<s>[INST] 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.
 [/INST] $$L=\lim_{x \rightarrow 0} \frac{x^n \sin^{n} x}{x^n-\sin^n x}$$ $$= \lim_{x \rightarrow 0} \frac{x^{2n}(1-x^2/6)^n}{x^{n}-x^n(1-x^2/6)^n}$$
$$=\lim_{x \rightarrow 0} \frac{x^n (1-nx^2/6)}{nx^2/6}=\lim_{x \rightarrow 0}\frac {6x^{n-2}}{n}= 3,~ if~ n =2,$$
$$= 0, ~if~ n > 2, $$ $$=\infty ~~if~~n= n <2 $$
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