Question

$\underset{x\to 0}{lim}{\left(\frac{\sin x}{x}\right)}^{1/x^{{}^2}}=e^{-1/a}$.
Find the value of a.

• A. $a=2$
• B. $a=3$
• C. $a=5$
• D. $a=6$

Answer 1

The correct option is D $a=6$

$\underset{x\to 0}{lim}{\left(\frac{\sin x}{x}\right)}^{\frac{1}{x^2}}=\underset{x\to 0}{lim}\left(exp\left(\log {\left(\frac{\sin x}{x}\right)}^{\frac{1}{x^2}}\right)\right)=exp\underset{x\to 0}{lim}\left(\frac{\log \left(\frac{\sin x}{x}\right)}{x^2}\right)$

Applying L-Hospital's rule
$=exp\underset{x\to 0}{lim}\left(\frac{x\cos x-\sin x}{2x^2\sin x}\right)$

Again applying L-Hospital's rule
$=exp\underset{x\to 0}{lim}(-\frac{\sin x}{2(x\cos x+2\sin x)})$

Again applying L-Hospital's rule
$=exp\underset{x\to 0}{lim}(-\frac{\cos x}{2(3\cos x-x\sin x)})=exp\left(\frac{-1}{6}\right)=\frac{1}{\sqrt[6]{6e}}$

$a=6$