If \(\lim _{x \rightarrow 0} \frac{\left(e^{k x}-1\right) \sin k x}{x^{2}}=4\), then \(k\) is equal to

\(\begin{aligned} & \lim _{x \rightarrow 0} \frac{\left(e^{k x}-1\right) \sin k x}{x^{2}}=4 \\ & \Rightarrow \quad \lim _{x \rightarrow 0}\left[\frac{e^{k x} \sin k x}{x^{2}}-\frac{\sin k x}{x^{2}}\right]=4 \\ & \Rightarrow \lim _{x \rightarrow 0}\left[\frac{e^{k x} \cos k x \cdot k+k e^{k x} \sin k x}{2 x}-\frac{k \cos k x}{2 x}\right]=4 \\ & \Rightarrow \lim _{x \rightarrow 0}\left[\frac{k^{2} e^{k x} \cos k x-k^{2} e^{k x} \sin k x+k^{2} e^{k x} \sin k x}{+k^{2} e^{k x} \cos k x}{2}\right.\\ & \Rightarrow \frac{k^{2}-0+0+k^{2}}{2}+0=4 \\ & \Rightarrow \quad k^{2}=4 \\ & \Rightarrow \quad k=\pm 2 \end{aligned}\)