Question

$\underset{x \rightarrow 0} {\text{Lim}} \frac{1}{\sin x} \int\limits_0^{\ln (1+x)}(1-\tan 2 y)^{1 / y} d y$ equals

• A. $e ^{-2}$
• B. $e$
• C. $e^2$
• D. $e^4$

Answer 1

Answer: (A)

$l=\underset{x \rightarrow 0} {\text{Lim}} \frac{\int\limits_0^{\ln (1+x)}(1-\tan 2 y)^{1 / y} d y}{\frac{\sin x}{x} \cdot x}$
Using L'Hospital's Rule
$l=\underset{x \rightarrow 0} {\text{Lim}} \frac{[1-\tan 2(\ln (1+x))]^{\frac{1}{\ln (1+x)}}}{(1+x)} \quad\left(1^{\infty}\right)$
$= e ^{-\underset{ x \rightarrow 0}{\text{Lim}} \frac{1}{\ln (1+ x )} \tan 2(\ln (1+ x ))}= e ^{-\underset{ x \rightarrow 0} {\text{Lim}} \frac{\tan (2 \ln (1+ x ))}{2 \ln (1+ x )} \cdot 2}= e ^{-2}\left( \text{using} \underset{x \rightarrow 0} {\text{Lim}}\frac{\tan \theta}{\theta}=1\right)$