Question

$\underset{x\to 0}{lim}{\left(\frac{sinx}{x}\right)}^{\left(\frac{sinx}{x-sinx}\right)}$ equals

• A. 1
• B. e
• C. $e^{-1}$
• D. $e^{-2}$

Answer 1

The correct option is C

$e^{-1}$

$\underset{x\to 0}{lim}{\left(\frac{sinx}{x}\right)}^{\left(\frac{sinx}{x-sinx}\right)}$ $\left(form{1}^{\infty }\right)$

= $e^{-\underset{x\to 0}{lim}(\frac{sinx}{x}-1)\frac{sinx}{(x-sinx)}}$
= $e^{-\underset{x\to 0}{lim}\frac{-sinx}{x}}$
= $e^1$