SBMPTN Zone : Limit Trigonometri [2]

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

Tipe:

Standar SBMPTN HOTS

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No. 11

Nilai \displaystyle\lim_{x\to\frac{\pi}2}\dfrac{\sin\left(x-\dfrac{\pi}2\right)}{\sqrt{\dfrac{x}2}-\sqrt{\dfrac{\pi}4}} adalah

1. 4\sqrt{\pi}
2. 2\sqrt{\pi}
3. \sqrt{\pi}

4. \dfrac12\sqrt{\pi}
5. \dfrac14\sqrt{\pi}

\(\begin{aligned} \displaystyle\lim_{x\to\frac{\pi}2}\dfrac{\sin\left(x-\dfrac{\pi}2\right)}{\sqrt{\dfrac{x}2}-\sqrt{\dfrac{\pi}4}}\color{red}{\cdot\dfrac{\sqrt{\dfrac{x}2}+\sqrt{\dfrac{\pi}4}}{\sqrt{\dfrac{x}2}+\sqrt{\dfrac{\pi}4}}}&=\displaystyle\lim_{x\to\frac{\pi}2}\dfrac{\left(\sqrt{\dfrac{x}2}+\sqrt{\dfrac{\pi}4}\right)\sin\left(x-\dfrac{\pi}2\right)}{\dfrac{x}2-\dfrac{\pi}4}\\[20pt] &=\displaystyle\lim_{x\to\frac{\pi}2}\dfrac{\left(\sqrt{\dfrac{x}2}+\sqrt{\dfrac{\pi}4}\right)\sin\left(x-\dfrac{\pi}2\right)}{\dfrac12\left(x-\dfrac{\pi}2\right)}\\[20pt] &=\displaystyle\lim_{x\to\frac{\pi}2}2\left(\sqrt{\dfrac{x}2}+\sqrt{\dfrac{\pi}4}\right)\cdot\dfrac{\sin\left(x-\dfrac{\pi}2\right)}{x-\dfrac{\pi}2}\\[20pt] &=2\left(\sqrt{\dfrac{\dfrac{\pi}2}2}+\sqrt{\dfrac{\pi}4}\right)\cdot1\\[20pt] &=2\left(\sqrt{\dfrac{\pi}4}+\sqrt{\dfrac{\pi}4}\right)\\[8pt] &=2\left(\dfrac12\sqrt{\pi}+\dfrac12\sqrt{\pi}\right)\\ &=\boxed{\boxed{2\sqrt{\pi}}} \end{aligned}\)

Jadi, {\displaystyle\lim_{x\to\frac{\pi}2}\dfrac{\sin\left(x-\dfrac{\pi}2\right)}{\sqrt{\dfrac{x}2}-\sqrt{\dfrac{\pi}4}}=2\sqrt{\pi}}.
JAWAB: B

No. 12

\displaystyle\lim_{x\to0}\dfrac{3x}{\left(1-\sqrt{\cos2x}\right)\csc2x}=

1. 8
2. 6
3. 4

4. 3
5. 2

SKMP September 2020

\(\eqalign{ \displaystyle\lim_{x\to0}\dfrac{3x}{\left(1-\sqrt{\cos2x}\right)\csc2x}&=\displaystyle\lim_{x\to0}\dfrac{3x\sin2x}{1-\sqrt{\cos2x}}\cdot{\color{red}\dfrac{1+\sqrt{\cos2x}}{1+\sqrt{\cos2x}}}\\ &=\displaystyle\lim_{x\to0}\dfrac{3x\sin2x\left(1+\sqrt{\cos2x}\right)}{1-\cos2x}\\ &=\displaystyle\lim_{x\to0}\dfrac{3x\sin2x\left(1+\sqrt{\cos2x}\right)}{1-\left(1-2\sin^2x\right)}\\ &=\displaystyle\lim_{x\to0}\dfrac{3x\sin2x\left(1+\sqrt{\cos2x}\right)}{1-1+2\sin^2x}\\ &=\displaystyle\lim_{x\to0}\dfrac{3x\sin2x\left(1+\sqrt{\cos2x}\right)}{2\sin^2x}\\ &=\displaystyle\lim_{x\to0}\dfrac32\cdot\dfrac{x}{\sin x}\cdot\dfrac{\sin2x}{\sin x}\cdot\left(1+\sqrt{\cos2x}\right)\\ &=\dfrac32\cdot\dfrac11\cdot\dfrac21\cdot\left(1+\sqrt{\cos2(0)}\right)\\ &=3\left(1+\sqrt{\cos0}\right)\\ &=3\left(1+\sqrt1\right)\\ &=3(1+1)\\ &=3(2)\\ &=\boxed{\boxed{6}} }\)

Jadi, \displaystyle\lim_{x\to0}\dfrac{3x}{\left(1-\sqrt{\cos2x}\right)\csc2x}=6.
JAWAB: B

