MARETONG: Soal dan Pembahasan Limit Fungsi Trigonometri

Friday, June 28, 2019

Soal dan Pembahasan Limit Fungsi Trigonometri


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.

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]
$cos\ 4x = cos(2x + 2x)$
$cos\ 4x = cos^2\ 2x - sin^2\ 2x$
$cos\ 4x = 1 - sin^2\ 2x - sin^2\ 2x$
$cos\ 4x = 1 - 2sin^2\ 2x$
$2sin^2\ 2x = 1 - cos\ 4x$ . . . . *

$sin\ 4x = sin(2x + 2x)$
$sin\ 4x = 2sin\ 2xcos\ 2x$ . . . . **

$\displaystyle \lim_{x \to 0}\dfrac{1 - cos\ 4x}{2x\ sin\ 4x} = \displaystyle \lim_{x \to 0}\dfrac{2sin^2\ 2x}{2x.2sin\ 2xcos\ 2x}$
$= \displaystyle \lim_{x \to 0}\dfrac{sin\ 2x}{2x.cos\ 2x}$
$= \displaystyle \lim_{x \to 0}\dfrac{sin\ 2x}{2x}.\displaystyle \lim_{x \to 0}\dfrac{1}{cos\ 2x}$
$= 1.1$
$= 1$
$Jawab:\ A.$

$2.\ Nilai\ \displaystyle \lim_{x \to 0}\dfrac{xtan\ 3x}{1 - cos^2\ x}\ adalah$ . . . .
$A.\ 0$
$B.\ \dfrac14$
$C.\ \dfrac24$
$D.\ \dfrac34$
$E.\ 1$
[Soal dan Pembahasan Limit Fungsi Trigonometri UN 2015 MtkIPA]
$1 - cos^2\ x = sin^2\ x$ . . . . *

$\displaystyle \lim_{x \to 0}\dfrac{xtan\ 3x}{1 - cos^2\ x} = \displaystyle \lim_{x \to 0}\dfrac{xtan\ 3x}{sin^2\ x}$
$= \displaystyle \lim_{x \to 0}\dfrac{x}{sin\ x}.\displaystyle \lim_{x \to 0}\dfrac{tan\ 3x}{sin\ x}$
$= 1.3$
$= 3$
$Jawab:\ -$

$3.\ Nilai\ \displaystyle \lim_{x \to 0}\dfrac{4xcos\ x}{sin\ x + sin\ 3x} =$ . . . .
$A.\ 4$
$B.\ 3$
$C.\ \dfrac43$
$D.\ 1$
$E.\ \dfrac34$
[Soal dan Pembahasan Limit Fungsi Trigonometri UN 2014 MtkIPA]
$sin\ A + sin\ B = 2sin\dfrac12(A + B)cos\dfrac12(A - B)$
$sin\ x + sin\ 3x = 2sin\dfrac12(x + 3x)cos\dfrac12(x - 3x)$
$= 2sin\ 2xcos\ (-x)$
$= 2sin\ 2xcos\ x → ingat:\ cos\ (-x) = cos\ x$

$\displaystyle \lim_{x \to 0}\dfrac{4xcos\ x}{sin\ x + sin\ 3x} = \displaystyle \lim_{x \to 0}\dfrac{4xcos\ x}{2sin\ 2xcos\ x}$
$=\displaystyle \lim_{x \to 0}\dfrac{4x}{2sin\ 2x} $
$= \dfrac12\displaystyle \lim_{x \to 0}\dfrac{4x}{sin\ 2x}$
$= \dfrac12.2$
$= 1$
$Jawab:\ D.$

$4.\ Nilai\ \displaystyle \lim_{x \to 1}\dfrac{sin^2(x - 1)}{x^2 - 2x + 1} =$ . . . .
$A.\ 0$
$B.\ 1$
$C.\ 2$
$D.\ 4$
$E.\ \infty$
[Soal dan Pembahasan Limit Fungsi Trigonometri UN 2013 MtkIPA]
$\displaystyle \lim_{x \to 1}\dfrac{sin^2(x - 1)}{x^2 - 2x + 1} = \displaystyle \lim_{x \to 1}\dfrac{sin^2(x - 1)}{(x - 1)^2}$
$= \displaystyle \lim_{(x - 1) \to 0}\left(\dfrac{sin\ (x - 1)}{(x - 1)}\right)^2$
$Misalkan\ x - 1 = p$
$= \displaystyle \lim_{p \to 0}\left(\dfrac{sin\ p}{p}\right)^2$
$= \left(\displaystyle \lim_{p \to 0}\dfrac{sin\ p}{p}\right)^2$
$= 1^2$
$= 1$
$Jawab:\ B.$

$5.\ Nilai\ \displaystyle \lim_{x \to 0}\dfrac{cos\ 4x - 1}{x tan\ 2x} =$ . . . .
$A.\ 4$
$B.\ 2$
$C.\ -1$
$D.\ -2$
$E.\ -4$
[Soal dan Pembahasan Limit Fungsi Trigonometri UN 2012 MtkIPA]
$2sin^2\ 2x = 1 - cos\ 4x$ (lihat pembahasan soal nomor 1.)
$-2sin^2\ 2x = cos\ 4x - 1$ . . . . *

$\displaystyle \lim_{x \to 0}\dfrac{cos\ 4x - 1}{x tan\ 2x} = \displaystyle \lim_{x \to 0}\dfrac{-2sin^2\ 2x}{x tan\ 2x}$
$= -2.\displaystyle \lim_{x \to 0}\dfrac{sin\ 2x}{x}.\displaystyle \lim_{x \to 0}\dfrac{sin\ 2x}{tan\ 2x}$
$= -2.2.1$
$= -4$
$Jawab:\ E.$

$6.\ Nilai\ \displaystyle \lim_{x \to 0}\dfrac{1 - cos\ 2x}{1 - cos\ 4x} =$ . . . .
$A.\ -\dfrac12$
$B.\ -\dfrac14$
$C.\ 0$
$D.\ \dfrac{1}{16}$
$E.\ \dfrac14$
[Soal dan Pembahasan Limit Fungsi Trigonometri UN 2011 MtkIPA]
$1 - cos\ 4x = 2sin^2\ 2x$ . . . . * Lihat pembahasan soal no 1.
$1 - cos\ 2x = 2sin^2\ x$ . . . . *

$\displaystyle \lim_{x \to 0}\dfrac{1 - cos\ 2x}{1 - cos\ 4x} = \displaystyle \lim_{x \to 0}\dfrac{2sin^2\ x}{2sin^2\ 2x}$
$= \displaystyle \lim_{x \to 0}\left(\dfrac{sin\ x}{sin\ 2x}\right)^2$
$= \left(\displaystyle \lim_{x \to 0}\dfrac{sin\ x}{sin\ 2x}\right)^2$
$= \left(\dfrac12 \right)^2$
$= \dfrac14$
$Jawab:\ E.$

$7.\ Nilai\ \displaystyle \lim_{x \to 0}\left(\dfrac{sin\ x + sin\ 5x}{6x} \right) =$ . . . .
$A.\ 2$
$B.\ 1$
$C.\ \dfrac12$
$D.\ \dfrac13$
$E.\ -1$
[Soal dan Pembahasan Limit Fungsi Trigonometri UN 2010 MtkIPA]
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:
$Bagi\ pembilang\ dan\ penyebut\ dengan\ x$
$\displaystyle \lim_{x \to 0}\left(\dfrac{sin\ x + sin\ 5x}{6x} \right) = \displaystyle \lim_{x \to 0}\left(\dfrac{\dfrac{sin\ x}{x} + \dfrac{sin\ 5x}{x}}{\dfrac{6x}{x}} \right)$
$= \left(\dfrac{\displaystyle \lim_{x \to 0}\dfrac{sin\ x}{x} + \displaystyle \lim_{x \to 0}\dfrac{sin\ 5x}{x}}{\displaystyle \lim_{x \to 0}\dfrac{6x}{x}} \right)$
$= \dfrac{1 + 5}{6}$
$= 1$
$Jawab:\ B.$

