If \(f(x)=\left\{\begin{array}{cl}\frac{1-\cos K x}{x \sin x}, & \text { if } x \neq 0 \\ \frac{1}{2}, & \text { if } x=0\end{array}\right.\) is continuous at \(x=0\), then the value of \(K\) is

We have,
\(f(x)=\left\{\begin{array}{cc}
\frac{1-\cos k x}{x \sin x} & , x \neq 0 \\
\frac{1}{2} & , x=0
\end{array}\right.\)
\(f(x)\) is continuous at \(x=0\)
\(\begin{aligned}
&\therefore \quad \lim _{x \rightarrow 0} \frac{1-\cos k x}{x \sin x}=\frac{1}{2} \\
&\Rightarrow \quad \lim _{x \rightarrow 0} \frac{2 \sin ^{2} \frac{k}{2} x}{x \sin x}=\frac{1}{2} \\
&\Rightarrow \lim _{x \rightarrow 0} 2\left(\frac{\sin \frac{k x}{2}}{\frac{k x}{2}}\right)^{2} \times \lim _{x \rightarrow 0} \frac{1}{\sin x} \times \frac{k^{2}}{4}=\frac{1}{2} \\
&\Rightarrow \quad 2 \times \frac{k^{2}}{4}=\frac{1}{2}
\end{aligned}\)
\(\Rightarrow \quad \begin{aligned}
k^{2} &=1 \\
k &=\pm 1
\end{aligned}\)