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I've no clue how to solve this limit. I tried to solve the inner limit first, but the fact that I have another variable there(n) really makes things difficult(never had to solve such a limit before). How do you go about solving these types of limits?$$\lim_{n\to \infty }\left[\lim_{x\to \:0}\left(1+\sin^2x+\sin^22x+\ldots+\sin^2nx\right)^{1/(n^3x^2)}\right]$$

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    $\begingroup$ People are giving answers here too quickly. This would have been a good one for some well chosen hints. $\endgroup$ – zhw. Nov 25 '15 at 18:49
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Since $\displaystyle\sin^2 x=\frac{1-\cos 2x}{2}$, we have (easily verified by using $2\cos x=e^{ix}+e^{-ix}$): $$ 1+\sum_{i=1}^n\sin(i x)^2=1+\frac{n}{2}-\sum_{i=1}^n\cos(2kx)=1+\frac{n}{2}-\frac{\cos((n+1)x)\sin(n x)}{2\sin x} $$ Thus we have: $$ \lim_{n\to\infty}\lim_{x\to 0}\left(1+\frac{n}{2}-\frac{\cos((n+1)x)\sin(n x)}{2\sin x}\right)^{\frac{1}{n^3x^2}} $$ We will calculate limit of the logarithm of the sum, as it is strictly positive, then use taylor series expansion around $0$: $$ \lim_{n\to\infty}\lim_{x\to 0}\frac{\displaystyle\ln\left(1+\frac{n}{2}-\frac{\cos((n+1)x)\sin(n x)}{2\sin x}\right)}{n^3x^2}=$$ $$=\lim_{n\to\infty}\lim_{x\to 0}\left(\frac{2 n^2+3 n+1}{6 n^2}+\mathcal{O}(x^2)\right)=\lim_{n\to\infty}\frac{2 n^2+3 n+1}{6 n^2}=\frac{1}{3} $$ Thus the original limit is $e^{1/3}$.

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$$ \begin{aligned} &\lim_{x\rightarrow 0} [1 + \sin^2 x + \sin^2(2x) + \cdots + \sin^2(nx)]^{1/(n^3 x^2)} \\ &= \lim_{x\rightarrow 0} [1 + x^2 + (2x)^2 + \cdots + (nx)^2]^{1/(n^3 x^2)} \\ &= \lim_{x\rightarrow 0} \left[1 + \frac{n(n+1)(2n+1)}{6} x^2\right]^{1/(n^3 x^2)} \\ &= \exp\left[\frac{(n+1)(2n+1)}{6 n^2} \right]. \\ \end{aligned} $$ Then, taking the limit of $n\rightarrow\infty$ yields $e^{1/3}$.

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    $\begingroup$ How do you just remove the $sin$? $\endgroup$ – MikhaelM Nov 25 '15 at 18:44
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    $\begingroup$ since $\sin x$'s taylor series expansion around $0$ is $x-\frac{x^3}{6}+\mathcal{O}(x^5)$, thus $\sin x\approx x$ for small $x$ $\endgroup$ – Ákos Somogyi Nov 25 '15 at 18:46
  • $\begingroup$ @ÁkosSomogyi. Thanks for answering this for me. This step does require some explanation. Therefore I like you answer better. $\endgroup$ – hbp Nov 25 '15 at 18:50
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Heuristically, $\sin x\sim x$ for small $x$.

So,

$$\sum_{k=1}^n \sin ^2(kx)\sim\sum_{k=1}^n k^2x^2=\frac{n(n+1)(2n+1)x^2}{6}$$

and

$$\log \left(1+\sum_{k=1}^n \sin ^2(kx)\right)\sim \frac{n(n+1)(2n+1)x^2}{6}$$

Therefore,

$$\lim_{n\to \infty}\lim_{x\to 0}\frac{\log \left(1+\sum_{k=1}^n \sin ^2(kx)\right)}{n^3x^2}=1/3$$

and we conclude that

$$\lim_{n\to \infty}\lim_{x\to 0} \left(1+\sum_{k=1}^n \sin ^2(kx)\right)^{1/n^3x^2}=e^{1/3}$$

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