$8.$ 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]
$\displaystyle \lim_{x \to -3}\dfrac{x^2 + 6x + 9}{2 - 2cos(2x + 6)}$
$= \displaystyle \lim_{x \to -3}\dfrac{(x + 3)^2}{2(1 - cos2(x + 3))}$

$Analog:$
$1 - cos\ 2x = 2sin^2\ x$
$1 - cos\ 2(x + 3) = 2sin^2\ (x + 3)$

$= \displaystyle \lim_{(x + 3) \to 0}\dfrac{(x + 3)^2}{2.2sin^2\ (x + 3)}$
$= \dfrac14.\displaystyle \lim_{(x + 3) \to 0}\left(\dfrac{(x + 3)}{sin\ (x + 3)}\right)^2$
$= \dfrac14.\left(\displaystyle \lim_{(x + 3) \to 0}\dfrac{(x + 3)}{sin\ (x + 3)}\right)^2$
$Misalkan\ x + 3 = p$
$= \dfrac14.\left(\displaystyle \lim_{p \to 0}\dfrac{p}{sin\ p}\right)^2$
$= \dfrac14.1^2$
$= \dfrac14$
$Jawab:\ E.$

$9.\ Nilai\ \displaystyle \lim_{x \to 0}\dfrac{2x sin\ 3x}{1 - cos\ 6x} =$ . . . .
$A.\ -1$
$B.\ -\dfrac13$
$C.\ 0$
$D.\ \dfrac13$
$E.\ 1$
[Soal dan Pembahasan Limit Fungsi Trigonometri UN 2007 MtkIPA]
$1 - cos\ 6x = 2sin^2\ 3x$ . . . . *

$\displaystyle \lim_{x \to 0}\dfrac{2x sin\ 3x}{1 - cos\ 6x} = \displaystyle \lim_{x \to 0}\dfrac{2x sin\ 3x}{2sin^2\ 3x}$
$= \displaystyle \lim_{x \to 0}\dfrac{x}{sin\ 3x}.\displaystyle \lim_{x \to 0}\dfrac{sin\ 3x}{sin\ 3x}$
$= \dfrac13.1$
$= \dfrac13$
$Jawab:\ D.$

$10.$ 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]
$\displaystyle \lim_{x \to 0}\dfrac{tan\ 2xcos\ 8x - tan\ 2x}{16x^3}$
$= \displaystyle \lim_{x \to 0}\dfrac{tan\ 2x(cos\ 8x - 1)}{16x^3}$

$1 - cos\ 8x = 2sin^2\ 4x$
$cos\ 8x - 1 = -2sin^2\ 4x$

$=\displaystyle \lim_{x \to 0}\dfrac{tan\ 2x.(-2sin^2\ 4x)}{x.16x^2}$
$= -2.\displaystyle \lim_{x \to 0}\dfrac{tan\ 2x}{x}.\displaystyle \lim_{x \to 0}\dfrac{sin^2\ 4x}{16x^2}$
$= -2.\displaystyle \lim_{x \to 0}\dfrac{tan\ 2x}{x}.\displaystyle \lim_{x \to 0}\left(\dfrac{sin\ 4x}{4x}\right)^2$
$= -2.\displaystyle \lim_{x \to 0}\dfrac{tan\ 2x}{x}.\left(\displaystyle \lim_{x \to 0}\dfrac{sin\ 4x}{4x}\right)^2$
$= -2.2.1^2$
$= -4$
$Jawab:\ A.$

$11.\ Nilai\ \displaystyle \lim_{x \to -2}\dfrac{(x + 6)sin(x + 2)}{x^2 - 3x - 10} =$ . . . .
$A.\ -\dfrac43$
$B.\ -\dfrac47$
$C.\ -\dfrac25$
$D.\ 0$
$E.\ 1$
[Soal dan Pembahasan Limit Fungsi Trigonometri UN 2004 MtkIPA]
$\displaystyle \lim_{x \to -2}\dfrac{(x + 6)sin(x + 2)}{x^2 - 3x - 10}$
$= \displaystyle \lim_{x \to -2}\dfrac{(x + 6)sin(x + 2)}{(x - 5)(x + 2)}$
$= \displaystyle \lim_{x \to -2}\dfrac{(x + 6)}{(x - 5)}.\displaystyle \lim_{(x + 2) \to 0}\dfrac{sin(x + 2)}{(x + 2)}$
$= \dfrac{-2 + 6}{-2 - 5}.1$
$= -\dfrac47$
$Jawab:\ B.$

$12.\ \displaystyle \lim_{x \to 2}\dfrac{sin(2x - 4)}{2 - \sqrt{6 - x}} =$ . . . .
$A.\ -8$
$B.\ -2$
$C.\ 0$
$D.\ 2$
$E.\ 8$
[Soal dan Pembahasan Limit Fungsi Trigonometri SBMPTN 2018 MtkIPA]
$\displaystyle \lim_{x \to 2}\dfrac{sin(2x - 4)}{2 - \sqrt{6 - x}}$
$= \displaystyle \lim_{x \to 2}\dfrac{sin2(x - 2)}{2 - \sqrt{6 - x}}.\dfrac{2 + \sqrt{6 - x}}{2 + \sqrt{6 - x}}$
$= \displaystyle \lim_{x \to 2}\dfrac{sin2(x - 2)}{(x - 2)}.(2 + \sqrt{6 - x})$
$= \displaystyle \lim_{(x - 2) \to 0}\dfrac{sin2(x - 2)}{(x - 2)}.\displaystyle \lim_{x \to 2}(2 + \sqrt{6 - x})$
$= 2.(2 + \sqrt{6 - 2})$
$= 2.4$
$= 8$
$Jawab:\ E.$

$13.$ 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]
$\displaystyle \lim_{x \to \dfrac{\pi}{2}}\dfrac{xcot^2\ x}{1 - sin\ x} = \displaystyle \lim_{x \to \dfrac{\pi}{2}}\dfrac{xcos^2\ x}{sin^2\ x(1 - sin\ x)}$
$= \displaystyle \lim_{x \to \dfrac{\pi}{2}}\dfrac{x(1 - sin^2\ x)}{sin^2\ x(1 - sin\ x)}$
$= \displaystyle \lim_{x \to \dfrac{\pi}{2}}\dfrac{x(1 - sin\ x)(1 + sin\ x)}{sin^2\ x(1 - sin\ x)}$
$= \displaystyle \lim_{x \to \dfrac{\pi}{2}}\dfrac{x(1 + sin\ x)}{sin^2\ x}$
$= \dfrac{\dfrac{\pi}{2}(1 + sin\ \dfrac{\pi}{2})}{sin^2\ \dfrac{\pi}{2}}$
$= \dfrac{\dfrac{\pi}{2}(1 + 1)}{1}$
$= \pi$
$Jawab:\ E.$

$14.\ \displaystyle \lim_{x \to \infty}\dfrac{2x^2 tan\left(\dfrac{1}{x}\right) - xsin\left(\dfrac{1}{x} \right) + \dfrac{1}{x}}{xcos\left(\dfrac{2}{x}\right)} =$ . . . .
$A.\ 2$
$B.\ 1$
$C.\ 0$
$D.\ -1$
$E.\ -2$
[Soal dan Pembahasan Limit Fungsi Trigonometri SBMPTN 2017 MtkIPA]
$\displaystyle \lim_{x \to \infty}\dfrac{2x^2 tan\left(\dfrac{1}{x}\right) - xsin\left(\dfrac{1}{x} \right) + \dfrac{1}{x}}{xcos\left(\dfrac{2}{x}\right)}$
$= \displaystyle \lim_{x \to \infty}\dfrac{2x tan\left(\dfrac{1}{x}\right) - sin\left(\dfrac{1}{x} \right) + \dfrac{1}{x^2}}{cos\left(\dfrac{2}{x}\right)}$
$= \displaystyle \lim_{\dfrac{1}{x} \to 0}\dfrac{2tan\left(\dfrac{1}{x}\right) - sin\left(\dfrac{1}{x} \right) + \dfrac{1}{x^2}}{\dfrac{1}{x}.cos2\left(\dfrac{1}{x}\right)}$

