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

\(\begin{aligned} \displaystyle\lim_{x\to0}\dfrac{x}{2\csc x\left(1-\sqrt{\cos x}\right)}&=\displaystyle\lim_{x\to0}\dfrac{x}{2\dfrac1{\sin x}\left(1-\sqrt{\cos x}\right)}\color{red}{\cdot\dfrac{1+\sqrt{\cos x}}{1+\sqrt{\cos x}}}\\ &=\displaystyle\lim_{x\to0}\dfrac{x\sin x\left(1+\sqrt{\cos x}\right)}{2\left(1-\cos x\right)}\color{red}{\cdot\dfrac{1+\cos x}{1+\cos x}}\\ &=\displaystyle\lim_{x\to0}\dfrac{x\sin x\left(1+\sqrt{\cos x}\right)(1+\cos x)}{2\left(1-\cos^2x\right)}\\ &=\displaystyle\lim_{x\to0}\dfrac{x\sin x\left(1+\sqrt{\cos x}\right)(1+\cos x)}{2\sin^2x}\\ &=\displaystyle\lim_{x\to0}\dfrac{x\left(1+\sqrt{\cos x}\right)(1+\cos x)}{2\sin x}\\ &=\displaystyle\lim_{x\to0}\dfrac{x}{\sin x}\cdot\displaystyle\lim_{x\to0}\dfrac{\left(1+\sqrt{\cos x}\right)(1+\cos x)}2\\ &=1\cdot\dfrac{\left(1+\sqrt{\cos 0}\right)(1+\cos 0)}2\\ &=\dfrac{\left(1+\sqrt1\right)(1+1)}2\\ &=\dfrac{\left(1+1\right)(\cancel{2})}{\cancel{2}}\\ &=\boxed{\boxed{2}} \end{aligned}\)

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

\(\begin{aligned} \displaystyle\lim_{x\to2}\dfrac{\left(x^2-5x+6\right)\sin(x-2)}{\left(x^2-x-2\right)^2}&=\displaystyle\lim_{x\to2}\dfrac{(x-3)(x-2)\sin(x-2)}{\left((x+1)(x-2)\right)^2}\\[8pt] &=\displaystyle\lim_{x\to2}\dfrac{(x-3)(x-2)\sin(x-2)}{(x+1)^2(x-2)^2}\\[8pt] &=\displaystyle\lim_{x\to2}\dfrac{(x-3)\sin(x-2)}{(x+1)^2(x-2)}\\[8pt] &=\displaystyle\lim_{x\to2}\dfrac{x-3}{(x+1)^2}\cdot\dfrac{\sin(x-2)}{x-2}\\[8pt] &=\dfrac{2-3}{(2+1)^2}\cdot1\\[8pt] &=\dfrac{-1}{(3)^2}\\ &=\boxed{\boxed{-\dfrac19}} \end{aligned}\)

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}=}

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\(\begin{aligned} \displaystyle\lim_{x\to0}\dfrac{\sin3x+\sin5x}{2x\cos4x}&=\displaystyle\lim_{x\to0}\dfrac{\dfrac{\sin3x}x+\dfrac{\sin5x}x}{\dfrac{2x\cos4x}x}\\ &=\displaystyle\lim_{x\to0}\dfrac{\dfrac{\sin3x}x+\dfrac{\sin5x}x}{2\cos4x}\\ &=\dfrac{3+5}{2\cos4(0)}\\ &=\dfrac8{2(1)}\\ &=\boxed{\boxed{4}} \end{aligned}\)

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

\(\begin{aligned} \displaystyle\lim_{x\to0}\dfrac{x\tan3x}{\sin^2x-\cos2x+1}&=\displaystyle\lim_{x\to0}\dfrac{x\tan3x}{\sin^2x-\left(1-2\sin^2x\right)+1}\\[8pt] &=\displaystyle\lim_{x\to0}\dfrac{x\tan3x}{\sin^2x-1+2\sin^2x+1}\\[8pt] &=\displaystyle\lim_{x\to0}\dfrac{x\tan3x}{3\sin^2x}\\[8pt] &=\displaystyle\lim_{x\to0}\dfrac13\cdot\dfrac{x}{\sin x}\cdot\dfrac{\tan3x}{\sin x}\\[8pt] &=\dfrac13\cdot1\cdot3\\ &=\boxed{\boxed{1}} \end{aligned}\)

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

\(\eqalign{ \displaystyle\lim_{x\to0}\dfrac{2-2\cos2x}{x^2}&=\displaystyle\lim_{x\to0}\dfrac{2-2\left(1-2\sin^2x\right)}{x^2}\\ &=\displaystyle\lim_{x\to0}\dfrac{2-2+4\sin^2x}{x^2}\\ &=\displaystyle\lim_{x\to0}\dfrac{4\sin^2x}{x^2}\\ &=\displaystyle\lim_{x\to0}4\cdot\dfrac{\sin x}x\cdot\dfrac{\sin x}x\\ &=4\cdot1\cdot1\\ &=\boxed{\boxed{4}} }\)

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

\(\eqalign{ \displaystyle\lim_{x\to4}\dfrac{x^2-16}{\sin(2x-8)}&=\displaystyle\lim_{x\to4}\dfrac{(x+4)(x-4)}{\sin2(x-4)}\\ &=\displaystyle\lim_{x\to4}(x+4)\cdot\displaystyle\lim_{x\to4}\dfrac{x-4}{\sin2(x-4)} }\)
Misal t=x-4
Jika x\to4 maka t=x-4\to4-4=0

\(\eqalign{ \displaystyle\lim_{x\to4}(x+4)\cdot\displaystyle\lim_{x\to4}\dfrac{x-4}{\sin2(x-4)}&=(4+4)\cdot\displaystyle\lim_{t\to0}\dfrac{t}{\sin2t}\\ &=(8)\cdot\dfrac12\\ &=\boxed{\boxed{4}} }\)

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

\(\eqalign{ \displaystyle\lim_{x\to0}\dfrac{1-\cos4x}{x\sin x}&=\displaystyle\lim_{x\to0}\dfrac{1-\left(1-2\sin^22x\right)}{x\sin x}\\ &=\displaystyle\lim_{x\to0}\dfrac{1-1+2\sin^22x}{x\sin x}\\ &=\displaystyle\lim_{x\to0}\dfrac{2\sin^22x}{x\sin x}\\ &=\displaystyle\lim_{x\to0}\left(2\cdot\dfrac{\sin2x}x\cdot\dfrac{\sin2x}{\sin x}\right)\\ &=2\cdot2\cdot2\\ &=\boxed{\boxed{8}} }\)

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

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