$\lim_{n \rightarrow \infty}\left(1+\dfrac{2}{n}\right)^{n^2}e^{-2n}$
$\lim_{n \rightarrow \infty}\left(e^{-2}\left(1+\dfrac{2}{n}\right)^n\right)^{n}$
It is indeterminate form $1^{\infty}$
I solve this like this $e^{\lim_{n \rightarrow \infty}}\frac{\ln(e^{-2}(1+\frac{2}{n})^n)}{\frac{1}{n}}$
I can't apply L'Hôpital's rule now how can I solve this problem quickly?
You have :
$$\left(1+\dfrac{2}{n}\right)^{n^2}=\exp\left(n^2\log\left(1+\frac{2}{n}\right)\right)=\exp\left(n^2\left(\frac{2}{n}-\frac{4}{2n^2}+o\left(\frac{1}{n^2}\right)\right)\right)\\=\exp\left(2n-2+o\left(1\right)\right)$$
So you get :
$$\left(1+\dfrac{2}{n}\right)^{n^2}e^{-2n}=e^{-2+o(1)} $$
So you get the final result :
$$\lim_{n \rightarrow \infty}\left(1+\dfrac{2}{n}\right)^{n^2}e^{-2n}=e^{-2} $$
Just a brief explanation about the answer above:
Observe that
$$\log(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \ldots .$$
Then, it's clear that (for $x = \frac{2}{n}$)
$$\log\left(1 + \frac{2}{n}\right) = \frac{2}{n} - \frac{4}{2n^2} + \ldots .$$