$Misalkan\ \dfrac{1}{x} = p$

$= \displaystyle \lim_{p \to 0}\dfrac{2tan\ p - sin\ p + p^2}{p.cos\ 2p}$
$= \displaystyle \lim_{p \to 0}\left(\dfrac{2tan\ p}{p} - \dfrac{sin\ p}{p} + p\right).\displaystyle \lim_{p \to 0}\dfrac{1}{cos\ 2p}$
$= (2 - 1 + 0).\dfrac{1}{1}$
$= 1$
$Jawab:\ B.$

$15.\ \displaystyle \lim_{x \to 0}\dfrac{x(\sqrt{x + 1} - 1)}{1 - cos\ x} =$ . . . .
$A.\ 2$
$B.\ 1$
$C.\ \dfrac12$
$D.\ -\dfrac12$
$E.\ -1$
[Soal dan Pembahasan Limit Fungsi Trigonometri SBMPTN 2016 MtkIPA]
$\displaystyle \lim_{x \to 0}\dfrac{x(\sqrt{x + 1} - 1)}{1 - cos\ x}$
$= \displaystyle \lim_{x \to 0}\dfrac{x(\sqrt{x + 1} - 1)}{1 - cos\ x}.\dfrac{(\sqrt{x + 1} + 1)}{(\sqrt{x + 1} + 1)}$

$1 - cos\ x = 2sin^2\ \dfrac12x$

$= \displaystyle \lim_{x \to 0}\dfrac{x^2}{2sin^2\ \dfrac12x}.\dfrac{1}{(\sqrt{x + 1} + 1)}$
$= \dfrac12.\displaystyle \lim_{x \to 0}\dfrac{x^2}{sin^2\ \dfrac12x}.\displaystyle \lim_{x \to 0}\dfrac{1}{(\sqrt{x + 1} + 1)}$
$= \dfrac12.\displaystyle \lim_{x \to 0}\left(\dfrac{x}{sin\ \dfrac12x}\right)^2.\displaystyle \lim_{x \to 0}\dfrac{1}{(\sqrt{x + 1} + 1)}$
$= \dfrac12.\left(\displaystyle \lim_{x \to 0}\dfrac{x}{sin\ \dfrac12x}\right)^2.\displaystyle \lim_{x \to 0}\dfrac{1}{(\sqrt{x + 1} + 1)}$
$= \dfrac12.\displaystyle \left( 2\right)^2.\dfrac{1}{(\sqrt{0 + 1} + 1)}$
$= \dfrac12.4.\dfrac12$
$= 1$
$Jawab:\ B.$

$16.\ \displaystyle \lim_{x \to a}\dfrac{sin(x - a)tan\sqrt[3]{x - a}} {2\sqrt[3]{x - a} - \sqrt[3]{(x - a)^2}} =$ . . . .
$A.\ -a$
$B.\ -\dfrac{1}{a}$
$C.\ 0$
$D.\ \dfrac{1}{a}$
$E.\ a$
[Soal dan Pembahasan Limit Fungsi Trigonometri SIMAK UI 2017 MtkIPA]
$\displaystyle \lim_{x \to a}\dfrac{sin(x - a)tan\sqrt[3]{x - a}} {2\sqrt[3]{x - a} - \sqrt[3]{(x - a)^2}}$
$= \displaystyle \lim_{x \to a}\dfrac{sin(x - a)tan\sqrt[3]{x - a}} {\sqrt[3]{x - a}(1 - \sqrt[3]{(x - a)})}$
$= \displaystyle \lim_{x \to a}\dfrac{sin(x - a)}{(1 - \sqrt[3]{(x - a)}} .\displaystyle \lim_{(x - a) \to 0}\dfrac{ tan\sqrt[3]{x - a}}{\sqrt[3]{x - a}}$
$= \dfrac{sin(a - a)}{(1 - \sqrt[3]{(a - a)}}.1$
$= \dfrac{sin\ 0}{(1 - \sqrt[3]{0})}$
$= \dfrac{0}{1}$
$= 0$
$Jawab:\ C.$

$17.\ \displaystyle \lim_{x \to 0}\dfrac{sin\ xcos\ x}{\sqrt{\pi + 2sin\ x} - \sqrt{\pi}} =$ . . . .
$A.\ -2\sqrt{\pi}$
$B.\ -\sqrt{\pi}$
$C.\ 0$
$D.\ \sqrt{\pi}$
$E.\ 2\sqrt{\pi}$
[Soal dan Pembahasan Limit Fungsi Trigonometri SBMPTN 2018 MtkIPA]
$\displaystyle \lim_{x \to 0}\dfrac{sin\ xcos\ x}{\sqrt{\pi + 2sin\ x} - \sqrt{\pi}}$
$= \displaystyle \lim_{x \to 0}\dfrac{sin\ xcos\ x}{\sqrt{\pi + 2sin\ x} - \sqrt{\pi}}.\dfrac{(\sqrt{\pi + 2sin\ x} + \sqrt{\pi})}{(\sqrt{\pi + 2sin\ x} + \sqrt{\pi})}$
$= \displaystyle \lim_{x \to 0}\dfrac{sin\ xcos\ x}{2sin\ x}.(\sqrt{\pi + 2sin\ x} + \sqrt{\pi})$
$= \dfrac12.\displaystyle \lim_{x \to 0}(cos\ x).(\sqrt{\pi + 2sin\ x} + \sqrt{\pi})$
$= \dfrac12.(cos\ 0).(\sqrt{\pi + 2sin\ 0} + \sqrt{\pi})$
$= \dfrac12.1.(\sqrt{\pi} + \sqrt{\pi})$
$= \dfrac12\sqrt{\pi}$
$= \sqrt{\pi}$
$Jawab:\ D.$

$18.\ \displaystyle \lim_{x \to \pi}\dfrac{sin\ x}{\sqrt{\pi + tan\ x} - \sqrt{\pi - tan\ x}} =$ . . . .
$A.\ -2\sqrt{\pi}$
$B.\ -\sqrt{\pi}$
$C.\ 0$
$D.\ \sqrt{\pi}$
$E.\ 2\sqrt{\pi}$
[Soal dan Pembahasan Limit Fungsi Trigonometri SBMPTN 2018 MtkIPA]
$\displaystyle \lim_{x \to \pi}\dfrac{sin\ x}{\sqrt{\pi + tan\ x} - \sqrt{\pi - tan\ x}}$
$= \displaystyle \lim_{x \to \pi}\dfrac{sin\ x}{\sqrt{\pi + tan\ x} - \sqrt{\pi - tan\ x}}.\dfrac{\sqrt{\pi + tan\ x} + \sqrt{\pi - tan\ x}}{\sqrt{\pi + tan\ x} + \sqrt{\pi - tan\ x}}$
$= \displaystyle \lim_{x \to \pi}\dfrac{sin\ x}{2tan\ x}.(\sqrt{\pi + tan\ x} + \sqrt{\pi - tan\ x})$
$= \dfrac12.\displaystyle \lim_{x \to \pi}\dfrac{sin\ x}{tan\ x}.(\sqrt{\pi + tan\ x} + \sqrt{\pi - tan\ x})$
$= \dfrac12.\displaystyle \lim_{x \to \pi}\dfrac{sin\ x}{\dfrac{sin\ x}{cos\ x}}.(\sqrt{\pi + tan\ x} + \sqrt{\pi - tan\ x})$
$= \dfrac12.\displaystyle \lim_{x \to \pi}cos\ x.(\sqrt{\pi + tan\ x} + \sqrt{\pi - tan\ x})$
$= \dfrac12.cos\ \pi.(\sqrt{\pi + tan\ \pi} + \sqrt{\pi - tan\ \pi})$
$= \dfrac12.-1.(\sqrt{\pi + 0} + \sqrt{\pi - 0})$
$= -\dfrac12.2\sqrt{\pi}$
$= -\sqrt{\pi}$
$Jawab:\ B.$

