Soal dan Pembahasan Limit Fungsi Trigonometri

Soal 1: SPMB 2006 Kode 310
$\lim_{x\rightarrow \frac{1}{2} \pi} \frac{sin \ x \ tan \ (2x - \pi)}{2 \pi-4x}=. . .$
(A) $-\frac{1}{2}$
(B) $\frac{1}{2}$
(C) $\frac{1}{3} \sqrt{3}$
(D) $1$
(E) $\sqrt{3}$

PEMBAHASAN:
$\lim_{x\rightarrow \frac{1}{2} \pi} \frac{sin \ x \ tan \ (2x - \pi)}{2 \pi-4x}$
$=\lim_{x\rightarrow \frac{1}{2} \pi} \frac{sin \ x \ -(tan \ (\pi-2x))}{2 \pi-4x}$
$=\lim_{x\rightarrow \frac{1}{2} \pi} \frac{-sin \ x \ tan \ (\pi-2x)}{2 (\pi-2x)}$
$=\lim_{x\rightarrow \frac{1}{2} \pi} \left (\frac{-sin \ x \ }{2} \times \frac{tan \ (\pi-2x) \ }{(\pi-2x)} \right )$
$=\frac{-sin (\frac{1}{2} \pi)}{2} \times 1$
$=-\frac{1}{2}$

Pilihan jawabannya adalah (A)

Soal 2: SPMB 2006 Kode 402
$\lim_{x\rightarrow 0} \frac{x^2 \sqrt{4-x^3}}{cos \ x \ - cos \ 3x \ }=. . .$
(A) $-\frac{3}{2}$
(B) $-\frac{1}{2}$
(C) $0$
(D) $\frac{1}{2}$
(E) $\frac{3}{2}$

PEMBAHASAN:
Kita gunakan identitas trigonometri:

$cos \ A \ - sin \ B \ = -2 sin \left (\frac{A+B}{2} \right )sin \left (\frac{A-B}{2} \right )$
$cos \ x \ - sin \ 3x \ = -2 sin \left (\frac{x+3x}{2} \right )sin \left (\frac{x-3x}{2} \right )$
$cos \ x \ - sin \ 3x \ = -2 sin \ 2x \ sin \ (-x) \ = 2 sin \ 2x \ sin \ x \$

$\lim_{x\rightarrow 0} \frac{x^2 \sqrt{4-x^3}}{cos \ x \ - cos \ 3x \ }$
$=\lim_{x\rightarrow 0} \frac{x^2 \sqrt{4-x^3}}{2 sin \ 2x \ sin \ x \ }$
$= \lim_{x\rightarrow 0} \left (\frac{x^2}{2 sin \ 2x \ sin \ x \ } \times \sqrt{4-x^3} \right )$
$= \frac{1}{2.2} \times \sqrt{4-0^3}$

$=\frac{1}{4} \times 2 = \frac{1}{2}$

Pilihan jawabannya adalah (D)

Soal 3: UM UGM 2005 Kode 812

$\lim_{x\rightarrow 0} \frac{x tan \ 5x \ }{cos \ 2x \ - cos \ 7x \ }=. . .$

(A) $\frac{1}{9}$
(B) $-\frac{1}{9}$
(C) $\frac{2}{9}$
(D) $-\frac{2}{9}$
(E) $0$

PEMBAHASAN:
$cos \ A \ - sin \ B \ = -2 sin \left (\frac{A+B}{2} \right )sin \left (\frac{A-B}{2} \right )$
$cos \ 2x \ - sin \ 7x \ = -2 sin \left (\frac{2x+7x}{2} \right )sin \left (\frac{2x-7x}{2} \right )$
$cos \ 2x \ - sin \ 7x \ = -2 sin \ (\frac{9x}{2}) \ sin \ (\frac{-5x}{2}) \$
$cos \ 2x \ - sin \ 7x \ = 2 sin \ (\frac{9x}{2}) \ sin \ (\frac{5x}{2}) \$

