$ \lim_{n->\infty} {(1+2/n)^{n^2} e^{-2n}} $
Answer is not $\ e^0$ . OR atleast it was not in the options.
My approach:
$ \lim_{n->\infty} {(1+2/n)^{n^2} .e^{-2n}} $ = $ \lim_{n->\infty} {(1+2/n)^{n^2}}/{e^{2n}} $
=$ \lim_{n->\infty} {(1+2/n)^{n^2}}/{(e^2)^{n}} $
= $ \lim_{n->\infty} {(1+2/n)^{n^2}}/{{((1+2/n)^{n})}^n} $
=$ \lim_{n->\infty} {e^{2{n}}}/{{e}^{2n}} $
=$ e^0 $
But this was not in the options. But My method seems to be pretty correct as well.
Note that
$$ (1+2/n)^{n^2}=e^{n^2\log(1+2/n)}=e^{n^2(2/n-2/n^2+o(1/n^2))}=e^{2n-2+o(1)}$$
thus
$$ (1+2/n)^{n^2} e^{-2n}=e^{-2+o(1)}\to \frac1{e^2}$$