$19.\ \displaystyle \lim_{x \to 0}\dfrac{4x + 3xcos\ 2x}{sin\ xcos\ x} =$ . . . .
$A.\ 8$
$B.\ 7$
$C.\ 6$
$D.\ 5$
$E.\ 2$
[Soal dan Pembahasan Limit Fungsi Trigonometri SBMPTN 2017 MtkIPA]
$Bagi\ pembilang\ dan\ penyebut\ dengan\ x\ !$

$\displaystyle \lim_{x \to 0}\dfrac{4x + 3xcos\ 2x}{sin\ xcos\ x} = \displaystyle \lim_{x \to 0}\dfrac{\dfrac{4x}{x} + \dfrac{3xcos\ 2x}{x}}{\dfrac{sin\ xcos\ x}{x}}$
$= \dfrac{\displaystyle \lim_{x \to 0}4 + \displaystyle \lim_{x \to 0}3cos\ 2x}{\displaystyle \lim_{x \to 0}\dfrac{sin\ x}{x}.\displaystyle \lim_{x \to 0}cos\ x}$
$= \dfrac{4 + 3.cos\ 2.0}{1.cos\ 0}$
$= \dfrac{4 + 3.1}{1.1}$
$= 7$
$Jawab:\ B.$

$20.\ \displaystyle \lim_{x \to \infty}\left(sec\ \dfrac{1}{\sqrt{x}} - 1 \right) =$ . . . .
$A.\ 1$
$B.\ \dfrac12$
$C.\ 0$
$D.\ -\dfrac12$
$E.\ -1$
[Soal dan Pembahasan Limit Fungsi Trigonometri SBMPTN 2017 Matematika IPA]
$\displaystyle \lim_{x \to \infty}\left(sec\ \dfrac{1}{\sqrt{x}} - 1 \right) = \displaystyle \lim_{\dfrac{1}{x} \to 0}\left(sec\ \sqrt{\dfrac{1}{x}} - 1 \right)$

$Misalkan\ \dfrac{1}{x} = p$

$= \displaystyle \lim_{p \to 0}\left(sec\ \sqrt{p} - 1 \right)$
$= sec\ \sqrt{0} - 1$
$= sec\ 0 - 1$
$= 1 - 1$
$= 0$
$Jawab:\ C.$

$21.\ Nilai\ dari\ \displaystyle \lim_{x \to 2}\dfrac{\sqrt{1 - cos(x - 2)}}{\sqrt{x^2 - 2x}}\ adalah$ . . . .
$A.\ 0$
$B.\ \dfrac12$
$C.\ \dfrac34$
$D.\ 1$
$E.\ \infty$
[Soal dan Pembahasan Limit Fungsi Trigonometri SBMPTN 2016 MtkIPA]
$\displaystyle \lim_{x \to 2}\dfrac{\sqrt{1 - cos(x - 2)}}{\sqrt{x^2 - 2x}} = \displaystyle \lim_{x \to 2}\sqrt{\dfrac{1 - cos(x - 2)}{x^2 - 2x}}$
$= \sqrt{\displaystyle \lim_{x \to 2}\dfrac{1 - cos(x - 2)}{x^2 - 2x}}$
$= \sqrt{\displaystyle \lim_{x \to 2}\dfrac{2sin^2\dfrac12(x - 2)}{x(x - 2)}}$
$= \sqrt{\displaystyle \lim_{(x - 2) \to 0}\dfrac{2sin\dfrac12(x - 2)}{(x - 2)}.\displaystyle \lim_{x \to 2}\dfrac{sin\dfrac12(x - 2)}{x}}$
$= \sqrt{2.\dfrac12.\dfrac{sin\dfrac12(2 - 2)}{2}}$
$= \sqrt{1.\dfrac{sin\ 0}{2}}$
$= \sqrt{\dfrac{0}{2}}$
$= \sqrt{0}$
$= 0$
$Jawab:\ A.$

$22.\ \displaystyle \lim_{x \to 0}\dfrac{\sqrt{2x^2 + 1} - 1}{\sqrt{3sin^5\ x + x^4}} =$ . . . .
$A.\ 0$
$B.\ \dfrac{\sqrt{2}}{\sqrt{3}}$
$C.\ \dfrac{\sqrt{3}}{\sqrt{4}}$
$D.\ \dfrac12$
$E.\ 1$
[Soal dan Pembahasan Limit Fungsi Trigonometri SBMPTN 2016 MtkIPA]
$\displaystyle \lim_{x \to 0}\dfrac{\sqrt{2x^2 + 1} - 1}{\sqrt{3sin^5\ x + x^4}}$
$= \displaystyle \lim_{x \to 0}\dfrac{\sqrt{2x^2 + 1} - 1}{\sqrt{3sin^5\ x + x^4}}.\dfrac{(\sqrt{2x^2 + 1} + 1)}{(\sqrt{2x^2 + 1} + 1)}$
$= \displaystyle \lim_{x \to 0}\dfrac{2x^2}{\sqrt{3sin^5\ x + x^4}}.\dfrac{1}{(\sqrt{2x^2 + 1} + 1)}$
$= \displaystyle \lim_{x \to 0}\dfrac{2x^2}{\sqrt{3sin^5\ x + x^4}}.\dfrac{1}{(\sqrt{2x^2 + 1} + 1)}\ \times\ \dfrac{\dfrac{1}{x^2}}{\dfrac{1}{x^2}}$
$= \displaystyle \lim_{x \to 0}\dfrac{2}{\sqrt{\dfrac{3sin^5\ x + x^4}{x^4}}}.\dfrac{1}{(\sqrt{2x^2 + 1} + 1)}$
$= \displaystyle \lim_{x \to 0}\dfrac{2}{\sqrt{3\dfrac{sin^4\ x}{x^4}.sin\ x + 1}}.\dfrac{1}{(\sqrt{2x^2 + 1} + 1)}$
$= \displaystyle \lim_{x \to 0}\dfrac{2}{\sqrt{3.sin\ x + 1}}.\dfrac{1}{(\sqrt{2x^2 + 1} + 1)}$
$= \dfrac{2}{\sqrt{3.sin\ 0 + 1}}.\dfrac{1}{(\sqrt{2.0^2 + 1} + 1)}$
$= 2.\dfrac12$
$= 1$
$Jawab:\ E.$

