Calculus: calculate the limit $ (1+\frac 1 {n^2})^{n^3}$ I am doing my homework in calculus and I got this problem that I got to find the limit for  $(1+1/n^2)^{n^3}$ as n goes to infinity.
I can rewrite the expression as $((1+1/n^2)^{n^2})^n$ but from what I know I can't take the limit inside the () before I take the limit of the outside exponent (in which case I would get e^n which goes to infinity).
Can anyone explain to me how I can calculate the limit rigorously? 
 A: Hints: there exists $\;N\in\Bbb N\;$ such that
$$n>N\implies 2\le\left(1+\frac1{n^2}\right)^{n^2}\le3\implies2^n\le\left(1+\frac1{n^2}\right)^{n^3}\le3^n$$
and now apply the squeeze theorem...
A: HINT:
Successive applications of Bernoulli's inequality, $(1+x)^n\ge 1+nx$, reveals
$$\left(1+\frac1{n^2}\right)^{n^3}\ge \left(1+\frac{n}{n^2}\right)^{n^2}\ge \left(1+\frac{n^2}{n^2}\right)^n=2^n$$
A: Since$$\lim_{n\to\infty}\left(1+\frac1{n^2}\right)^{n^2}=e,$$we have$$\lim_{n\to\infty}\left(\left(1+\frac1{n^2}\right)^{n^2}\right)^n=+\infty$$and therefore your limit is $+\infty$.
A: \begin{align*}
\lim_{n\rightarrow\infty}\ln(1+1/n^{2})^{n^{3}}&=\lim_{n\rightarrow\infty}n^{3}\ln(1+1/n^{2})\\
&=\lim_{n\rightarrow\infty}\dfrac{\ln\left(1+\dfrac{1}{n^{2}}\right)}{\dfrac{1}{n^{3}}}\\
&=\lim_{n\rightarrow\infty}\dfrac{1}{1+\dfrac{1}{n^{2}}}\dfrac{-2}{n^{3}}\dfrac{1}{\dfrac{-3}{n^{4}}}\\
&=\infty,
\end{align*}
so
\begin{align*}
\lim_{n\rightarrow\infty}\left(1+\dfrac{1}{n^{2}}\right)^{n^{3}}=\lim_{n\rightarrow\infty}\exp\left(\ln\left(1+\dfrac{1}{n^{2}}\right)^{n^{3}}\right)=\infty.
\end{align*}
