Question

If $f(x)=\frac{e^{x^2}-\cos x}{x^2}$ , for $x\ne 0$ 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 $x=0$
$\therefore \underset{x\to 0}{\lim }f(x)=f(0)$
$\Rightarrow f(0)=\underset{x\to 0}{\lim }\frac{e^{x^2}-\cos x}{x^2}$
$\Rightarrow f(0)=\underset{x\to 0}{\lim }\frac{2x{e}^{x^2}+\sin x}{2x}$
[apply L'Hospital rule]
$\Rightarrow f(0)=\underset{x\to 0}{\lim }\frac{2x{e}^{x^2}}{2x}+\underset{x\to 0}{\lim }\frac{\sin x}{2x}$
$=1+\frac{1}{2}=\frac{3}{2}$