Exercise Zone : Limit Trigonometri [2]

Berikut ini adalah kumpulan soal mengenai Limit Trigonometri tipe standar. Jika ada jawaban yang salah, mohon dikoreksi melalui komentar. Terima kasih.

Tipe:

Standar SBMPTN HOTS

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No. 11

\displaystyle\lim_{x\to0}\dfrac{x}{2\csc x\left(1-\sqrt{\cos x}\right)}=

1. -2
2. -1
3. 0

4. 1
5. 2

$\begin{array}{rl}{\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{x}{2\csc x(1-\sqrt{\cos x})}}}& ={\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{x}{2{\displaystyle \frac{1}{\sin x}}(1-\sqrt{\cos x})}}\cdot {\displaystyle \frac{1+\sqrt{\cos x}}{1+\sqrt{\cos x}}}}\\ & ={\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{x\sin x(1+\sqrt{\cos x})}{2(1-\cos x)}}\cdot {\displaystyle \frac{1+\cos x}{1+\cos x}}}\\ & ={\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{x\sin x(1+\sqrt{\cos x})(1+\cos x)}{2(1-{\cos }^{2}x)}}}\\ & ={\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{x\sin x(1+\sqrt{\cos x})(1+\cos x)}{2{\sin }^{2}x}}}\\ & ={\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{x(1+\sqrt{\cos x})(1+\cos x)}{2\sin x}}}\\ & ={\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{x}{\sin x}}\cdot {\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{(1+\sqrt{\cos x})(1+\cos x)}{2}}}}\\ & =1\cdot {\displaystyle \frac{(1+\sqrt{\cos 0})(1+\cos 0)}{2}}\\ & ={\displaystyle \frac{(1+\sqrt{1})(1+1)}{2}}\\ & ={\displaystyle \frac{(1+1)(\cancel{2})}{\cancel{2}}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 2}}}}\end{array}$

Jadi, \displaystyle\lim_{x\to0}\dfrac{x}{2\csc x\left(1-\sqrt{\cos x}\right)}=2.
JAWAB: E

No. 12

Nilai \displaystyle\lim_{x\to2}\dfrac{\left(x^2-5x+6\right)\sin(x-2)}{\left(x^2-x-2\right)^2} adalah

1. 0
2. -\dfrac19
3. \dfrac12

4. 2
5. 4

$\begin{array}{rl}{\displaystyle \underset{x\to 2}{lim}{\displaystyle \frac{(x^2-5x+6)\sin (x-2)}{{(x^2-x-2)}^2}}}& ={\displaystyle \underset{x\to 2}{lim}{\displaystyle \frac{(x-3)(x-2)\sin (x-2)}{{((x+1)(x-2))}^2}}}\\ & ={\displaystyle \underset{x\to 2}{lim}{\displaystyle \frac{(x-3)(x-2)\sin (x-2)}{(x+1{)}^2(x-2{)}^2}}}\\ & ={\displaystyle \underset{x\to 2}{lim}{\displaystyle \frac{(x-3)\sin (x-2)}{(x+1{)}^2(x-2)}}}\\ & ={\displaystyle \underset{x\to 2}{lim}{\displaystyle \frac{x-3}{(x+1{)}^2}}\cdot {\displaystyle \frac{\sin (x-2)}{x-2}}}\\ & ={\displaystyle \frac{2-3}{(2+1{)}^2}}\cdot 1\\ & ={\displaystyle \frac{-1}{(3{)}^2}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle -{\displaystyle \frac{1}{9}}}}}}\end{array}$

Jadi, \displaystyle\lim_{x\to2}\dfrac{\left(x^2-5x+6\right)\sin(x-2)}{\left(x^2-x-2\right)^2}=-\dfrac19.
JAWAB: B

No. 13

Nilai dari {\displaystyle\lim_{x\to0}\dfrac{\sin3x+\sin5x}{2x\cos4x}=}

brainly.co.id

$\begin{array}{rl}{\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{\sin 3x+\sin 5x}{2x\cos 4x}}}& ={\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{{\displaystyle \frac{\sin 3x}{x}}+{\displaystyle \frac{\sin 5x}{x}}}{{\displaystyle \frac{2x\cos 4x}{x}}}}}\\ & ={\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{{\displaystyle \frac{\sin 3x}{x}}+{\displaystyle \frac{\sin 5x}{x}}}{2\cos 4x}}}\\ & ={\displaystyle \frac{3+5}{2\cos 4(0)}}\\ & ={\displaystyle \frac{8}{2(1)}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 4}}}}\end{array}$

Jadi, {\displaystyle\lim_{x\to0}\dfrac{\sin3x+\sin5x}{2x\cos4x}=4}.

