We have, $\quad f(x)= \begin{cases}\frac{1-\cos x}{x}, & x \neq 0 \\ k, & x=0\end{cases}$ Since, $f(x)$ is $\operatorname{continuous~at~} x=0$ $\therefore \quad f(0)=\lim _{x \rightarrow 0} f(x)$ $\Rightarrow \quad k=\lim _{x \rightarrow 0} \frac{1-\cos x}{x}$ $\Rightarrow \quad k=\lim _{x \rightarrow 0} \frac{2 \sin 2(x / 2)}{x}$ $\Rightarrow \quad k=\lim _{x \rightarrow 0} \frac{\sin (x / 2)}{(x / 2)} \times \sin (x / 2)$ $\Rightarrow \quad k=0$
Since, $f(x)$ is continuous at $x=0$