Question

$\underset{x\to 0}{lim}{\left[\frac{x-1+\cos x}{x}\right]}^{\frac{1}{x}}=$

• A. 11
• B. $0$
• C. $e^{-1{/}_2}$
• D. $e^{1{/}_2}$

Answer 1

The correct option is C $e^{-1{/}_2}$

$\underset{x\to 0}{lim}{\left[\frac{x-1+\cos x}{x}\right]}^{\frac{1}{x}}$
$=e^{\underset{x\to 0}{lim}(\frac{x-1+\cos x}{x}-1)\frac{1}{x}}$
$=e^{-\underset{x\to 0}{lim}\frac{1-\cos x}{x^2}}$
$=e^{-\frac{1}{2}}$

Hence, option C.