find $\lim_{n \to \infty} \frac{(n^2+1)(n^2+2) \cdots (n^2+n)}{(n^2-1)(n^2-2) \cdots (n^2-n)}$ 
Find $$\lim_{n \to \infty} \frac{(n^2+1)(n^2+2) \cdots (n^2+n)}{(n^2-1)(n^2-2) \cdots (n^2-n)}$$

I tried to apply the squeeze theorem, yet none of my attempts led me to the solution.

The squeeze theorem fails because the limit is not $1$ which in turn happens because the number of factors is unbounded when $n\to\infty$. This suggests that more precise asymptotics might be needed. To get these, consider that the logarithm of the $n$th ratio is
$$
S_n=\sum_{k=1}^n\log\left(1+\frac{k}{n^2}\right)-\log\left(1-\frac{k}{n^2}\right).
$$
As everybody knows, when $x\to0$, $\log(1+x)=x-\frac12x^2+o(x^2)$. Choose $n$ large enough such that, for every $|x|\leqslant\frac1n$,
$$
x-x^2\leqslant\log(1+x)\leqslant x.
$$
(It happens that $n=2$ suffices, but this is irrelevant.) Then, applying this double inequality to each $\log\left(1\pm\frac{k}{n^2}\right)$ in $S_n$, one gets
$$
\sum_{k=1}^n2\frac{k}{n^2}-\frac{k^2}{n^4}\leqslant S_n\leqslant\sum_{k=1}^n2\frac{k}{n^2}+\frac{k^2}{n^4},
$$
that is, using the exact value $\sum\limits_{k=1}^nk=\frac12n(n+1)$ and the crude estimate $\sum\limits_{k=1}^nk^2\leqslant n^3$,
$$
\frac{n(n+1)}{n^2}-\frac1{n^4}n^3\leqslant S_n\leqslant\frac{n(n+1)}{n^2}+\frac1{n^4}n^3.
$$
In other words,
$$
1\leqslant S_n\leqslant1+\frac2n,
$$
from which the limit of $\mathrm e^{S_n}$ should be easy to write down.
