Limit of type $\frac{0}{0}$: $\lim_{x\to0} \frac{\sin(x^n)-\sin^n(x)}{x^{n+2}}$ $$\lim_{x\to 0}\ \frac{\sin\left(x^n\right)-\sin^n\left(x\right)}{x^{n+2}} ~~~~~~~ \mbox{for} ~~~ n \geq 2$$
It's a multiple choice problem and the answer is $\dfrac{n}{6}$. 
I tried it for $n=2$ and i got the answer $\dfrac{2}{6}$ which fits, but i had to apply L'Hospital multiple times and it was kinda annoying. 
I wonder if this can be solved for the general case (without using Taylor) or even for $n=2$ but in a simpler way.

Write it
$$\frac{\sin (x^n) - (\sin x)^n}{x^{n+2}} = \underbrace{x^{2n-2}}_{\to 0}\underbrace{\frac{\sin (x^n) - x^n}{x^{3n}}}_{\to - \frac{1}{6}} + \frac{x^n - (\sin x)^n}{x^{n+2}},$$
and then
$$\frac{x^n - (\sin x)^n}{x^{n+2}} = \frac{x-\sin x}{x^3}\sum_{k = 0}^{n-1} \frac{x^{n-1-k}(\sin x)^k}{x^{n-1}}$$
using $a^n - b^n = (a-b)(a^{n-1} + a^{n-2}b + \dotsc + ab^{n-2} + b^{n-1})$. With $\lim\limits_{x\to 0} \frac{\sin x}{x} = 1$, the limit of the last sum is easily determined as $n$.
