Pembahasan Soal UN Limit Fungsi Trigonometri

Pembahasan soal-soal Ujian Nasional (UN) SMA bidang studi Matematika IPA untuk pokok bahasan Limit Fungsi Trigonometri.

1. UN 2005
Nilai

_{x \to 0}^{l i m} \frac{s i n 3 x - s i n 3 x c o s 2 x}{2 x^{3}}

$\mathrm{_{x \to 0}^{lim}\frac{sin\,3x-sin\,3x\:cos\,2x}{2x^{3}}}$ = ...
A.

\frac{1}{2}

$\frac{1}{2}$
B.

\frac{2}{3}

$\frac{2}{3}$
C.

\frac{3}{2}

$\frac{3}{2}$
D. 2
E. 3

Pembahasan :

_{x \to 0}^{l i m} \frac{s i n 3 x - s i n 3 x c o s 2 x}{2 x^{3}}

$\mathrm{_{x \to 0}^{lim}\frac{sin\,3x-sin\,3x\:cos\,2x}{2x^{3}}}$

_{x \to 0}^{l i m} \frac{s i n 3 x (1 - c o s 2 x)}{2 x^{3}}

$\mathrm{_{x \to 0}^{lim}\frac{sin\,3x(1-cos\,2x)}{2x^{3}}}$

_{x \to 0}^{l i m} \frac{s i n 3 x (2 s i n^{2} x)}{2 x^{3}}

$\mathrm{_{x \to 0}^{lim}\frac{sin\,3x\,(2\,sin^{2}x)}{2x^{3}}}$

_{x \to 0}^{l i m} \frac{s i n 3 x s i n^{2} x}{x^{3}}

$\mathrm{_{x \to 0}^{lim}\frac{sin\,3x\,sin^{2}x}{x^{3}}}$

_{x \to 0}^{l i m} \frac{s i n 3 x}{x}

$\mathrm{_{x \to 0}^{lim}\frac{sin\,3x}{x}}$ ×

_{x \to 0}^{l i m} \frac{s i n x}{x}

$\mathrm{_{x \to 0}^{lim}\frac{sin\,x}{x}}$ ×

_{x \to 0}^{l i m} \frac{s i n x}{x}

$\mathrm{_{x \to 0}^{lim}\frac{sin\,x}{x}}$

= 3 × 1 × 1 = 3

Jawaban : E

2. UN 2006
Nilai

_{x \to \frac{π}{4}}^{l i m} \frac{c o s 2 x}{c o s x - s i n x}

$\mathrm{_{x \to \frac{\pi }{4}}^{lim}\frac{cos\,2x}{cos\,x-sin\,x}}$ = ...
A. 0
B.

\frac{1}{2}

$\frac{1}{2}$ √2
C. 1
D. √2
E. ∞

Pembahasan :

_{x \to \frac{π}{4}}^{l i m} \frac{c o s 2 x}{c o s x - s i n x}

$\mathrm{_{x \to \frac{\pi }{4}}^{lim}\frac{cos\,2x}{cos\,x-sin\,x}}$

_{x \to \frac{π}{4}}^{l i m} \frac{c o s^{2} x - s i n^{2} x}{c o s x - s i n x}

$\mathrm{_{x \to \frac{\pi }{4}}^{lim}\frac{cos^{2}x-sin^{2}x}{cos\,x-sin\,x}}$

_{x \to \frac{π}{4}}^{l i m} \frac{(c o s x - s i n x) (c o s x + s i n x)}{c o s x - s i n x}

$\mathrm{_{x \to \frac{\pi }{4}}^{lim}\frac{(cos\,x-sin\,x)(cos\,x+sin\,x)}{cos\,x-sin\,x}}$

_{x \to \frac{π}{4}}^{l i m} (c o s x + s i n x)

$\mathrm{_{x \to \frac{\pi }{4}}^{lim}(cos\,x+sin\,x)}$

= cos

\frac{π}{4}

$\frac{\pi}{4}$ + sin

\frac{π}{4}

$\frac{\pi}{4}$

=

\frac{1}{2}

$\frac{1}{2}$ √2 +

\frac{1}{2}

$\frac{1}{2}$ √2 = √2

Jawaban : D

3. UN 2007
Nilai

_{x \to 0}^{l i m} \frac{1 - c o s 2 x}{x t a n \frac{1}{2} x}

$\mathrm{_{x \to 0}^{lim}\frac{1-cos\,2x}{x\,tan\,\frac{1}{2}x}}$ = ...
A. −4
B. −2
C. 1
D. 2
E. 4

