value of $n$ in limits 
If $\displaystyle \lim_{x\rightarrow 0}\frac{x^n\sin^{n}(x)}{x^{n}-(\sin x)^{n}}$ is a finite non zero number.
Then value of $n$ is equals

What I try:
$$\sin x = x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots $$
$$\lim_{x\rightarrow 0}\frac{x^n\bigg(x-\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots \bigg)^n}{x^{n}-\bigg(x-\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots \bigg)^n}$$
$$\lim_{x\rightarrow 0}\frac{x^{n+1}\bigg(1-\frac{x^2}{3!}+\frac{x^4}{5!}+\cdots \bigg)^n}{x^{n}-\bigg(x-\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots \bigg)^n}$$
How do I solve it? Help me please.

For each $n\in\mathbb N$, the first non-null term of the Taylor series of $x^n\sin^n(x)$ centered at $0$ is $x^{2n}$, whereas the first non-null term of the Taylor series of $x^n-\sin^n(x)$ centered at $0$ is $kx^{n+2}$, for some $k\neq0$. So, your limit will be finite and non zero if and only if $2n=n+2$. And this occurs if and only if $n=2$.