No. 14

Nilai {\displaystyle\lim_{x\to0}\dfrac{x\tan3x}{\sin^2x-\cos2x+1}=}

1. \dfrac12
2. -1
3. -2

4. 1
5. 2

$\begin{array}{rl}{\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{x\tan 3x}{{\sin }^{2}x-\cos 2x+1}}}& ={\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{x\tan 3x}{{\sin }^{2}x-(1-2{\sin }^{2}x)+1}}}\\ & ={\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{x\tan 3x}{{\sin }^{2}x-1+2{\sin }^{2}x+1}}}\\ & ={\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{x\tan 3x}{3{\sin }^{2}x}}}\\ & ={\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{1}{3}}\cdot {\displaystyle \frac{x}{\sin x}}\cdot {\displaystyle \frac{\tan 3x}{\sin x}}}\\ & ={\displaystyle \frac{1}{3}}\cdot 1\cdot 3\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 1}}}}\end{array}$

Jadi, {\displaystyle\lim_{x\to0}\dfrac{x\tan3x}{\sin^2x-\cos2x+1}=1}.
JAWAB: D

No. 15

\displaystyle\lim_{x\to0}\dfrac{\sin3x}{5x}

Grup Telegram

\displaystyle\lim_{x\to0}\dfrac{\sin3x}{5x}=\boxed{\boxed{\dfrac35}}

Jadi, \displaystyle\lim_{x\to0}\dfrac{\sin3x}{5x}=\dfrac35.
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No. 16

\displaystyle\lim_{x\to0}\dfrac{4\tan5x}{3x}

Grup Telegram

\displaystyle\lim_{x\to0}\dfrac{4\tan5x}{3x}=\dfrac{4\cdot5}3=\boxed{\boxed{\dfrac{20}3}}

Jadi, \displaystyle\lim_{x\to0}\dfrac{4\tan5x}{3x}=\dfrac{20}3.
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No. 17

Tentukan nilai dari \displaystyle\lim_{x\to0}\dfrac{2-2\cos2x}{x^2}

Grup Telegram

$\begin{array}{rl}{\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{2-2\cos 2x}{x^2}}}& ={\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{2-2(1-2{\sin }^{2}x)}{x^2}}}\\ & ={\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{2-2+4{\sin }^{2}x}{x^2}}}\\ & ={\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{4{\sin }^{2}x}{x^2}}}\\ & ={\displaystyle \underset{x\to 0}{lim}4\cdot {\displaystyle \frac{\sin x}{x}}\cdot {\displaystyle \frac{\sin x}{x}}}\\ & =4\cdot 1\cdot 1\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 4}}}}\end{array}$

Jadi, \displaystyle\lim_{x\to0}\dfrac{2-2\cos2x}{x^2}=4.
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No. 18

Nilai dari \displaystyle\lim_{x\to4}\dfrac{x^2-16}{\sin(2x-8)}

Grup Telegram

$\begin{array}{rl}{\displaystyle \underset{x\to 4}{lim}{\displaystyle \frac{x^2-16}{\sin (2x-8)}}}& ={\displaystyle \underset{x\to 4}{lim}{\displaystyle \frac{(x+4)(x-4)}{\sin 2(x-4)}}}\\ & ={\displaystyle \underset{x\to 4}{lim}(x+4)\cdot {\displaystyle \underset{x\to 4}{lim}{\displaystyle \frac{x-4}{\sin 2(x-4)}}}}\end{array}$
Misal t=x-4
Jika x\to4 maka t=x-4\to4-4=0

$\begin{array}{rl}{\displaystyle \underset{x\to 4}{lim}(x+4)\cdot {\displaystyle \underset{x\to 4}{lim}{\displaystyle \frac{x-4}{\sin 2(x-4)}}}}& =(4+4)\cdot {\displaystyle \underset{t\to 0}{lim}{\displaystyle \frac{t}{\sin 2t}}}\\ & =(8)\cdot {\displaystyle \frac{1}{2}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 4}}}}\end{array}$

Jadi, \displaystyle\lim_{x\to4}\dfrac{x^2-16}{\sin(2x-8)}=4.

No. 19

\displaystyle\lim_{x\to0}\dfrac{1-\cos4x}{x\sin x}=

Grup Telegram

$\begin{array}{rl}{\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{1-\cos 4x}{x\sin x}}}& ={\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{1-(1-2{\sin }^{2}2x)}{x\sin x}}}\\ & ={\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{1-1+2{\sin }^{2}2x}{x\sin x}}}\\ & ={\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{2{\sin }^{2}2x}{x\sin x}}}\\ & ={\displaystyle \underset{x\to 0}{lim}(2\cdot {\displaystyle \frac{\sin 2x}{x}}\cdot {\displaystyle \frac{\sin 2x}{\sin x}})}\\ & =2\cdot 2\cdot 2\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 8}}}}\end{array}$

Jadi, \displaystyle\lim_{x\to0}\dfrac{1-\cos4x}{x\sin x}=8.

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