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?

$$ \lim\limits_{x\to \infty  } \lim\limits_{n\to \infty} \sum\limits_{r=1}^n \frac{r^2(\sin x)^x}{n^3}$$
$$ =\lim\limits_{x\to \infty  } \lim\limits_{n\to \infty} \frac{(\sin x)^x}{n^3}\sum\limits_{r=1}^n r^2$$
$$ =\lim\limits_{x\to \infty  } \lim\limits_{n\to \infty} \frac{(\sin x)^x}{n^3}\left(\frac{n(n+1)(2n+1)}{6}\right)$$
$$ =\frac16\left(\lim\limits_{x\to \infty} (\sin x)^x\left(\lim\limits_{n\to \infty} \frac{2n^3+3n^2+n}{n^3}\right)\right)$$
$$ =\frac16\left(\lim\limits_{x\to \infty} (\sin x)^x\left(\lim\limits_{n\to \infty} \left(2+\frac3n+\frac1{n^2}\right)\right)\right)$$
$$ =\frac13\lim\limits_{x\to \infty} (\sin x)^x=\mbox{non existent}$$
However
$$\frac13\lim\limits_{x\to 0} (\sin x)^x=\frac13$$
Perhaps this is what you intended to type.
