f(x)= [(e^kx -1) (sin kx ) ] / 4x²

x is nt equal to 0

f(0)=9

x=0value of k

the given function is:
$$\begin{gathered} f\left(x\right)=\frac{(e^{kx}-1)\left(\sin kx\right)}{4x^2} for x\ne 0 and \\ =9 for x=0 \end{gathered}$$therefore
$$\begin{gathered} \underset{x\to 0}{lim}\frac{(e^{kx}-1)\left(\sin kx\right)}{4x^2}=9 if the function is continuous at x=0 \\ \underset{h\to 0}{lim}\frac{(e^{kh}-1)\left(\sin kh\right)}{4h^2}=9 \\ \underset{h\to 0}{lim} \frac{e^{kh}-1}{kh}.\underset{h\to 0}{lim}\frac{\sin kh}{kh}.\frac{k^2}{4}=9 \\ \Rightarrow 1.1.\frac{k^2}{4}=9 \\ \Rightarrow k^2=36 \\ k=\pm 6 \end{gathered}$$
the relation is satisfied only for k=6

thus k =6

hope this helps you