Question

If $f(x) = \frac{e^{x^2} - \cos \, x}{x^2}$ , for $x \neq 0$ is continuous at $x = 0$, then value of $f(0)$ is

• A. $\frac{2}{3}$
• B. $\frac{5}{2}$
• C. 1
• D. $\frac{3}{2}$

Answer 1

Answer: (D)

We have,
$f(x)=\frac{e^{x^{2}}-\cos x}{x^{2}}$ is continuous at $x=0$
$\therefore \displaystyle\lim _{x \rightarrow 0} f(x)=f(0) $
$\Rightarrow f(0) = \displaystyle\lim _{x \rightarrow 0} \frac{e^{x^{2}}-\cos x}{x^{2}}$
$\Rightarrow f(0)= \displaystyle\lim _{x \rightarrow 0} \frac{2 x e^{x^{2}}+\sin x}{2 x}$
[apply L'Hospital rule]
$\Rightarrow f(0)=\displaystyle\lim _{x \rightarrow 0} \frac{2 x e^{x^{2}}}{2 x}+\displaystyle\lim _{x \rightarrow 0} \frac{\sin x}{2 x}$
$=1+\frac{1}{2}=\frac{3}{2}$