$23.\ \displaystyle \lim_{x \to 0}\sqrt{\dfrac{xtan\ x}{sin^2\ x - cos\ 2x + 1}} =$ . . . .
$A.\ 3$
$B.\ \sqrt{3}$
$C.\ \dfrac{\sqrt{3}}{3}$
$D.\ \dfrac13$
$E.\ \dfrac{\sqrt{3}}{2}$
[Soal dan Pembahasan Limit Fungsi Trigonometri SBMPTN 2013 MtkIPA]
$\displaystyle \lim_{x \to 0}\sqrt{\dfrac{xtan\ x}{sin^2\ x - cos\ 2x + 1}}$
$= \displaystyle \lim_{x \to 0}\sqrt{\dfrac{xtan\ x}{sin^2\ x - (1 - 2sin^2\ x) + 1}}$
$= \displaystyle \lim_{x \to 0}\sqrt{\dfrac{xtan\ x}{3sin^2\ x}}$
$= \sqrt{\displaystyle \lim_{x \to 0}\dfrac{xtan\ x}{3sin^2\ x}}$
$= \sqrt{\dfrac13.\displaystyle \lim_{x \to 0}\dfrac{x}{sin\ x}.\displaystyle \lim_{x \to 0}\dfrac{tan\ x}{sin\ x}}$
$= \sqrt{\dfrac13.1.1}$
$= \sqrt{\dfrac13}$
$= \dfrac{\sqrt{3}}{3}$
$Jawab:\ C.$

$24.\ \displaystyle \lim_{x \to 0}\dfrac{sin^2\ x - cos\ x + 1}{xtan\ x} =$ . . . .
$A.\ \dfrac32$
$B.\ \dfrac12$
$C.\ -\dfrac12$
$D.\ -1$
$E.\ -2$
[Soal dan Pembahasan Limit Fungsi Trigonometri SBMPTN 2013 MtkIPA]
$\displaystyle \lim_{x \to 0}\dfrac{sin^2\ x - cos\ x + 1}{xtan\ x}$
$= \displaystyle \lim_{x \to 0}\dfrac{sin^2\ x - (1 - 2sin^2\ \dfrac12x) + 1}{xtan\ x}$
$= \displaystyle \lim_{x \to 0}\dfrac{sin^2\ x + 2sin^2\ \dfrac12x}{xtan\ x}$
$= \displaystyle \lim_{x \to 0}\left(\dfrac{sin^2\ x}{xtan\ x} + \dfrac{2sin^2\ \dfrac12x}{xtan\ x}\right)$
$= \left(\displaystyle \lim_{x \to 0}\dfrac{sin^2\ x}{xtan\ x} + \displaystyle \lim_{x \to 0}\dfrac{2sin^2\ \dfrac12x}{xtan\ x}\right)$
$= \left(\displaystyle \lim_{x \to 0}\dfrac{sin\ x.sin\ x}{x.tan\ x} + \displaystyle \lim_{x \to 0}\dfrac{2sin\ \dfrac12x.sin\ \dfrac12x}{x.tan\ x}\right)$
$= 1.1 + 2.\dfrac12.\dfrac12$
$= 1 + \dfrac12$
$= \dfrac32$
$Jawab:\ A.$

$25.\ \displaystyle \lim_{x \to 0}\dfrac{1 - cos^2\ 2x}{x^2tan\left(x + \dfrac{\pi}{4}\right)} =$ . . . .
$A.\ -2$
$B.\ 0$
$C.\ \sqrt{2}$
$D.\ \sqrt{3}$
$E.\ 4$
[Soal dan Pembahasan Limit Fungsi Trigonometri SNMPTN 2012 MtkIPA]
$sin^2\ 2x + cos^2\ 2x = 1$
$sin^2\ 2x = 1 - cos^2\ 2x$ . . . . *

$\displaystyle \lim_{x \to 0}\dfrac{1 - cos^2\ 2x}{x^2tan\left(x + \dfrac{\pi}{4}\right)} = \displaystyle \lim_{x \to 0}\dfrac{sin^2\ 2x}{x^2tan\left(x + \dfrac{\pi}{4}\right)}$
$= \displaystyle \lim_{x \to 0}\dfrac{sin^2\ 2x}{x^2}.\displaystyle \lim_{x \to 0}\dfrac{1}{tan\left(x + \dfrac{\pi}{4}\right)}$
$= \displaystyle \lim_{x \to 0}\dfrac{sin\ 2x.sin\ 2x}{x.x}.\displaystyle \lim_{x \to 0}\dfrac{1}{tan\left(x + \dfrac{\pi}{4}\right)}$
$= 2.2.\dfrac{1}{tan\left(0 + \dfrac{\pi}{4}\right)}$
$= 2.2.\dfrac{1}{tan\left(0 + \dfrac{\pi}{4}\right)}$
$= 2.2.1$
$= 4$
$Jawab:\ E.$

$26.\ \displaystyle \lim_{x \to -y}\dfrac{tan\ x + tan\ y}{\left(\dfrac{x^2 - y^2}{-2y^2}\right)(1 - tan\ xtan\ y)} =$ . . . .
$A.\ -1$
$B.\ 1$
$C.\ 0$
$D.\ y$
$E.\ -y$
[Soal dan Pembahasan Limit Fungsi Trigonometri Utul 2017 MtkIPA]
$tan(x + y) = \dfrac{tan\ x + tan\ y}{1 - tan\ xtan\ y}$

$\displaystyle \lim_{x \to -y}\dfrac{tan\ x + tan\ y}{\left(\dfrac{x^2 - y^2}{-2y^2}\right)(1 - tan\ xtan\ y)}$
$= \displaystyle \lim_{x \to -y}\dfrac{tan( x + y)}{\left(\dfrac{x^2 - y^2}{-2y^2}\right)}$
$= \displaystyle \lim_{x \to -y}\dfrac{-2y^2tan( x + y)}{x^2 - y^2}$
$= \displaystyle \lim_{x \to -y}\dfrac{-2y^2tan( x + y)}{(x - y)(x + y)}$
$= \displaystyle \lim_{x \to -y}\dfrac{-2y^2}{(x - y)}.\displaystyle \lim_{(x + y) \to 0}\dfrac{tan( x + y)}{(x + y)}$
$= \dfrac{-2y^2}{(-y - y)}.1$
$= \dfrac{-2y^2}{-2y}$
$= y$
$Jawab:\ D.$

$27.\ \displaystyle \lim_{x \to -4}\dfrac{1 - cos(x + 4)}{x^2 + 8x + 16} =$ . . . .
$A. -2$
$B.\ -\dfrac12$
$C.\ \dfrac13$
$D.\ \dfrac12$
$E.\ 2$
[Soal dan Pembahasan Limit Fungsi Trigonometri Utul 2017 MtkIPA]
$\displaystyle \lim_{x \to -4}\dfrac{1 - cos(x + 4)}{x^2 + 8x + 16} = \displaystyle \lim_{x \to -4}\dfrac{2sin^2\ \dfrac12(x + 4)}{(x + 4)^2}$
$= 2.\left(\displaystyle \lim_{(x + 4) \to 0}\dfrac{sin\ \dfrac12(x + 4)}{(x + 4)}\right)^2$
$= 2.\left(\dfrac12\right)^2$
$= 2.\dfrac14$
$= \dfrac12$
$Jawab:\ D.$

$28.\ \displaystyle \lim_{x \to 3}\dfrac{(x + 6)tan(2x - 6)}{x^2 - x - 6} =$ . . . .
$A.\ -\dfrac{18}{5}$
$B.\ -\dfrac95$
$C.\ \dfrac95$
$D.\ \dfrac{18}{5}$
$E.\ \dfrac{27}{5}$
[Soal dan Pembahasan Limit Fungsi Trigonometri Utul 2016 MtkIPA]
$\displaystyle \lim_{x \to 3}\dfrac{(x + 6)tan(2x - 6)}{x^2 - x - 6}$
$= \displaystyle \lim_{x \to 3}\dfrac{(x + 6)tan2(x - 3)}{(x + 2)(x - 3)}$
$= \displaystyle \lim_{x \to 3}\dfrac{(x + 6)}{(x + 2)}.\displaystyle \lim_{(x - 3) \to 0}\dfrac{tan2(x - 3)}{(x - 3)}$
$= \dfrac{(3 + 6)}{(3 + 2)}.2$
$= \dfrac{18}{5}$
$Jawab:\ D.$

