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$\lim _{x \rightarrow \frac{\pi}{2}} \frac{1+\cos 2 x}{\cot 3 x\left(3^{\sin 2 x}-1\right)}=$
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Verified Answer
The correct answer is:
$\frac{1}{3 \log 3}$
$\lim _{x \rightarrow \frac{\pi}{2}} \frac{1+\cos 2 x}{\cot 3 x\left(3^{\sin 2 x}-1\right)} \Rightarrow \lim _{x \rightarrow \frac{\pi}{2}} \frac{2 \cos ^2 x}{\frac{\cos 3 x}{\sin 3 x} \times\left(3^{\sin 2 x}-1\right)}$
Let $x=\frac{\pi}{2}-h$, then $\lim _{h \rightarrow 0} \frac{2 \sin ^2 h}{\frac{\sin 3 h}{\cos 3 h}\left(3^{\sin 2 h}-1\right)}$
$$
\begin{aligned}
& =\lim _{h \rightarrow 0} \frac{2 \frac{\sin ^2 h}{h^2} \times h^2(\cos 3 h)}{\left(\frac{\sin 3 h}{3 h} \times 3 h\right) \frac{3^{\sin 2 h}-1}{\sin 2 h} \times \frac{\sin 2 h}{2 h} \times 2 h} \\
& =\frac{2 \times 1 \times 1}{(1 \times 3) \times\left[\log _e(3)\right] \times 1 \times 2}=\frac{1}{3 \log _e 3}
\end{aligned}
$$
Let $x=\frac{\pi}{2}-h$, then $\lim _{h \rightarrow 0} \frac{2 \sin ^2 h}{\frac{\sin 3 h}{\cos 3 h}\left(3^{\sin 2 h}-1\right)}$
$$
\begin{aligned}
& =\lim _{h \rightarrow 0} \frac{2 \frac{\sin ^2 h}{h^2} \times h^2(\cos 3 h)}{\left(\frac{\sin 3 h}{3 h} \times 3 h\right) \frac{3^{\sin 2 h}-1}{\sin 2 h} \times \frac{\sin 2 h}{2 h} \times 2 h} \\
& =\frac{2 \times 1 \times 1}{(1 \times 3) \times\left[\log _e(3)\right] \times 1 \times 2}=\frac{1}{3 \log _e 3}
\end{aligned}
$$
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