Answer:
Option D
Explanation:
We have,
f(x) = $\frac{e^{x^{2}}-\cos x}{x^{2}}$ is continuous at x=0
$\therefore$ $\lim_{x \rightarrow 0}f(x)=f(0)$
$\Rightarrow$ $f(0)=\lim_{x \rightarrow0}\frac{e^{x^{2}}-\cos x}{x^{2}}$
$\Rightarrow$ $f(0)=\lim_{x \rightarrow0}\frac{2xe^{x^{2}}+\sin x}{2x}$
[ apply L' Hosptial rule]
$\Rightarrow$ $f(0)=\lim_{x \rightarrow0}\frac{2xe^{x^{2}}}{2x}+\lim_{x \rightarrow 0}\frac{\sin x}{2x}=1+\frac{1}{2}=\frac{3}{2}$