$29.\ \displaystyle \lim_{x \to 0}\dfrac{1 - cos^3\ x}{xtan\ x} =$ . . . .
$A.\ 0$
$B.\ \dfrac12$
$C.\ \dfrac34$
$D.\ \dfrac32$
$E.\ 3$
[Soal dan Pembahasan Limit Fungsi Trigonometri Utul 2013 MtkIPA]
$\left(1 - cos\ x\right)^3 = 1 - 3cos\ x + 3cos^2\ x - cos^3\ x$
$\left(1 - cos\ x\right)^3 + 3cos\ x - 3cos^2\ x = 1 - cos^3\ x$
$\left(1 - cos\ x\right)^3 + 3cos\ x(1 - cos\ x) = 1 - cos^3\ x$
$\left(2sin^2\ \dfrac12x \right)^3 + 3cos\ x\left(2sin^2\ \dfrac12x\right) = 1 - cos^3\ x$
$\left(8sin^6\ \dfrac12x \right) + 6cos\ x\left(sin^2\ \dfrac12x\right) = 1 - cos^3\ x$

$\displaystyle \lim_{x \to 0}\dfrac{1 - cos^3\ x}{xtan\ x} = \displaystyle \lim_{x \to 0}\dfrac{\left(8sin^6\ \dfrac12x \right) + 6cos\ x(sin^2\ \dfrac12x)}{xtan\ x}$

$= \displaystyle \lim_{x \to 0}\left(\dfrac{8sin^6\ \dfrac12x}{xtan\ x} \right) + \displaystyle \lim_{x \to 0}6cos\ x\left(\dfrac{2sin^2\ \dfrac12x}{xtan\ x}\right)$

$= 8\displaystyle \lim_{x \to 0}\left(\dfrac{sin^2\ \dfrac12x}{xtan\ x}.sin^4\ \dfrac12x \right) + 6\displaystyle \lim_{x \to 0}cos\ x\left(\dfrac{sin^2\ \dfrac12x}{xtan\ x}\right)$

$= 8.\dfrac14.0 + 6.1.\dfrac14$

$= 0 + \dfrac32$

$= \dfrac32$

$Jawab:\ D.$

$30.$ 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]
$Ingat:\ sudut-sudut\ berelasi\ !$
$sin(\dfrac{\pi}{4} - x) = cos(\dfrac{\pi}{2} - (\dfrac{\pi}{4} - x))$
$sin(\dfrac{\pi}{4} - x) = cos(x + \dfrac{\pi}{4})$ . . . . *

$\displaystyle \lim_{x \to \dfrac{\pi}{4}}sin(\dfrac{\pi}{4} - x)tan(x + \dfrac{\pi}{4})$
$= \displaystyle \lim_{x \to \dfrac{\pi}{4}}cos(x + \dfrac{\pi}{4})\dfrac{sin(x + \dfrac{\pi}{4})}{cos(x + \dfrac{\pi}{4})}$
$= \displaystyle \lim_{x \to \dfrac{\pi}{4}}sin(x + \dfrac{\pi}{4})$
$= sin(\dfrac{\pi}{4} + \dfrac{\pi}{4})$
$= sin(\dfrac{\pi}{2})$
$= 1$
$Jawab:\ B.$

$31.\ \displaystyle \lim_{x \to 0}\dfrac{5x - tan\ 5x}{x^3} =$ . . . .
$A.\ \dfrac{125}{3}$
$B.\ \dfrac{115}{3}$
$C.\ \dfrac{125}{6}$
$D.\ \dfrac{-125}{6}$
$E.\ \dfrac{-125}{3}$
[Soal dan Pembahasan Limit Fungsi Trigonometri SIMAK UI 2017 MtkIPA]
$Gunakan\ Aturan\ L'Hospital$
$\displaystyle \lim_{x \to 0}\dfrac{5x - tan\ 5x}{x^3} = \displaystyle \lim_{x \to 0}\dfrac{5 - 5sec^2\ 5x}{3x^2}$
$= \displaystyle \lim_{x \to 0}\dfrac{5(1 - sec^2\ 5x)}{3x^2}$

$Ingat:\ 1 - sec^2\ ax = -tan^2\ ax$

$= \displaystyle \lim_{x \to 0}\dfrac{5(-tan^2\ 5x)}{3x^2}$
$= -\dfrac53\displaystyle \lim_{x \to 0}\dfrac{(tan^2\ 5x)}{x^2}$
$= -\dfrac53\displaystyle \lim_{x \to 0}\dfrac{tan\ 5x}{x}.\displaystyle \lim_{x \to 0}\dfrac{tan\ 5x}{x}$
$= -\dfrac53.5.5$
$= -\dfrac{125}{3}$
$Jawab:\ E.$

$32.\ \displaystyle \lim_{x \to 0}\dfrac{\displaystyle \int_{0}^{x}\sqrt{1 + cos\ t}\ dt}{x} =$ . . . .
$A.\ 0$
$B.\ 1$
$C.\ \sqrt{2}$
$D.\ \sqrt{3}$
$E.\ \dfrac12\sqrt{2}$
[Soal dan Pembahasan Limit Fungsi Trigonometri SIMAK UI 2016 MtkIPA]
$1 + cos\ t = 2cos^2\ \dfrac12t$

$\displaystyle \int_{0}^{x}\sqrt{2cos^2\ \dfrac12t}\ dt = \displaystyle \sqrt{2}\int_{0}^{x}cos\ \dfrac12t\ dt$
$= \left[\displaystyle 2\sqrt{2}sin\ \dfrac12t\ \right]_{0}^{x}$
$= \displaystyle 2\sqrt{2}sin\ \dfrac12x$

$\displaystyle \lim_{x \to 0}\dfrac{\displaystyle \int_{0}^{x}\sqrt{1 + cos\ t}\ dt}{x} = \displaystyle \lim_{x \to 0}\dfrac{\displaystyle 2\sqrt{2}sin\ \dfrac12x}{x}$
$= 2\sqrt{2}.\dfrac12$
$= \sqrt{2}$
$Jawab:\ C.$

$33.\ Jika\ f(x) = sin\ 2x,\ maka$
$\displaystyle \lim_{h \to 0}\dfrac{f\left(x + \dfrac h2 \right) - 2f(x) + f\left(x - \dfrac h2 \right)}{h^2} =$ . . . .
$A.\ 2sin\ 2x$
$B.\ sin\ 2x$
$C.\ 0$
$D.\ -sin\ 2x$
$E.\ -2sin\ 2x$
[Soal dan Pembahasan Limit Fungsi Trigonometri SIMAK UI 2015 MtkIPA]
$f(x) = sin\ 2x$
$f\left(x + \dfrac h2 \right) = sin(2x + h)$
$= sin\ 2xcos\ h + cos\ 2xsin\ h$

$f\left(x - \dfrac h2 \right) = sin(2x - h)$
$= sin\ 2xcos\ h - cos\ 2xsin\ h$

$f\left(x + \dfrac h2 \right) - 2f(x) + f\left(x - \dfrac h2 \right)$
$= sin\ 2xcos\ h + cos\ 2xsin\ h - 2sin\ 2x $
$\ \ \ + sin\ 2xcos\ h - cos\ 2xsin\ h$
$= 2sin\ 2xcos\ h - 2sin\ 2x$
$= 2sin\ 2x(cos\ h - 1)$
$= 2sin\ 2x(-2sin^2\ \dfrac h2)$
$= -4sin\ 2x(sin^2\ \dfrac h2)$

$\displaystyle \lim_{h \to 0}\dfrac{f\left(x + \dfrac h2 \right) - 2f(x) + f\left(x - \dfrac h2 \right)}{h^2}$
$= \displaystyle \lim_{h \to 0}\dfrac{-4sin\ 2x(sin^2\ \dfrac h2)}{h^2}$
$= -4sin\ 2x.\displaystyle \lim_{h \to 0}\dfrac{(sin^2\ \dfrac h2)}{h^2}$
$= -4sin\ 2x.\dfrac 12. \dfrac 12$
$= -sin\ 2x$
$Jawab:\ D.$

