Question

$ \displaystyle \lim_{x \to 0} \frac{e^{x^2} - \cos x}{\sin^{2}x}$ is equal to :

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

Answer 1

Answer: (B)

$\displaystyle \lim _{x \rightarrow 0} \frac{e^{x^{2}}-\cos x}{\sin ^{2} x}$ ($\frac{0}{0}$ form)
$= \displaystyle \lim _{x \rightarrow 0} \frac{2 x e^{x^{2}}+\sin x}{\sin 2 x}$ ($\frac{0}{0}$ form)
$= \displaystyle \lim _{x \rightarrow 0} \frac{2 x\left(2 x e^{x^{2}}\right)+2 e^{x^{2}}+\cos x}{2 \cos 2 x}=\frac{3}{2}$
(Using L-Hospital Rule)