Pembahasan :

_{x \to 0}^{l i m} \frac{1 - c o s 2 x}{x t a n \frac{1}{2} x}

$\mathrm{_{x \to 0}^{lim}\frac{1-cos\,2x}{x\,tan\,\frac{1}{2}x}}$

_{x \to 0}^{l i m} \frac{2 s i n^{2} x}{x t a n \frac{1}{2} x}

$\mathrm{_{x \to 0}^{lim}\frac{2\,sin^{2}x}{x\,tan\,\frac{1}{2}x}}$

2 ×

_{x \to 0}^{l i m} \frac{s i n x}{x}

$\mathrm{_{x \to 0}^{lim}\frac{sin\,x}{x}}$ ×

_{x \to 0}^{l i m} \frac{s i n x}{t a n \frac{1}{2} x}

$\mathrm{_{x \to 0}^{lim}\frac{sin\,x}{tan\,\frac{1}{2}x}}$

= 2 × 1 ×

\frac{1}{\frac{1}{2}}

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

Jawaban : E

4. UN 2009
Nilai

_{x \to \frac{π}{3}}^{l i m} \frac{t a n (3 x - π) c o s 2 x}{s i n (3 x - π)}

$\mathrm{_{x \to \frac{\pi }{3}}^{lim}\frac{tan(3x-\pi )\,cos\,2x}{sin(3x-\pi )}}$ = ...
A.

- \frac{1}{2}

$-\frac{1}{2}$
B.

\frac{1}{2}

$\frac{1}{2}$
C.

\frac{1}{2}

$\frac{1}{2}$ √2
D.

\frac{1}{2}

$\frac{1}{2}$ √3
E.

\frac{3}{2}

$\frac{3}{2}$

Pembahasan :

_{x \to \frac{π}{3}}^{l i m} \frac{t a n (3 x - π) c o s 2 x}{s i n (3 x - π)}

$\mathrm{_{x \to \frac{\pi }{3}}^{lim}\frac{tan(3x-\pi )\,cos\,2x}{sin(3x-\pi )}}$

_{x \to \frac{π}{3}}^{l i m} c o s 2 x

$\mathrm{_{x \to \frac{\pi }{3}}^{lim}cos\,2x}$ ×

_{x \to \frac{π}{3}}^{l i m} \frac{t a n (3 x - π)}{s i n (3 x - π)}

$\mathrm{_{x \to \frac{\pi }{3}}^{lim}\frac{tan(3x-\pi )}{sin(3x-\pi )}}$

Misalkan u = 3x − π
Jika x →

\frac{π}{3}

$\frac{\pi}{3}$ maka u → 0

_{x \to \frac{π}{3}}^{l i m} c o s 2 x

$\mathrm{_{x \to \frac{\pi }{3}}^{lim}cos\,2x}$ ×

_{u \to 0}^{l i m} \frac{t a n u}{s i n u}

$\mathrm{_{u \to 0}^{lim}\frac{tan\,u}{sin\,u}}$

= cos 2(

\frac{π}{3}

$\frac{\pi}{3}$ ) × 1

= cos (

\frac{2 π}{3}

$\frac{2\pi}{3}$ ) =

- \frac{1}{2}

$-\frac{1}{2}$

Jawaban : A

5. UN 2010
Nilai

_{x \to 0}^{l i m} (\frac{c o s 4 x s i n 3 x}{5 x})

$\mathrm{_{x \to 0}^{lim}\left ( \frac{cos\,4x\,sin\,3x}{5x} \right )}$ = ...
A.

\frac{5}{3}

$\frac{5}{3}$
B. 1
C.

\frac{3}{5}

$\frac{3}{5}$
D.

\frac{1}{5}

$\frac{1}{5}$
E. 0

Pembahasan :

_{x \to 0}^{l i m} \frac{c o s 4 x s i n 3 x}{5 x}

$\mathrm{_{x \to 0}^{lim} \frac{cos\,4x\,sin\,3x}{5x}}$

$\mathrm{_{x \to 0}^{lim} cos\,4x}$ ×

$\mathrm{_{x \to 0}^{lim} \frac{sin\,3x}{5x}}$

= cos (4.0) ×

$\frac{3}{5}$

= 1 ×

$\frac{3}{5}$ =

$\frac{3}{5}$

Jawaban : C

6. UN 2010
Nilai

$\mathrm{_{x \to 0}^{lim}\left ( \frac{sin\,4x-sin\,2x}{6x} \right )}$ = ...
A. 1
B.

