The value of k (k<0) for which  the function  f  defined as $\mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{l}\frac{1-\cos \mathrm{kx}}{\mathrm{xsin}\mathrm{x}},\mathrm{x}\ne 0\\ \frac{1}{2},\mathrm{x}=0\end{array}\right.$ is continuous at x= 0 is 

Given that the function $\mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{l}\frac{1-\cos \mathrm{kx}}{\mathrm{xsin}\mathrm{x}},\mathrm{x}\ne 0\\ \frac{1}{2},\mathrm{x}=0\end{array}\right.$ is continuous at $x=0$

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

Hence, 

$\begin{array}{l}\underset{x\to 0}{lim}\left(\frac{1-\cos \mathrm{kx}}{\mathrm{xsin}\mathrm{x}}\right)=\frac{1}{2}\\ \underset{x\to 0}{lim}\left(\frac{2{\sin }^2\frac{\mathrm{kx}}{2}}{\mathrm{xsin}\mathrm{x}}\right)=\frac{1}{2}\\ \underset{\mathrm{x}\to 0}{\lim }2{\left(\frac{\mathrm{k}}{2}\right)}^2{\left(\frac{\sin \frac{\mathrm{kx}}{2}}{\frac{\mathrm{kx}}{2}}\right)}^2\left(\frac{\mathrm{x}}{\sin \mathrm{x}}\right)=\frac{1}{2}\\ \Rightarrow {\mathrm{k}}^2=1\Rightarrow \mathrm{k}=\pm 1\text{ but }\mathrm{k}<0\Rightarrow \mathrm{k}=-1\end{array}$