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}

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

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

$\begin{array}{rl}{\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{3x}{(1-\sqrt{\cos 2x})\csc 2x}}}& ={\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{3x\sin 2x}{1-\sqrt{\cos 2x}}}\cdot {\displaystyle \frac{1+\sqrt{\cos 2x}}{1+\sqrt{\cos 2x}}}}\\ & ={\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{3x\sin 2x(1+\sqrt{\cos 2x})}{1-\cos 2x}}}\\ & ={\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{3x\sin 2x(1+\sqrt{\cos 2x})}{1-(1-2{\sin }^{2}x)}}}\\ & ={\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{3x\sin 2x(1+\sqrt{\cos 2x})}{1-1+2{\sin }^{2}x}}}\\ & ={\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{3x\sin 2x(1+\sqrt{\cos 2x})}{2{\sin }^{2}x}}}\\ & ={\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{3}{2}}\cdot {\displaystyle \frac{x}{\sin x}}\cdot {\displaystyle \frac{\sin 2x}{\sin x}}\cdot (1+\sqrt{\cos 2x})}\\ & ={\displaystyle \frac{3}{2}}\cdot {\displaystyle \frac{1}{1}}\cdot {\displaystyle \frac{2}{1}}\cdot (1+\sqrt{\cos 2(0)})\\ & =3(1+\sqrt{\cos 0})\\ & =3(1+\sqrt{1})\\ & =3(1+1)\\ & =3(2)\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 6}}}}\end{array}$

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

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

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

$\begin{array}{rl}{\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{3x}{(1-\sqrt{\cos 2x})\csc 2x}}}& ={\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{3x\sin 2x}{1-\sqrt{\cos 2x}}}\cdot {\displaystyle \frac{1+\sqrt{\cos 2x}}{1+\sqrt{\cos 2x}}}}\\ & ={\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{3x\sin 2x(1+\sqrt{\cos 2x})}{1-\cos 2x}}\cdot {\displaystyle \frac{1+\cos 2x}{1+\cos 2x}}}\\ & ={\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{3x\sin 2x(1+\sqrt{\cos 2x})(1+\cos 2x)}{1-{\cos }^{2}2x}}}\\ & ={\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{3x\sin 2x(1+\sqrt{\cos 2x})(1+\cos 2x)}{{\sin }^{2}2x}}}\\ & ={\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{3x}{\sin 2x}}\cdot {\displaystyle \frac{\sin 2x}{\sin 2x}}\cdot (1+\sqrt{\cos 2x})\cdot (1+\cos 2x)}\\ & ={\displaystyle \frac{3}{2}}\cdot {\displaystyle \frac{2}{2}}\cdot (1+\sqrt{\cos 0})\cdot (1+\cos 0)\\ & ={\displaystyle \frac{3}{2}}\cdot 1\cdot (1+\sqrt{1})\cdot (1+1)\\ & ={\displaystyle \frac{3}{2}}\cdot (1+1)\cdot 2\\ & ={\displaystyle \frac{3}{2}}\cdot 2\cdot 2\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 6}}}}\end{array}$

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

$\begin{array}{rl}{\displaystyle \underset{a\to b}{lim}{\displaystyle \frac{\tan a-\tan b}{1+(1-{\displaystyle \frac{a}{b}})\tan a\tan b-{\displaystyle \frac{a}{b}}}}}& ={\displaystyle \underset{a\to b}{lim}{\displaystyle \frac{\tan a-\tan b}{1-{\displaystyle \frac{a}{b}}+(1-{\displaystyle \frac{a}{b}})\tan a\tan b}}}\\ & ={\displaystyle \underset{a\to b}{lim}{\displaystyle \frac{\tan a-\tan b}{(1-{\displaystyle \frac{a}{b}})(1+\tan a\tan b)}}}\\ & ={\displaystyle \underset{a\to b}{lim}{\displaystyle \frac{\tan (a-b)}{{\displaystyle \frac{b-a}{b}}}}}\\ & ={\displaystyle \underset{a\to b}{lim}{\displaystyle \frac{\tan (a-b)}{{\displaystyle \frac{a-b}{-b}}}}}\\ & ={\displaystyle \underset{a\to b}{lim}{\displaystyle \frac{-b\tan (a-b)}{a-b}}}\end{array}$
Misal a-b=x
Jika a\to b maka {x=a-b\to b-b=0}

$\begin{array}{rl}{\displaystyle \underset{a\to b}{lim}{\displaystyle \frac{-b\tan (a-b)}{a-b}}}& ={\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{-b\tan x}{x}}}\\ & ={\displaystyle \underset{x\to 0}{lim}-b\cdot {\displaystyle \frac{tanx}{x}}}\\ & =-b\cdot 1\\ & =\boxed{{\displaystyle \boxed{{\displaystyle -b}}}}\end{array}$

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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