$\lim_{n\to\infty} \left(1+\frac{2}{n}\right)^{n^2} e^{-2n}$ 
$$\lim_{n\to\infty} (1+\frac{2}{n})^{n^2} e^{-2n} = ? $$
  
  
*
  
*$e$
  
*$e^2$
  
*$e^{-1}$
  
*$e^{-2}$
  

My answer is 1. Since $$\lim_{n \to \infty} \left(1+\frac{2}{n}\right)^{n} =e^{+2}.$$
Therefore $$\lim_{n\to\infty} \left(1+\frac{2}{n}\right)^{n^2} =e^{2n}.$$
Where am I wrong?
 A: The fact that $(1+\frac{2}{n})^n\to e^2$ as $n\to\infty$ does not mean that $(1+\frac{2}{n})^{n^2}$ acts like $e^{2n}$ for large $n$, although it is easy to see why you might think so!
Exponents have huge effects on behavior; they can take even the smallest deltas and make them explode.
So, instead, let's consider logarithms:
$$
\ln\left[\left(1+\frac{2}{n}\right)^{n^2}e^{-2n}\right]=n^2\ln\left(1+\frac{2}{n}\right)-2n.
$$
There are a few ways to go from here.  One way: Taylor series. You can show that $\ln\left(1+\frac{2}{n}\right)=\frac{2}{n}-\frac{2}{n^2}+O(\frac{1}{n^3})$ for $n$ large, which implies that
$$
n^2\ln\left(1+\frac{2}{n}\right)-2n=2n-2-2n+O\left(\frac{1}{n}\right)=-2+O(\frac{1}{n}),
$$
so that the logarithm of your value approaches $-2$.  Thus, the answer should be $e^{-2}$.
If you aren't comfortable with Taylor series, you could also rewrite this as
$$
\frac{\ln\left(1+\frac{2}{n}\right)-\frac{2}{n}}{\frac{1}{n^2}}
$$
which is a $\frac{0}{0}$ indeterminate form, and use L'Hopital's rule.
A: Note that
$$\left(1+\frac{2}{n}\right)^{n^2}=e^{n^2\log (1+\frac{2}{n} )}=e^{n^2\left(\frac{2}{n}-\frac{2}{n^2}+o(1/n^2)\right)}=e^{2n-2+o(1)}$$
then
$$\left(1+\frac{2}{n}\right)^{n^2} e^{-2n}=e^{-2+o(1)}\to e^{-2}$$