$\frac{2}{3}$
C.

$\frac{1}{2}$
D.

$\frac{1}{3}$
E.

$\frac{1}{6}$

Pembahasan :

$\mathrm{_{x \to 0}^{lim} \frac{sin\,4x-sin\,2x}{6x}}$

$\mathrm{_{x \to 0}^{lim} \frac{2\,cos\,\frac{1}{2}(4x+2x)\,sin\,\frac{1}{2}(4x-2x)}{6x} }$

$\frac{2}{6}$ ×

$\mathrm{_{x \to 0}^{lim} \frac{cos\,3x\,sin\,x}{x}}$

$\frac{2}{6}$ ×

$\mathrm{_{x \to 0}^{lim}}$ cos 3x ×

$\mathrm{_{x \to 0}^{lim} \frac{sin\,x}{x}}$

=

$\frac{2}{6}$ × cos (3.0) × 1 =

$\frac{1}{3}$

Jawaban : D

7. UN 2011
Nilai

$\mathrm{_{x \to 0}^{lim}\frac{1-cos\,2x}{2x\,sin\,2x}}$ = ...
A.

$\frac{1}{8}$
B.

$\frac{1}{6}$
C.

$\frac{1}{4}$
D.

$\frac{1}{2}$
E. 1

Pembahasan :

$\mathrm{_{x \to 0}^{lim}\frac{1-cos\,2x}{2x\,sin\,2x}}$

$\mathrm{_{x \to 0}^{lim}\frac{2\,sin^{2}x}{2x\,sin\,2x}}$

$\mathrm{_{x \to 0}^{lim}\frac{sin\,x}{x}}$ ×

$\mathrm{_{x \to 0}^{lim}\frac{sin\,x}{sin\,2x}}$

= 1 ×

$\frac{1}{2}$ =

$\frac{1}{2}$

Jawaban : D

8. UN 2012
Nilai

$\mathrm{_{x \to 0}^{lim}\frac{cos\,4x-1}{x\,tan\,2x}}$ = ...
A. 4
B. 2
C. −1
D. −2
E. −4

Pembahasan :

$\mathrm{_{x \to 0}^{lim}\frac{-(1-cos\,4x)}{x\,tan\,2x}}$

$\mathrm{_{x \to 0}^{lim}\frac{-(2sin^{2}2x)}{x\,tan\,2x}}$

−2 ×

$\mathrm{_{x \to 0}^{lim}\frac{sin\,2x}{x}\;\times\;_{x \to 0}^{lim}\frac{sin\,2x}{tan\,2x} }$

= −2 ×

$\frac{2}{1}$ ×

$\frac{2}{2}$ = −4

Jawaban : E

9. UN 2013
Nilai dari

$\mathrm{_{x \to -2}^{lim}\frac{(x^{2}-4)\;tan\,(x+2)}{sin^{2}(x+2)}=...}$
A. −4
B. −3
C. 0
D. 4
E. ∞

Pembahasan :

$\mathrm{_{x \to -2}^{lim}\frac{(x-2)(x+2)\;tan\,(x+2)}{sin^{2}(x+2)}}$

$\mathrm{_{x \to -2}^{lim}}$ (x − 2) ×

$\mathrm{_{x \to -2}^{lim}\frac{(x+2)}{sin(x+2)}}$ ×

$\mathrm{\frac{tan(x+2)}{sin(x+2)}}$

Misalkan u = x + 2
Jika x → −2 maka u → 0

$\mathrm{_{x \to -2}^{lim}}$ (x − 2) ×

$\mathrm{_{u \to 0}^{lim}\frac{u}{sin\,u}}$ ×

$\mathrm{_{u \to 0}^{lim}\frac{tan\,u}{sin\,u}}$

= (−2 − 2) × 1 × 1 = −4

Jawaban : A

10. UN 2013
Nilai dari

$\mathrm{_{x \to 3}^{lim}\frac{x\,tan\,(2x-6)}{sin\,(x-3)}=...}$
A. 0
B.

