pi}{4} \right)} {2(1 - sin\ 2x)}$
$= \displaystyle \lim_{x \to \dfrac{\pi}{4}} \dfrac{\left(x - \dfrac{\pi}{4} \right)tan\left(3x - \dfrac{3\pi}{4} \right)} {2(sin^2\ x + cos^2\ x - 2sin\ xcos\ x)}$
$\ \ Ingat:\ sin^2\ x + cos^2\ x = 1$
$\ sin\ 2x = 2sin\ xcos\ x$
$= \displaystyle \lim_{x \to \dfrac{\pi}{4}} \dfrac{\left(x - \dfrac{\pi}{4} \right)tan3\left(x - \dfrac{\pi}{4} \right)} {2\left(sin\ x - cos\ x\right)^2}$
$Ingat\ !$
$\ cos\ x = sin\ \left(\dfrac{\pi}{2} - x\right)$
$sin\ A - sin\ B = 2cos\dfrac12(A + B)sin\dfrac12(A - B)$
$= \displaystyle \lim_{x \to \dfrac{\pi}{4}} \dfrac{\left(x - \dfrac{\pi}{4} \right)tan3\left(x - \dfrac{\pi}{4} \right)} {2\left(sin\ x - sin\ \left(\dfrac{\pi}{2} - x \right) \right)^2}$
$= \displaystyle \lim_{x \to \dfrac{\pi}{4}} \dfrac{\left(x - \dfrac{\pi}{4} \right)tan3\left(x - \dfrac{\pi}{4} \right)} {2\left(2cos\ \dfrac{\pi}{4} sin\ \left(x - \dfrac{\pi}{4}\right) \right)^2}$
$= \displaystyle \lim_{x \to \dfrac{\pi}{4}} \dfrac{\left(x - \dfrac{\pi}{4} \right)tan3\left(x - \dfrac{\pi}{4} \right)} {2\left(4.\dfrac12. sin^2\ \left(x - \dfrac{\pi}{4}\right) \right)}$
$= \dfrac14.\displaystyle \lim_{x \to \dfrac{\pi}{4}} \dfrac{\left(x - \dfrac{\pi}{4} \right)tan3\left(x - \dfrac{\pi}{4} \right)} {\left( sin^2\ \left(x - \dfrac{\pi}{4}\right) \right)}$
$= \dfrac14.\displaystyle \lim_{\left(x - \dfrac{\pi}{4}\right) \to 0} \dfrac{\left(x - \dfrac{\pi}{4} \right)tan3\left(x - \dfrac{\pi}{4} \right)} {\left( sin\ \left(x - \dfrac{\pi}{4}\right).sin\ \left(x - \dfrac{\pi}{4}\right) \right)}$
$= \dfrac14.1.3$
$= \dfrac34$
$Jawab:\ E.$
$= \displaystyle \lim_{x \to \dfrac{\pi}{4}} \dfrac{\left(x - \dfrac{\pi}{4} \right)tan\left(3x - \dfrac{3\pi}{4} \right)} {2(sin^2\ x + cos^2\ x - 2sin\ xcos\ x)}$
$\ \ Ingat:\ sin^2\ x + cos^2\ x = 1$
$\ sin\ 2x = 2sin\ xcos\ x$
$= \displaystyle \lim_{x \to \dfrac{\pi}{4}} \dfrac{\left(x - \dfrac{\pi}{4} \right)tan3\left(x - \dfrac{\pi}{4} \right)} {2\left(sin\ x - cos\ x\right)^2}$
$Ingat\ !$
$\ cos\ x = sin\ \left(\dfrac{\pi}{2} - x\right)$
$sin\ A - sin\ B = 2cos\dfrac12(A + B)sin\dfrac12(A - B)$
$= \displaystyle \lim_{x \to \dfrac{\pi}{4}} \dfrac{\left(x - \dfrac{\pi}{4} \right)tan3\left(x - \dfrac{\pi}{4} \right)} {2\left(sin\ x - sin\ \left(\dfrac{\pi}{2} - x \right) \right)^2}$
$= \displaystyle \lim_{x \to \dfrac{\pi}{4}} \dfrac{\left(x - \dfrac{\pi}{4} \right)tan3\left(x - \dfrac{\pi}{4} \right)} {2\left(2cos\ \dfrac{\pi}{4} sin\ \left(x - \dfrac{\pi}{4}\right) \right)^2}$
$= \displaystyle \lim_{x \to \dfrac{\pi}{4}} \dfrac{\left(x - \dfrac{\pi}{4} \right)tan3\left(x - \dfrac{\pi}{4} \right)} {2\left(4.\dfrac12. sin^2\ \left(x - \dfrac{\pi}{4}\right) \right)}$
$= \dfrac14.\displaystyle \lim_{x \to \dfrac{\pi}{4}} \dfrac{\left(x - \dfrac{\pi}{4} \right)tan3\left(x - \dfrac{\pi}{4} \right)} {\left( sin^2\ \left(x - \dfrac{\pi}{4}\right) \right)}$
$= \dfrac14.\displaystyle \lim_{\left(x - \dfrac{\pi}{4}\right) \to 0} \dfrac{\left(x - \dfrac{\pi}{4} \right)tan3\left(x - \dfrac{\pi}{4} \right)} {\left( sin\ \left(x - \dfrac{\pi}{4}\right).sin\ \left(x - \dfrac{\pi}{4}\right) \right)}$
$= \dfrac14.1.3$
$= \dfrac34$
$Jawab:\ E.$
$40.\ \displaystyle \lim_{x \to \infty}x^2sin\ \dfrac 1x tan\ \dfrac 1x =$ . . . .
$A.\ -1$
$B.\ 0$
$C.\ \dfrac12$
$D.\ 1$
$E.\ 2$
[Soal dan Pembahasan Limit Fungsi Trigonometri SPMB 2005 MtkIPA]
$A.\ -1$
$B.\ 0$
$C.\ \dfrac12$
$D.\ 1$
$E.\ 2$
[Soal dan Pembahasan Limit Fungsi Trigonometri SPMB 2005 MtkIPA]
$\displaystyle \lim_{x \to \infty}x^2sin\ \dfrac 1x tan\ \dfrac 1x = \displaystyle \lim_{\dfrac 1x \to 0}x^2sin\ \dfrac 1x tan\ \dfrac 1x$
$= \displaystyle \lim_{\dfrac 1x \to 0}\dfrac{sin\ \dfrac 1x tan\ \dfrac 1x}{\left(\dfrac 1x\right)^2}$
$Misalkan\ \dfrac 1x = p$
$= \displaystyle \lim_{p \to 0}\dfrac{sin\ p\ tan\ p}{p^2}$
$= \displaystyle \lim_{p \to 0}\dfrac{sin\ p\ tan\ p}{p.p}$
$= 1.1$
$= 1$
$Jawab:\ D.$
$= \displaystyle \lim_{\dfrac 1x \to 0}\dfrac{sin\ \dfrac 1x tan\ \dfrac 1x}{\left(\dfrac 1x\right)^2}$
$Misalkan\ \dfrac 1x = p$
$= \displaystyle \lim_{p \to 0}\dfrac{sin\ p\ tan\ p}{p^2}$
$= \displaystyle \lim_{p \to 0}\dfrac{sin\ p\ tan\ p}{p.p}$
$= 1.1$
$= 1$
$Jawab:\ D.$
Demikianlah Pembahasan Soal UN / UNBK dan SBMPTN Limit Fungsi Trigonometri, semoga bermanfaat. Selamat belajar !
Disusun oleh:
Joslin Sibarani
Alumni Teknik Sipil ITB
www.maretong.com
mntp
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