Find the value of k for which

Given: $f(x)=\left\{\begin{array}{l}\frac{1-\cos 4 x}{8 x^{2}}, \text { when } x \neq 0 \\ k, \text { when } x=0\end{array}\right.$

If f(x) is continuous at x = 0, then

$\lim _{x \rightarrow 0} f(x)=f(0)$

$\Rightarrow \lim _{x \rightarrow 0} \frac{1-\cos 4 x}{8 x^{2}}=f(0)$

$\Rightarrow \lim _{x \rightarrow 0} \frac{2 \sin ^{2} 2 x}{8 x^{2}}=f(0)$

$\Rightarrow \frac{2}{2} \lim _{x \rightarrow 0} \frac{\sin ^{2} 2 x}{4 x^{2}}=f(0)$

$\Rightarrow \frac{2}{2} \lim _{x \rightarrow 0}\left(\frac{\sin 2 x}{2 x}\right)^{2}=f(0)$

$\Rightarrow 1 \times 1=f(0)$

$\Rightarrow k=1 \quad(\because f(0)=k)$