If $\underset{x\to 0}{lim} \frac{x^n-\sin x^n}{x-{\sin }^{n}x}$ is non –zero finite, then n must be equal to

For n = 0, we have $\underset{x\to 0}{lim} \frac{1-\sin 1}{x-1}=\sin 1-1$

$\text{ For }\mathrm{n}=1,\underset{x\to 0}{lim} \frac{x-\sin x}{x-\sin x}=1$

$$\begin{gathered} \text{ For }\mathrm{n}=2,\underset{x\to 0}{lim} \frac{x^2-{\sin }^{2}x}{x-{\sin }^{2}x} \\ =\underset{x\to 0}{lim} \frac{1-\frac{{\sin }^{2}x}{x^2}}{\frac{1}{x}-\frac{{\sin }^{2}x}{x^2}} \end{gathered}$$

This does not exist for n=3 also given

limit does not exist. Hence, n=0 or 1