$\frac{1}{2}$
C. 2
D. 3
E. 6

Pembahasan :

$\mathrm{_{x \to 3}^{lim}\frac{x\,tan\,2(x-3)}{sin\,(x-3)}}$

$\mathrm{_{x \to 3}^{lim}\,x\;\times\; _{x \to 3}^{lim}\frac{tan\,2(x-3)}{sin\,(x-3)}}$

Misalkan u = x − 3
Jika x → 3 maka u → 0

$\mathrm{_{x \to 3}^{lim}\,x\;\times\; _{u \to 0}^{lim}\frac{tan\,2u}{sin\,u}}$

= 3 × 2 = 6

Jawaban : E

11. UN 2014
Nilai

$\mathrm{_{x \to 0}^{lim}\frac{1-cos\,8x}{sin\,2x\,tan\,2x}}$ = ...
A. 16
B. 12
C. 8
D. 4
E. 2

Pembahasan :

$\mathrm{_{x \to 0}^{lim}\frac{1-cos\,8x}{sin\,2x\,tan\,2x}}$

$\mathrm{_{x \to 0}^{lim}\frac{2\,sin^{2}4x}{sin\,2x\,tan\,2x}}$

2 ×

$\mathrm{_{x \to 0}^{lim}\frac{sin\,4x}{sin\,2x}\;\times\;_{x \to 0}^{lim}\frac{sin\,4x}{tan\,2x} }$

= 2 ×

$\frac{4}{2}$ ×

$\frac{4}{2}$ = 8

Jawaban : C

12. UN 2015
Nilai dari

$\mathrm{_{x \to 0}^{lim}\frac{x\,tan\,x}{2\,cos^{2}x\,-2}}$ = ...
A.

$-\frac{1}{2}$
B.

$-\frac{1}{4}$
C. 0
D.

$\frac{1}{2}$
E. 1

Pembahasan :

$\mathrm{_{x \to 0}^{lim}\frac{x\,tan\,x}{2\,cos^{2}x\,-2}}$

$\mathrm{_{x \to 0}^{lim}\frac{x\,tan\,x}{-2(1-cos^{2}x)}}$

$\mathrm{_{x \to 0}^{lim}\frac{x\,tan\,x}{-2\,sin^{2}x}}$

$-\frac{1}{2}$ ×

$\mathrm{_{x \to 0}^{lim}\frac{x}{sin\,x}\;\times\;_{x \to 0}^{lim}\frac{tan\,x}{sin\,x} }$

=

$-\frac{1}{2}$ × 1 × 1 =

$-\frac{1}{2}$

Jawaban : A

13. UN 2015
Nilai

$\mathrm{_{x \to 0}^{lim}\frac{x\,tan\,3x}{1-cos^{2}2x}}$ = ...
A. 0
B.

$\frac{1}{4}$
C.

$\frac{2}{4}$
D.

$\frac{3}{4}$
E. 1

Pembahasan :

$\mathrm{_{x \to 0}^{lim}\frac{x\,tan\,3x}{1-cos^{2}2x}}$

$\mathrm{_{x \to 0}^{lim}\frac{x\,tan\,3x}{sin^{2}2x}}$

$\mathrm{_{x \to 0}^{lim}\frac{x}{sin\,2x}\;\times\;_{x \to 0}^{lim}\frac{tan\,3x}{sin\,2x} }$

=

$\frac{1}{2}$ ×

$\frac{3}{2}$ =

$\frac{3}{4}$

Jawaban : D

14. UN 2016
Nilai

$\mathrm{_{x \to 0}^{lim}\frac{1-cos\,4x}{2x\,sin\,4x}}$ = ...
A. 1
B.

$\frac{1}{2}$
C. 0
D.

$-\frac{1}{2}$
E. −1

Pembahasan :

$\begin{align} \mathrm{\lim_{x \to 0}\,\frac{1-cos\,4x}{2x\,sin\,4x}} & = \mathrm{\lim_{x \to 0}\,\frac{{\color{red}\not}2\,sin^{2}2x}{{\color{red}\not}2x\,sin\,4x}} \\ & = \mathrm{\lim_{x \to 0}\,\left (\frac{sin\,2x}{x}\cdot \frac{sin\,2x}{sin\,4x} \right )} \\ & = \frac{2}{1}\cdot \frac{2}{4} \\ & = 1 \end{align}$

Jawaban : A

Pembahasan Soal UN Limit Fungsi Trigonometri

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