$\lim_{x\rightarrow 0} \frac{x tan \ 5x \ }{2 sin \ (\frac{9x}{2}) \ sin \ (\frac{5x}{2}) \ }$
$=\lim_{x\rightarrow 0} \left (\frac{x}{2 sin \ (\frac{9x}{2}) \ } \times \frac{tan \ 5x \ }{sin \ (\frac{5x}{2}) \ } \right )$
$=\frac{1}{2 \ . \ \frac{9}{2}} \times \frac{5}{\frac{5}{2}}$
$=\frac{1}{9} \times 2 = \frac{2}{9}$

Pilihan jawabannya adalah (C)

Soal 4: SPMB 2005 Kode 780
$\lim_{x\rightarrow 1} \frac{(x^2 + x - 2)sin(x-1)}{x^2 - 2x +1}=. . .$

(A) $4$
(B) $3$
(C) $0$
(D) $-\frac{1}{4}$
(E) $-\frac{1}{2}$

PEMBAHASAN:
$\lim_{x\rightarrow 1} \frac{(x^2 + x - 2)sin(x-1)}{x^2 - 2x +1}$
$=\lim_{x\rightarrow 1} \frac{(x+ 2)(x-1)sin(x-1)}{(x-1}^2$
$=\lim_{x\rightarrow 1} \frac{(x+ 2)(x-1)sin(x-1)}{(x-1)(x-1)}$
$=\lim_{x\rightarrow 1} \left (\frac{(x+ 2)(x-1)}{(x-1)} \times \frac{sin(x-1)}{(x-1)} \right )$
$=\lim_{x\rightarrow 1} (x+ 2) \times \lim_{x\rightarrow 1}\frac{sin(x-1)}{(x-1)}$
$=1+ 2 \times 1 = 3$

Pilihan jawaban yang sesuai adalah (B)

Soal 5: UM UGM 2004 Kode 322
$\lim_{x\rightarrow 1} \frac{tan(x-1)sin(1- \sqrt{x})}{x^2-2x+1}=. . .$

(A) $-1$
(B) $-\frac{1}{2}$
(C) $0$
(D) $\frac{1}{2}$
(E) $1$

PEMBAHASAN:
$\lim_{x\rightarrow 1} \frac{tan(x-1)sin(1- \sqrt{x})}{x^2-2x+1}$
$=\lim_{x\rightarrow 1} \frac{tan(x-1)sin(1- \sqrt{x})}{(x-1)(x-1)}$
$=\lim_{x\rightarrow 1} \frac{tan(x-1)sin(1- \sqrt{x})}{-(x-1)(1-x)}$
$=\lim_{x\rightarrow 1} \frac{tan(x-1)sin(1- \sqrt{x})}{-(x-1)(1-\sqrt{x})(1+\sqrt{x})}$
$=\lim_{x\rightarrow 1} \left (\frac{tan(x-1)}{x-1} \times \frac{sin(1- \sqrt{x})}{(1- \sqrt{x})} \times \frac{1}{-(1+\sqrt{x})} \right )$
$= \lim_{x\rightarrow 1}\frac{tan(x-1)}{x-1} \times \lim_{x\rightarrow 1} \frac{sin(1- \sqrt{x})}{(1- \sqrt{x})} \times \lim_{x\rightarrow 1}\frac{1}{-(1+\sqrt{x})}$
$=1 \times 1 \times \frac{1}{-(1+\sqrt{1})}=-\frac{1}{2}$

Pilihan jawaban yang sesuai adalah (B)

Baca Juga

Soal dan Pembahasan Limit Fungsi Trigonometri

Soal 6: UM STIS 2011

Jika $\lim_{x\rightarrow 0} \frac{x^a \ sin^4x}{sin^6x}=1$ , maka nilai $a$ yang memenuhi adalah . . .

