If $f\left(x\right)=\{\begin{array}{cc}\frac{1-\cos x}{x},& x\ne 0\\ k,& x=0\end{array}$ is continuous at $x=0$, then the value of $k$ is

For continuity of $f\left(x\right)$ at $x=0$, we must have,
$f\left(0\right)=\underset{x\to 0}{lim}f\left(x\right)$
$=\underset{x\to 0}{lim}\frac{1-\cos x}{x}\left[\frac{0}{0}\text{form}\right]$
Applying L hospital's rule
$=\underset{x\to 0}{lim}\frac{\sin x}{1}=\sin 0=0$
$\therefore k=0$