Berikut ini adalah kumpulan soal mengenai Limit Trigonometri tingkat SBMPTN. Jika ada jawaban yang salah, mohon dikoreksi melalui komentar. Terima kasih.
No. 1
\displaystyle\lim_{x\to0^+}\left(\cot x\right)^\frac1{\ln x}= ....
Misal y=\left(\cot x\right)^\frac1{\ln x}
\(\begin{aligned}
\ln y&=\ln\left(\left(\cot x\right)^\frac1{\ln x}\right)\\
&=\dfrac1{\ln x}\cdot\ln(\cot x)\\
&=\dfrac{\ln(\cot x)}{\ln x}\\
\displaystyle\lim_{x\to0^+}\ln y&=\displaystyle\lim_{x\to0^+}\dfrac{\ln(\cot x)}{\ln x}\\
\ln\displaystyle\lim_{x\to0^+}y&=\displaystyle\lim_{x\to0^+}\dfrac{-\dfrac1{\sin x\cos x}}{\dfrac1x}\\
&=\displaystyle\lim_{x\to0^+}\dfrac{-x}{\sin x\cos x}\\
&=\displaystyle\lim_{x\to0^+}\left(-\dfrac{x}{\sin x}\right)\cdot\displaystyle\lim_{x\to0^+}\dfrac1{\cos x}\\
&=(-1)\cdot\dfrac11\\
&=-1\\
\displaystyle\lim_{x\to0^+}y&=e^{-1}\\
\displaystyle\lim_{x\to0^+}\left(\cot x\right)^\frac1{\ln x}&=\dfrac1e
\end{aligned}\)
No. 2
Nilai dari
\displaystyle\lim_{x\to0}\dfrac{\sin^{4030}2x+\tan^{4030}4x}{\sin^{4030}4x-(2x)^{4030}} yang paling mendekati adalah
\(\begin{aligned}
\displaystyle\lim_{x\to0}\dfrac{\sin^{4030}2x+\tan^{4030}4x}{\sin^{4030}4x-(2x)^{4030}}&=\displaystyle\lim_{x\to0}\dfrac{\dfrac{\sin^{4030}2x}{(4x)^{4030}}+\dfrac{\tan^{4030}4x}{(4x)^{4030}}}{\dfrac{\sin^{4030}4x}{(4x)^{4030}}-\dfrac{(2x)^{4030}}{(4x)^{4030}}}\\[25pt]
&=\displaystyle\lim_{x\to0}\dfrac{\dfrac1{2^{4030}}\cdot\dfrac{\sin^{4030}2x}{(2x)^{4030}}+\dfrac{\tan^{4030}4x}{(4x)^{4030}}}{\dfrac{\sin^{4030}4x}{(4x)^{4030}}-\dfrac1{2^{4030}}\cdot\dfrac{(2x)^{4030}}{(2x)^{4030}}}\\[25pt]
&=\dfrac{\dfrac1{2^{4030}}+1}{1-\dfrac1{2^{4030}}}\\[25pt]
&=\dfrac{2^{4030}+1}{2^{4030}-1}\\[10pt]
&=\dfrac{2^{4030}-1+2}{2^{4030}-1}\\[10pt]
&=1+\dfrac2{2^{4030}-1}\\
&=1+0{,}000...\\
&=1{,}000...\\
&\approx\boxed{\boxed{1}}
\end{aligned}\)
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