Question

If $f\left(x\right)=\frac{e^{x^2}-\cos x}{x^2}$, for $x\ne 0$, is continuous at $x=0$, find $f\left(0\right)$.

Answer 1

Let ${lim}_{x\to 0}f\left(x\right)=L={lim}_{x\to 0}\left[\frac{e^{x^2}-\cos x}{x^2}\right]={lim}_{x\to 0}[\frac{(e^{x^2}-1)}{x^2}-\left(\frac{\cos x-1}{x^2}\right)]$

$L=[1-\frac{1}{2}]=\frac{1}{2}$

So, $f\left(0\right)=\frac{1}{2}$