Olimpiade Zone : Limit Trigonometri

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

Tingkat Kesulitan:

Dasar SBMPTN Olimpiade

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No. 1

\displaystyle\lim_{x\to0^+}\left(\cot x\right)^\frac1{\ln x}= ....

Misal y=\left(\cot x\right)^\frac1{\ln x}
$\begin{array}{rl}\ln y& =\ln \left({(\cot x)}^{\frac{1}{\ln x}}\right)\\ & ={\displaystyle \frac{1}{\ln x}}\cdot \ln (\cot x)\\ & ={\displaystyle \frac{\ln (\cot x)}{\ln x}}\\ {\displaystyle \underset{x\to 0^{+}}{lim}\ln y}& ={\displaystyle \underset{x\to 0^{+}}{lim}{\displaystyle \frac{\ln (\cot x)}{\ln x}}}\\ \ln {\displaystyle \underset{x\to 0^{+}}{lim}y}& ={\displaystyle \underset{x\to 0^{+}}{lim}{\displaystyle \frac{-{\displaystyle \frac{1}{\sin x\cos x}}}{{\displaystyle \frac{1}{x}}}}}\\ & ={\displaystyle \underset{x\to 0^{+}}{lim}{\displaystyle \frac{-x}{\sin x\cos x}}}\\ & ={\displaystyle \underset{x\to 0^{+}}{lim}(-{\displaystyle \frac{x}{\sin x}})\cdot {\displaystyle \underset{x\to 0^{+}}{lim}{\displaystyle \frac{1}{\cos x}}}}\\ & =(-1)\cdot {\displaystyle \frac{1}{1}}\\ & =-1\\ {\displaystyle \underset{x\to 0^{+}}{lim}y}& =e^{-1}\\ {\displaystyle \underset{x\to 0^{+}}{lim}{(\cot x)}^{\frac{1}{\ln x}}}& ={\displaystyle \frac{1}{e}}\end{array}$

Jadi, \displaystyle\lim_{x\to0^+}\left(\cot x\right)^\frac1{\ln x}=\dfrac1e.

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No. 2

Nilai dari \displaystyle\lim_{x\to0}\dfrac{\sin^{4030}2x+\tan^{4030}4x}{\sin^{4030}4x-(2x)^{4030}} yang paling mendekati adalah

1. 0
2. 1
3. -1

4. -\infty
5. \infty

$\begin{array}{rl}{\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{{\sin }^{4030}2x+{\tan }^{4030}4x}{{\sin }^{4030}4x-(2x{)}^{4030}}}}& ={\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{{\displaystyle \frac{{\sin }^{4030}2x}{(4x{)}^{4030}}}+{\displaystyle \frac{{\tan }^{4030}4x}{(4x{)}^{4030}}}}{{\displaystyle \frac{{\sin }^{4030}4x}{(4x{)}^{4030}}}-{\displaystyle \frac{(2x{)}^{4030}}{(4x{)}^{4030}}}}}}\\ & ={\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{{\displaystyle \frac{1}{2^{4030}}}\cdot {\displaystyle \frac{{\sin }^{4030}2x}{(2x{)}^{4030}}}+{\displaystyle \frac{{\tan }^{4030}4x}{(4x{)}^{4030}}}}{{\displaystyle \frac{{\sin }^{4030}4x}{(4x{)}^{4030}}}-{\displaystyle \frac{1}{2^{4030}}}\cdot {\displaystyle \frac{(2x{)}^{4030}}{(2x{)}^{4030}}}}}}\\ & ={\displaystyle \frac{{\displaystyle \frac{1}{2^{4030}}}+1}{1-{\displaystyle \frac{1}{2^{4030}}}}}\\ & ={\displaystyle \frac{2^{4030}+1}{2^{4030}-1}}\\ & ={\displaystyle \frac{2^{4030}-1+2}{2^{4030}-1}}\\ & =1+{\displaystyle \frac{2}{2^{4030}-1}}\\ & =1+\mathrm{0,000.}..\\ & =\mathrm{1,000.}..\\ & \approx \boxed{{\displaystyle \boxed{{\displaystyle 1}}}}\end{array}$

Jadi, nilai dari \displaystyle\lim_{x\to0}\dfrac{\sin^{4030}2x+\tan^{4030}4x}{\sin^{4030}4x-(2x)^{4030}} yang paling mendekati adalah 1.
JAWAB: B