$34.\ \displaystyle \lim_{x \to 0}\dfrac{\sqrt{1 + tan\ x} - \sqrt{1 + sin\ x}}{x^3} =$ . . . .
$A.\ -1$
$B.\ -\dfrac 14$
$C.\ 0$
$D.\ \dfrac 14$
$E.\ 1$
[Soal dan Pembahasan Limit Fungsi Trigonometri SIMAK UI 2013 MtkIPA]
$\displaystyle \lim_{x \to 0}\dfrac{\sqrt{1 + tan\ x} - \sqrt{1 + sin\ x}}{x^3}$
$= \displaystyle \lim_{x \to 0}\dfrac{\sqrt{1 + tan\ x} - \sqrt{1 + sin\ x}}{x^3}.\dfrac{(\sqrt{1 + tan\ x} + \sqrt{1 + sin\ x})}{(\sqrt{1 + tan\ x} + \sqrt{1 + sin\ x})}$
$= \displaystyle \lim_{x \to 0}\dfrac{tan\ x - sin\ x}{x^3}.\dfrac{1}{(\sqrt{1 + tan\ x} + \sqrt{1 + sin\ x})}$
$= \displaystyle \lim_{x \to 0}\dfrac{\dfrac{sin\ x}{cos\ x} - sin\ x}{x^3}.\dfrac{1}{(\sqrt{1 + tan\ x} + \sqrt{1 + sin\ x})}$
$= \displaystyle \lim_{x \to 0}\dfrac{sin\ x - sin\ x.cos\ x}{(cos\ x)x^3}.\dfrac{1}{(\sqrt{1 + tan\ x} + \sqrt{1 + sin\ x})}$
$= \displaystyle \lim_{x \to 0}\dfrac{sin\ x(1 - cos\ x)}{(cos\ x)x^3}.\dfrac{1}{(\sqrt{1 + tan\ x} + \sqrt{1 + sin\ x})}$
$= \displaystyle \lim_{x \to 0}\dfrac{sin\ x(2sin^2\ \dfrac 12x)}{(cos\ x)x^3}.\dfrac{1}{(\sqrt{1 + tan\ x} + \sqrt{1 + sin\ x})}$
$= 2.\displaystyle \lim_{x \to 0}\dfrac{sin\ x.sin\ \dfrac 12x.sin\ \dfrac 12x}{x.x.x}.\dfrac{1}{cos\ x(\sqrt{1 + tan\ x} + \sqrt{1 + sin\ x})}$
$= 2.1.\dfrac12.\dfrac12.\dfrac{1}{cos\ 0(\sqrt{1 + tan\ 0} + \sqrt{1 + sin\ 0})}$
$= \dfrac12.\dfrac{1}{1.(\sqrt{1 + 0} + \sqrt{1 + 0})}$
$= \dfrac12.\dfrac12$
$= \dfrac14$
$Jawab:\ D.$

$35.\ \displaystyle \lim_{x \to 0}\dfrac{cos\ x\ sin\ x - tan\ x}{x^2sin\ x} =$ . . . .
$A.\ -1$
$B.\ -\dfrac12$
$C.\ 0$
$D.\ \dfrac12$
$E.\ 1$
[Soal dan Pembahasan Limit Fungsi Trigonometri SIMAK UI 2013 MtkIPA]
$\displaystyle \lim_{x \to 0}\dfrac{cos\ x\ sin\ x - tan\ x}{x^2sin\ x}$
$= \displaystyle \lim_{x \to 0}\dfrac{cos^2\ x\ sin\ x - sin\ x}{(cos\ x)x^2sin\ x}$
$= \displaystyle \lim_{x \to 0}\dfrac{sin\ x(cos^2\ x - 1)}{(cos\ x)x^2sin\ x}$
$= \displaystyle \lim_{x \to 0}\dfrac{(cos^2\ x - 1)}{(cos\ x)x^2}$
$= \displaystyle \lim_{x \to 0}\dfrac{-sin^2\ x}{(cos\ x)x^2}$
$= -\displaystyle \lim_{x \to 0}\dfrac{sin^2\ x}{x^2}.\displaystyle \lim_{x \to 0}\dfrac{1}{cos\ x}$
$= -\displaystyle \lim_{x \to 0}\dfrac{sin\ x.sin\ x}{x.x}.\displaystyle \lim_{x \to 0}\dfrac{1}{cos\ x}$
$= -1.1.\dfrac{1}{cos\ 0}$
$= -1$
$Jawab:\ A.$

$36.\ \displaystyle \lim_{a \to b}\dfrac{tan\ a - tan\ b}{1 + (1 - \dfrac ab)tan\ a\ tan\ b - \dfrac ab} =$ . . . .
$A.\ \dfrac 1b$
$B.\ b$
$C.\ -b$
$D.\ \dfrac {-1}{b}$
$E.\ 1$
[Soal dan Pembahasan Limit Fungsi Trigonometri SIMAK UI 2011 MtkIPA]
$\displaystyle \lim_{a \to b}\dfrac{tan\ a - tan\ b}{1 + (1 - \dfrac ab)tan\ a\ tan\ b - \dfrac ab}$
$= \displaystyle \lim_{a \to b}\dfrac{tan\ a - tan\ b}{1 + tan\ a\ tan\ b - \dfrac abtan\ a\ tan\ b - \dfrac ab}$
$= \displaystyle \lim_{a \to b}\dfrac{tan\ a - tan\ b}{1 + tan\ a\ tan\ b - \dfrac ab(1 + tan\ a\ tan\ b)}$
$= \displaystyle \lim_{a \to b}\dfrac{tan\ a - tan\ b}{(1 + tan\ a\ tan\ b)(1 - \dfrac ab)}$

$Ingat:\ tan(a - b) = \dfrac{tan\ a - tan\ b}{1 + tan\ a\ tan\ b}$

$= \displaystyle \lim_{a \to b}\dfrac{tan(a - b)}{1 - \dfrac ab}$
$= \displaystyle \lim_{a \to b}\dfrac{tan(a - b)}{\dfrac{b - a}{b}}$
$= b.\displaystyle \lim_{a \to b}\dfrac{tan(a - b)}{-(a - b)}$
$= -b.\displaystyle \lim_{(a - b) \to 0}\dfrac{tan(a - b)}{(a - b)}$
$= -b.1$
$= -b$
$Jawab:\ C.$

$37.\ \displaystyle \lim_{x \to \dfrac {\pi}{4}}\dfrac{\dfrac{1}{\sqrt{2}}sin(\dfrac{\pi}{4} - 2x) + \dfrac{1}{\sqrt{2}}cos(\dfrac{\pi}{4} - 2x) }{4x - \pi} =$ . . . .
$A.\ \dfrac 14$
$B.\ \dfrac 12$
$C.\ 0$
$D.\ -\dfrac 14$
$E.\ -\dfrac 12$
[Soal dan Pembahasan Limit Fungsi Trigonometri Utul 2009 MtkIPA]
$Ingat\ !$
$acos\ P + bsin\ P = kcos(P - \theta)$
$k = \sqrt{a^2 + b^2}$
$tan\ \theta = \dfrac ba$
$\theta = arc\ tan\left(\dfrac ba\right)$

$Soal\ !$
$\displaystyle \lim_{x \to \dfrac {\pi}{4}}\dfrac{\dfrac{1}{\sqrt{2}}sin(\dfrac{\pi}{4} - 2x) + \dfrac{1}{\sqrt{2}}cos(\dfrac{\pi}{4} - 2x) }{4x - \pi}$

