Solve $\lim\limits _{n\to \infty }\lim\limits_{x\to \:0}\left(1+\sin^2x+\sin^22x+\ldots+\sin^2nx\right)^{1/(n^3x^2)}$ 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?$$\lim_{n\to \infty }\left[\lim_{x\to \:0}\left(1+\sin^2x+\sin^22x+\ldots+\sin^2nx\right)^{1/(n^3x^2)}\right]$$
 A: 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 $e^{1/3}$.
A: $$
\begin{aligned}
&\lim_{x\rightarrow 0} [1 + \sin^2 x + \sin^2(2x) + \cdots + \sin^2(nx)]^{1/(n^3 x^2)} \\
&=
\lim_{x\rightarrow 0} [1 + x^2 + (2x)^2 + \cdots + (nx)^2]^{1/(n^3 x^2)} \\
&=
\lim_{x\rightarrow 0} \left[1 + \frac{n(n+1)(2n+1)}{6} x^2\right]^{1/(n^3 x^2)} \\
&=
\exp\left[\frac{(n+1)(2n+1)}{6 n^2} \right]. \\
\end{aligned}
$$
Then, taking the limit of $n\rightarrow\infty$ yields $e^{1/3}$.
A: 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 
$$\lim_{n\to \infty}\lim_{x\to 0} \left(1+\sum_{k=1}^n \sin ^2(kx)\right)^{1/n^3x^2}=e^{1/3}$$
