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 $\frac{3}{2}$

$\underset{x\to 0}{lim}\frac{e^{x^2}-\cos x}{{\sin }^{2}x}$
Use L'Hospital Rule
$\Rightarrow \underset{x\to 0}{lim}\frac{2x{e}^{x^2}+\sin x}{2\sin x\cos x}=\underset{x\to 0}{lim}\frac{2x{e}^{x^2}+\sin x}{\sin 2x}$
Again use L'Hospital's Rule
$\Rightarrow \Rightarrow \underset{x\to 0}{lim}\frac{4x^2{e}^{x^2}+2e^{x^2}+\cos x}{2\cos 2x}=\frac{3}{2}$