Question

Find :-
$\underset{x\to 0}{lim}\frac{e^{x^2}-\cos x}{x^2}$

• A. $\frac{1}{2}$
• B. $\frac{2}{3}$
• C. $\frac{3}{2}$
• D. $2$

Answer 1

The correct option is C $\frac{3}{2}$

We have,

${lim}_{x\to 0}\left(\frac{e^{x^2}-\cos x}{x^2}\right)$

Apply L-Hospital rule,

$\Rightarrow {lim}_{x\to 0}\left(\frac{e^{x^2}\times 2x+\sin x}{2x}\right)$

$\Rightarrow {lim}_{x\to 0}(e^{x^2}+\frac{\sin x}{2x})$

$\Rightarrow e^0+\frac{1}{2}\left(1\right)[\because {lim}_{x\to 0}\left(\frac{\sin x}{x}\right)]$

$\Rightarrow 1+\frac{1}{2}$

$\Rightarrow \frac{3}{2}$

Hence, this is the answer.