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$$\lim_{x\to 0}\ \frac{\sin\left(x^n\right)-\sin^n\left(x\right)}{x^{n+2}} ~~~~~~~ \mbox{for} ~~~ n \geq 2$$

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.

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    Taylor expansion should do the trick. Just don't try to compute the full Taylor expansion of $\sin^nx$! – 2017-01-15
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    @Wojowu The only problem is that i don't know anything about that... Here, it is only learnt at university and I'm in high school. – 2017-01-15
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    Ah, I see; I didn't know you are in a high school. – 2017-01-15
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    I'm in high school... – 2017-01-15
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    @Livua Have you learnt induction? Once you've done $n=2$ the rest could follow from there. – 2017-01-15
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    :-( Why can't I use Taylor, even if you can't understand it, its still a good method. – 2017-01-15
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    @ianMiller I can say the answer is $\frac{n}{6}$ even after the result for $n = 2$ because it's a multiple choise question, but i wondered how it can be solver in another way. – 2017-01-15
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    @SimpleArt It could be, but if i don't fully understand it, i can't apply it in another situations ... – 2017-01-15
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    Do you know (in the sense that you are allowed to use it) that $$\lim_{x\to 0} \frac{x-\sin x}{x^3} = \frac{1}{6}\,?$$ – 2017-01-15
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    @DanielFischer Yes – 2017-01-15
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    Then write $$\frac{\sin (x^n) - (\sin x)^n}{x^{n+2}} = \frac{\sin (x^n) - x^n}{x^{n+2}} + \frac{x^n - (\sin x)^n}{x^{n+2}}.$$ The first tends to $0$ because $n+2 < 3n$. For the second, use $a^n - b^n = (a-b)(a^{n-1} + a^{n-2}b + \dotsc + ab^{n-2} + b^{n-1})$. – 2017-01-15
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    @DanielFischer Should post as a full answer :-) – 2017-01-15

2 Answers 2

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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 $n$.

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Taylor expansions:

$$\sin(x)=x-\frac16x^3+\mathcal O(x^5)$$

$$\begin{align}\lim_{x\to0}\frac{\sin(x^n)-\sin^n(x)}{x^{n+2}}&=\lim_{x\to0}\frac{x^n-\frac16x^{3n}-x^n+\frac n6x^{n+2}+\mathcal O(x^{n+4})}{x^{n+2}}\\&=\lim_{x\to0}\frac n6-\frac16x^{2n-2}+\mathcal O(x^2)\\&=\frac n6\end{align}$$