# $\lim_{n \rightarrow \infty}\left(1+\dfrac{2}{n}\right)^{n^2}e^{-2n}$ [duplicate]

$$\lim_{n \rightarrow \infty}\left(1+\dfrac{2}{n}\right)^{n^2}e^{-2n}$$

$$\lim_{n \rightarrow \infty}\left(e^{-2}\left(1+\dfrac{2}{n}\right)^n\right)^{n}$$

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 seemingly had a typo in the title which I corrected. Let me know if it is not what you wanted. Feb 12, 2021 at 14:13

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 :

$$\lim_{n \rightarrow \infty}\left(1+\dfrac{2}{n}\right)^{n^2}e^{-2n}=e^{-2}$$

$$\log(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \ldots .$$
Then, it's clear that (for $$x = \frac{2}{n}$$)
$$\log\left(1 + \frac{2}{n}\right) = \frac{2}{n} - \frac{4}{2n^2} + \ldots .$$