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.

Assuming $n\neq 0$ (to make the expression under limit meaningful) the given expression can be written as $$\dfrac{\dfrac{x} {\sin x} - 1}{\left(\dfrac{x}{\sin x}\right)^n-1}\cdot\frac{x^3}{x-\sin x} \cdot x^{n-2}$$ The first factor tend to $1/n$ and second one tends to $6$ and hence the limiting behavior of the expression crucially depends on that of $x^{n-2}$. The desired limit is thus $6/n=3$ if $n=2$, is $0$ if $n>2$ and does not exist if $n<2$.
