Trouble with finding the limit of this sequence Well I was trying to find the limit of - 
$$ \lim_{x\rightarrow \infty  } \lim_{n\rightarrow \infty} \sum_{r=1}^{n} \frac{\left [ r^2(\sin x)^x \right ]}{n^3}$$
obviously $$  \sum_{r=1}^{n}\frac{r^2(\sin x)^x) -1}{n^3}< \sum_{r=1}^{n} \frac{\left [ r^2(\sin x)^x \right ]}{n^3} \leq  \sum_{r=1}^{n}\frac{r^2(\sin x)^x}{n^3} $$
summing them up and applying squeeze theorem gives-
$$ \lim_{x\rightarrow \infty  } \lim_{n\rightarrow \infty} \sum_{r=1}^{n} \frac{\left [ r^2(\sin x)^x \right ]}{n^3} = \frac{\sin x^x}{3}$$
now since $-1<\sin x<1$ so $\lim_{x\rightarrow \infty } \sin x^x = 0$ so the limit should be , according to me , $0$ but the answer is $1/3$.
Where did I go wrong?

$$\lim_{n\to +\infty}\sum_{r=1}^{n}\frac{r^2}{n^3}=\lim_{n\to +\infty}\frac{1}{n}\sum_{r=1}^{n}\left(\frac{r}{n}\right)^2 = \int_{0}^{1}u^2\,du = \frac{1}{3}$$
by a Riemann sum argument, but neither $\lim_{x\to +\infty}(\sin x)^x$ or $\lim_{x\to +\infty}|\sin x\,|^{x}$ exist, as pointed in the comments.