$Lihat\ soal\ dan\ perhatikan\ pembilangnya\ !$
$k^2 = \left(\dfrac{1}{\sqrt{2}} \right)^2 + \left(\dfrac{1}{\sqrt{2}} \right)^2$
$K^2 = 1$
$k = 1$
$tan\ \theta = \dfrac{\dfrac{1}{\sqrt{2}}}{\dfrac{1}{\sqrt{2}}} = 1$
$\theta = \dfrac{\pi}{4}$

$= \displaystyle \lim_{x \to \dfrac {\pi}{4}}\dfrac{cos\ \left(\dfrac{\pi}{4} - 2x - \dfrac{\pi}{4} \right) }{4x - \pi}$
$= \displaystyle \lim_{x \to \dfrac {\pi}{4}}\dfrac{cos\ \left(-2x \right) }{4x - \pi}$
$= \displaystyle \lim_{x \to \dfrac {\pi}{4}}\dfrac{cos\ \left(2x \right) }{4x - \pi}$

$Gunakan\ Aturan\ L'Hospital\ !$

$= \displaystyle \lim_{x \to \dfrac {\pi}{4}}\dfrac{-2sin\ \left(2x \right) }{4}$
$= \displaystyle \lim_{x \to \dfrac {\pi}{4}}\dfrac{-2sin\ \left(2.\dfrac{\pi}{4} \right) }{4}$
$= \dfrac{-2sin\ \left(\dfrac{\pi}{2} \right) }{4}$
$= -\dfrac12$
$Jawab:\ E.$

$38.\ \displaystyle \lim_{x \to \dfrac{\pi}{2}} \dfrac{sin\ xtan\ (2x - \pi)}{2\pi - 4x} =$ . . . .
$A.\ -\dfrac12$
$B.\ \dfrac12$
$C.\ \dfrac13\sqrt{3}$
$D. 1$
$E.\ \sqrt{3}$
[Soal dan Pembahasan Limit Fungsi Trigonometri SPMB 2006 MtkIPA]
$\displaystyle \lim_{x \to \dfrac{\pi}{2}} \dfrac{sin\ xtan\ (2x - \pi)}{2\pi - 4x}$
$= \displaystyle \lim_{x \to \dfrac{\pi}{2}} \dfrac{sin\ xtan\ 2(x - \dfrac{\pi}{2})}{-4(x - \dfrac{\pi}{2})}$
$= \left(-\dfrac 14\right).\left(\displaystyle \lim_{x \to \dfrac{\pi}{2}} sin\ x\right) .\left(\displaystyle \lim_{(x - \dfrac{\pi}{2}) \to 0}\dfrac{tan\ 2(x - \dfrac{\pi}{2})}{(x - \dfrac{\pi}{2})}\right)$
$= -\dfrac 14.sin\ \dfrac {\pi}{2}.2$
$= -\dfrac 14.1.2$
$= -\dfrac 12$
$Jawab:\ A.$

$39.\ \displaystyle \lim_{x \to \dfrac{\pi}{4}} \dfrac{\left(x - \dfrac{\pi}{4} \right)tan\left(3x - \dfrac{3\pi}{4} \right)} {2(1 - sin\ 2x)} =$ . . . .
$A.\ 0$
$B.\ -\dfrac32$
$C.\ \dfrac32$
$D.\ -\dfrac34$
$E.\ \dfrac34$
[Soal dan Pembahasan Limit Fungsi Trigonometri Utul 2005 MtkIPA]
$\displaystyle \lim_{x \to \dfrac{\pi}{4}} \dfrac{\left(x - \dfrac{\pi}{4} \right)tan\left(3x - \dfrac{3\pi}{4} \right)} {2(1 - sin\ 2x)}$
$= \displaystyle \lim_{x \to \dfrac{\pi}{4}} \dfrac{\left(x - \dfrac{\pi}{4} \right)tan\left(3x - \dfrac{3\pi}{4} \right)} {2(sin^2\ x + cos^2\ x - 2sin\ xcos\ x)}$

$\ \ Ingat:\ sin^2\ x + cos^2\ x = 1$
$\ sin\ 2x = 2sin\ xcos\ x$

$= \displaystyle \lim_{x \to \dfrac{\pi}{4}} \dfrac{\left(x - \dfrac{\pi}{4} \right)tan3\left(x - \dfrac{\pi}{4} \right)} {2\left(sin\ x - cos\ x\right)^2}$

$Ingat\ !$
$\ cos\ x = sin\ \left(\dfrac{\pi}{2} - x\right)$
$sin\ A - sin\ B = 2cos\dfrac12(A + B)sin\dfrac12(A - B)$

$= \displaystyle \lim_{x \to \dfrac{\pi}{4}} \dfrac{\left(x - \dfrac{\pi}{4} \right)tan3\left(x - \dfrac{\pi}{4} \right)} {2\left(sin\ x - sin\ \left(\dfrac{\pi}{2} - x \right) \right)^2}$
$= \displaystyle \lim_{x \to \dfrac{\pi}{4}} \dfrac{\left(x - \dfrac{\pi}{4} \right)tan3\left(x - \dfrac{\pi}{4} \right)} {2\left(2cos\ \dfrac{\pi}{4} sin\ \left(x - \dfrac{\pi}{4}\right) \right)^2}$
$= \displaystyle \lim_{x \to \dfrac{\pi}{4}} \dfrac{\left(x - \dfrac{\pi}{4} \right)tan3\left(x - \dfrac{\pi}{4} \right)} {2\left(4.\dfrac12. sin^2\ \left(x - \dfrac{\pi}{4}\right) \right)}$
$= \dfrac14.\displaystyle \lim_{x \to \dfrac{\pi}{4}} \dfrac{\left(x - \dfrac{\pi}{4} \right)tan3\left(x - \dfrac{\pi}{4} \right)} {\left( sin^2\ \left(x - \dfrac{\pi}{4}\right) \right)}$
$= \dfrac14.\displaystyle \lim_{\left(x - \dfrac{\pi}{4}\right) \to 0} \dfrac{\left(x - \dfrac{\pi}{4} \right)tan3\left(x - \dfrac{\pi}{4} \right)} {\left( sin\ \left(x - \dfrac{\pi}{4}\right).sin\ \left(x - \dfrac{\pi}{4}\right) \right)}$
$= \dfrac14.1.3$
$= \dfrac34$
$Jawab:\ E.$

$40.\ \displaystyle \lim_{x \to \infty}x^2sin\ \dfrac 1x tan\ \dfrac 1x =$ . . . .
$A.\ -1$
$B.\ 0$
$C.\ \dfrac12$
$D.\ 1$
$E.\ 2$
[Soal dan Pembahasan Limit Fungsi Trigonometri SPMB 2005 MtkIPA]
$\displaystyle \lim_{x \to \infty}x^2sin\ \dfrac 1x tan\ \dfrac 1x = \displaystyle \lim_{\dfrac 1x \to 0}x^2sin\ \dfrac 1x tan\ \dfrac 1x$
$= \displaystyle \lim_{\dfrac 1x \to 0}\dfrac{sin\ \dfrac 1x tan\ \dfrac 1x}{\left(\dfrac 1x\right)^2}$

$Misalkan\ \dfrac 1x = p$

$= \displaystyle \lim_{p \to 0}\dfrac{sin\ p\ tan\ p}{p^2}$
$= \displaystyle \lim_{p \to 0}\dfrac{sin\ p\ tan\ p}{p.p}$
$= 1.1$
$= 1$
$Jawab:\ D.$

Demikianlah Pembahasan Soal UN / UNBK dan SBMPTN Limit Fungsi Trigonometri, semoga bermanfaat. Selamat belajar !

Disusun oleh:
Joslin Sibarani
Alumni Teknik Sipil ITB

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