No. 13

\displaystyle\lim_{x\to0}\sqrt{\dfrac{x\cdot\tan x}{1+\sin^2x-\cos2x}}=

1. \dfrac13
2. \dfrac13\sqrt3
3. \dfrac12\sqrt3

4. \sqrt3
5. 3

SKMP Juli 2020

\(\eqalign{ \displaystyle\lim_{x\to0}\sqrt{\dfrac{x\cdot\tan x}{1+\sin^2x-\cos2x}}&=\displaystyle\lim_{x\to0}\sqrt{\dfrac{x\cdot\tan x}{1+\sin^2x-\left(1-2\sin^2x\right)}}\\ &=\displaystyle\lim_{x\to0}\sqrt{\dfrac{x\cdot\tan x}{1+\sin^2x-1+2\sin^2x}}\\ &=\displaystyle\lim_{x\to0}\sqrt{\dfrac{x\cdot\tan x}{3\sin^2x}}\\ &=\displaystyle\lim_{x\to0}\sqrt{\dfrac13\cdot\dfrac{x}{\sin x}\cdot\dfrac{\tan x}{\sin x}}\\ &=\sqrt{\dfrac13}\\ &=\dfrac1{\sqrt3}\\ &=\boxed{\boxed{\dfrac13\sqrt3}} }\)

Jadi, \displaystyle\lim_{x\to0}\sqrt{\dfrac{x\cdot\tan x}{1+\sin^2x-\cos2x}}=\dfrac13\sqrt3.
JAWAB: B

No. 14

\displaystyle\lim_{x\to0}\dfrac{3x}{\left(1-\sqrt{\cos2x}\right)\csc2x}=

1. 8
2. 6
3. 4

4. 3
5. 2

SKMP September 2020

\(\eqalign{ \displaystyle\lim_{x\to0}\dfrac{3x}{\left(1-\sqrt{\cos 2x}\right)\csc2x}&=\displaystyle\lim_{x\to0}\dfrac{3x\sin2x}{1-\sqrt{\cos 2x}}{\color{red}\cdot\dfrac{1+\sqrt{\cos 2x}}{1+\sqrt{\cos 2x}}}\\ &=\displaystyle\lim_{x\to0}\dfrac{3x\sin2x\left(1+\sqrt{\cos 2x}\right)}{1-\cos 2x}{\color{red}\cdot\dfrac{1+\cos 2x}{1+\cos 2x}}\\ &=\displaystyle\lim_{x\to0}\dfrac{3x\sin2x\left(1+\sqrt{\cos 2x}\right)(1+\cos 2x)}{1-\cos^2 2x}\\ &=\displaystyle\lim_{x\to0}\dfrac{3x\sin2x\left(1+\sqrt{\cos 2x}\right)(1+\cos 2x)}{\sin^2 2x}\\ &=\displaystyle\lim_{x\to0}\dfrac{3x}{\sin 2x}\cdot\dfrac{\sin2x}{\sin 2x}\cdot\left(1+\sqrt{\cos 2x}\right)\cdot(1+\cos 2x)\\ &=\dfrac32\cdot\dfrac22\cdot\left(1+\sqrt{\cos 0}\right)\cdot(1+\cos 0)\\ &=\dfrac32\cdot1\cdot\left(1+\sqrt1\right)\cdot(1+1)\\ &=\dfrac32\cdot\left(1+1\right)\cdot2\\ &=\dfrac32\cdot2\cdot2\\ &=\boxed{\boxed{6}} }\)

Jadi, \displaystyle\lim_{x\to0}\dfrac{3x}{\left(1-\sqrt{\cos2x}\right)\csc2x}=6.
JAWAB: B

No. 15

\displaystyle\lim_{a\to b}\dfrac{\tan a-\tan b}{1+\left(1-\dfrac{a}b\right)\tan a\tan b-\dfrac{a}b}= ...

1. \dfrac1b
2. b
3. -b

4. -\dfrac1b
5. 1

SIMAK UI 2011

\(\eqalign{ \displaystyle\lim_{a\to b}\dfrac{\tan a-\tan b}{1+\left(1-\dfrac{a}b\right)\tan a\tan b-\dfrac{a}b}&=\displaystyle\lim_{a\to b}\dfrac{\tan a-\tan b}{1-\dfrac{a}b+\left(1-\dfrac{a}b\right)\tan a\tan b}\\ &=\displaystyle\lim_{a\to b}\dfrac{\tan a-\tan b}{\left(1-\dfrac{a}b\right)\left(1+\tan a\tan b\right)}\\ &=\displaystyle\lim_{a\to b}\dfrac{\tan( a-b)}{\dfrac{b-a}b}\\ &=\displaystyle\lim_{a\to b}\dfrac{\tan( a-b)}{\dfrac{a-b}{-b}}\\ &=\displaystyle\lim_{a\to b}\dfrac{-b\tan( a-b)}{a-b} }\)
Misal a-b=x
Jika a\to b maka {x=a-b\to b-b=0}

\(\eqalign{ \displaystyle\lim_{a\to b}\dfrac{-b\tan( a-b)}{a-b}&=\displaystyle\lim_{x\to0}\dfrac{-b\tan x}x\\ &=\displaystyle\lim_{x\to0}-b\cdot\dfrac{tan x}x\\ &=-b\cdot1\\ &=\boxed{\boxed{-b}} }\)

Jadi, \displaystyle\lim_{a\to b}\dfrac{\tan a-\tan b}{1+\left(1-\dfrac{a}b\right)\tan a\tan b-\dfrac{a}b}=-b.
JAWAB: C

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