A typical limit

First of all note that the denominator can be replaced with just $\large x^{n+2}$ because $\displaystyle \lim_{x\to 0} \frac{\sin(x)}{x} =1$ .

We shall use the taylor series expansion for $\displaystyle \sin(x) = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} +.......$

So we have to evaluate:-

$\displaystyle \lim_{x\to 0} \frac{x^{n} - \left(x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} +.......\right)^{n}}{x^{n+2}}$

Taking $\large x^{n}$ common from numerator and denominator we have:-

$\displaystyle \lim_{x\to 0}\frac{1 - \left(1 - \frac{x^{2}}{3!} + \frac{x^{4}}{5!} +.......\right)^{n}}{x^{2}}$

Now expanding $\left(1 - \frac{x^{2}}{3!} + \frac{x^{4}}{5!} +.......\right)^{n}$ binomially we have :-

$\displaystyle \lim_{x\to 0}\frac{1 - 1 +n\left(\frac{x^{2}}{3!}+\frac{x^{4}}{5!}+....\right) + O(x^4)}{x^{2}}$

So we have :-

$\displaystyle \frac{n}{3!} = 369$

So $\displaystyle n=2214$

Here $O(.)$ denotes the big O notation .