(A) $1$
(B) $2$
(C) $3$
(D) $4$
(E) $5$

PEMBAHASAN:
Kita gunakan konsep limit trogonometri dasar
$\lim_{x\rightarrow 0} \frac{sin \ ax \ }{bx}=\frac{a}{b}$ atau
$\lim_{x\rightarrow 0} \frac{ax}{sin \ bx \ }=\frac{a}{b}$ , maka

$\lim_{x\rightarrow 0} \frac{x^a \ sin^4x}{sin^2x sin^4x}=1$
$\lim_{x\rightarrow 0} \frac{x^a \ }{sin^2x}=1$

Agar nilai limit fungsi di atas benar adalah $1$ , maka nilai $a = 2$

Pilihan jawabannya adalah (B)

Soal 7: UTBK-SBMPTN 2019
Nilai $\lim_{x\rightarrow 0} \frac{cot \ 2x \ - csc \ 2x \ }{cos \ 3x \ tan\frac{1}{3}x}=. . .$

(A) $3$
(B) $2$
(C) $0$
(D) $-2$
(E) $-3$

PEMBAHASAN:
Kita gunakan konsep limit trogonometri dasar

$\lim_{x\rightarrow 0} \frac{tan \ ax \ }{bx}=\frac{a}{b}$
$\lim_{x\rightarrow 0} \frac{tan \ ax \ }{sin \ bx \ }=\frac{a}{b}$
$\lim_{x\rightarrow 0} \frac{sin \ ax \ }{sin \ bx \ }=\frac{a}{b}$

$\lim_{x\rightarrow 0} \frac{cot \ 2x \ - csc \ 2x \ }{cos \ 3x \ tan\frac{1}{3}x}$

$=\lim_{x\rightarrow 0} \frac{\frac{cos \ 2x \ }{sin \ 2x \ } - \frac{1}{sin \ 2x \ }}{cos \ 3x \ tan\frac{1}{3}x}$

$=\lim_{x\rightarrow 0} \frac{\frac{cos \ 2x \ -1}{sin \ 2x \ }}{cos \ 3x \ tan\frac{1}{3}x}$

$=\lim_{x\rightarrow 0} \frac{cos \ 2x \ -1}{cos \ 3x \ tan\frac{1}{3}x sin \ 2x \ }$

$=\lim_{x\rightarrow 0} \frac{1-sin^2 \ x \ -1}{cos \ 3x \ tan\frac{1}{3}x sin \ 2x \ }$

$=\lim_{x\rightarrow 0} \frac{-2sin^2 \ x \ }{cos \ 3x \ tan\frac{1}{3}x sin \ 2x \ }$

$=\lim_{x\rightarrow 0} \frac{-2sin \ x \ sin \ x \ }{cos \ 3x \ tan\frac{1}{3}x sin \ 2x \ }$

$=\lim_{x\rightarrow 0} \frac{-2}{cos3x} \times \lim_{x\rightarrow 0} \frac{sin \ x \ }{tan\frac{1}{3}x} \times \frac{sin \ x \ }{sin \ 2x \ }$

$=\frac{-2.1.1}{cos \ 0 \ .\frac{1}{3}.2}$

$=\frac{-2}{\frac{2}{3}}=-3$

Pilihan jawabannya adalah (E)

Soal 8: SNMPTN 2010 Kode 546
$\lim_{x\rightarrow 0} \frac{\sqrt{4x}}{\sqrt{sin \ 2x \ }}=. . .$

(A) $\sqrt{2}$
(B) $1$
(C) $\frac{1}{2}$
(D) $\frac{1}{4}$
(E) $0$

PEMBAHASAN:

Kita gunakan konsep limit trogonometri dasar $\lim_{x\rightarrow 0} \frac{ax}{sin \ bx \ }=\frac{a}{b}$ dan teorema limit $\lim_{x\rightarrow c} \sqrt[n]{f(x)} = \sqrt[n]{\lim_{x\rightarrow c}f(x)}$ , maka

$\lim_{x\rightarrow 0} \frac{\sqrt{4x}}{\sqrt{sin \ 2x \ }}=\lim_{x\rightarrow 0} \sqrt{\frac{4x}{sin \ 2x \ }}$
$=\sqrt{\lim_{x\rightarrow 0} \frac{4x}{sin \ 2x \ }}$
$=\sqrt{\frac{4}{2}}$
$=\sqrt{2}$

