Question

lf the function $f\left(x\right)=\frac{e^{x^2}-\cos x}{x^2}$ for $x\ne 0$ is continuous at $x=0$ then $f\left(0\right)=$

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

Answer 1

The correct option is B $\frac{3}{2}$

$f\left(x\right)=\frac{e^{x^2}-cosx}{x^2}$

Given f(x) is continuous at $x=0$

$\underset{x\to 0}{lim}f\left(x\right)=f\left(0\right)$

$=\underset{x\to 0}{lim}\frac{e^{x^2}-cosx}{x^2}$

It is of the form $\frac{0}{0}$ , so applying L-Hospital's rule

$=\underset{x\to 0}{lim}\frac{2x{e}^{x^2}+\sin x}{2x}$

$\underset{x\to 0}{lim}{e}^{x^2}+\underset{x\to 0}{lim}\frac{\sin x}{2x}$

$=1+\frac{1}{2}=\frac{3}{2}$