The value of a for which  $\frac{\sin 2x+a\sin x}{x^3}$ tends to a finite limit as $x\to 0$ is.

Since numerator as well as denominator both tends to zero as  $x\to 0$ so applying L' hospital Rule, we have

$$\begin{gathered} \underset{\mathrm{x}\to 0}{lim} \frac{\sin 2\mathrm{x}+\mathrm{asin}\mathrm{x}}{{\mathrm{x}}^3} \\ =\underset{\mathrm{x}\to 0}{lim} \frac{2\cos 2\mathrm{x}+\mathrm{acos}\mathrm{x}}{3{\mathrm{x}}^2} \end{gathered}$$

But the numerator tends to $2+a\text{ as }x\to 0\text{ and }{x}^2\to 0$ as $x\to 0$ So the last limit can be finite only if $2+a=0\text{, i.e., }$ $a=-2$ In this case the last limit will be equal to

$$\begin{gathered} -\underset{x\to 0}{lim} \frac{4\sin 2x+a\sin x}{6x} \\ =-\left(\frac{4}{3}+\frac{a}{6}\right) \\ =-1 \end{gathered}$$