Pilihan jawaban yang sesuai adalah (A)

Soal 9: SPMB 2006 Kode 111
$\lim_{x\rightarrow \frac{1}{2} \pi} \frac{\left (x- \frac{1}{2} \pi \right )^2 sin \ x \ }{cos^2 \ x \ }=. . .$

(A) $-1$
(B) $-\frac{1}{2}$
(C) $0$
(D) $1$
(E) $2$

PEMBAHASAN:
$\lim_{x\rightarrow \frac{1}{2} \pi} \frac{\left (x- \frac{1}{2} \pi \right )^2 sin \ x \ }{cos^2 \ x \ }$
$=\lim_{x\rightarrow \frac{1}{2} \pi} \frac{\left (\frac{1}{2} \pi -x \right )^2 sin \ x \ }{sin^2 \ \left (\frac{1}{2} \pi -x \right ) \ }$
$=\lim_{x\rightarrow \frac{1}{2} \pi} \left [\frac{\left (\frac{1}{2} \pi -x \right )^2 sin \ x \ }{sin^2 \ \left (\frac{1}{2} \pi -x \right ) \ } \right ]$
$=1 \times sin\frac{1}{2}\pi$
$=1 \times 1 = 1$

Pilihan jawaban yang sesuai adalah (D)

Soal 10: UM UGM 2005 Kode 812
$\lim_{x\rightarrow \frac{1}{4} \pi} \frac{\left (x - \frac{\pi}{4} \right )tan \left ( 3x-\frac{3 \pi}{4} \right )}{2(1-sin \ 2x \ )}=. . .$

(A) $0$
(B) $-\frac{3}{2}$
(C) $\frac{3}{2}$
(D) $-\frac{3}{4}$
(E) $\frac{3}{4}$

PEMBAHASAN:

Kita gunakan konsep identitas trogonometri dasar

$cos(\frac{1}{2}\pi - x) = sin \ x \$
$cos \ 2x \ = cos^2 \ x \ - sin^2 \ x \$
$cos \ 2x \ = 1 - 2 sin^2 \ x \$
$cos \ x \ = 1 - 2 sin^2 \ \frac{1}{2}x \$

$\lim_{x\rightarrow \frac{1}{4} \pi} \frac{\left (x - \frac{\pi}{4} \right )tan \left ( 3x-\frac{3 \pi}{4} \right )}{2(1-sin \ 2x \ )}$

$=\lim_{x\rightarrow \frac{1}{4} \pi} \frac{\left (x - \frac{\pi}{4} \right )\left (-tan \left ( \frac{3 \pi}{4}-3x \right ) \right )}{2\left (1-cos\left (\frac{1}{2}\pi-2x \right ) \right )}$

$=\lim_{x\rightarrow \frac{1}{4} \pi} \frac{-\left (x - \frac{\pi}{4} \right )tan \ 3 \left ( \frac{\pi}{4}-x \right )}{2\left [2sin^2 \left ( \frac{1}{2}\left (\frac{1}{2}\pi-2x \right ) \right ) \right ]}$

$=\lim_{x\rightarrow \frac{1}{4} \pi} \frac{\left (\frac{\pi}{4}-x \right )tan \ 3 \left ( \frac{\pi}{4}-x \right )}{4sin^2 \left (\frac{1}{4}\pi-x \right ) }$

$=\lim_{x\rightarrow \frac{1}{4} \pi} \left [\frac{\left (\frac{\pi}{4}-x \right )}{4sin \left (\frac{1}{4}\pi-x \right )} \times \frac{tan \ 3 \left ( \frac{\pi}{4}-x \right )}{sin \left (\frac{1}{4}\pi-x \right ) } \right ]$

$=\frac{1}{4} \times \frac{3}{1}$

$=\frac{3}{4}$

Pilihan jawaban yang sesuai adalah (E)

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