Question

$\underset{x\to 0}{lim}\frac{e^{x^2}-\cos x}{{\sin }^{2}x}$ is equal to

• A. $2$
• B. $3$
• C. $\frac{5}{4}$
• D. $\frac{3}{2}$

Answer 1

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

$={lim}_{x\to 0}\frac{e^{x^2}-\cos x}{{\sin }^{2}x}$

put $x=0$

$\frac{0}{0}$ form

by L'Hopital rule

$={lim}_{x\to 0}$ $\frac{2x{e}^{x^2}+\sin x}{2\sin x\cos x}$

$={lim}_{x\to 0}$ $\frac{2x{e}^{x^2}}{2\sin x\cos x}+$ ${lim}_{x\to 0}$ $\frac{\sin x}{2\sin x\cos x}$

$={lim}_{x\to 0}$ $\frac{e^{x^2}}{\cos x}+$ ${lim}_{x\to 0}$ $\frac{1}{2\cos x}$

since ,${lim}_{x\to 0}$ $\frac{x}{\sin x}=1$

$=$ $\frac{e^0}{\cos 0}+$ $\frac{1}{2\cos 0}$

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

$=\frac{3}{2}$