# Find the limit of $\left(1+ \frac{2}{n}\right)^{n^{2}} \exp(-2n)$ as $n \to \infty$.

Find the limit of $$\left(1+ \frac{2}{n}\right)^{n^{2}} \exp(-2n)$$ as $$n \to \infty$$.

By expansion - $$\lim\limits_{n \to \infty} \left[1+(n^{2})(2/n) + (n^{2})(n^{2}-1)/2 \dots ]/[1+2n+(2n)^{3}/3! \dots\right]$$

I didn't get any result. By applying limit directly, I'm getting indeterminate form. How to find this limit?

Hint. Note that $$\left(1+ \frac{2}{n}\right)^{n^{2}} \exp(-2n)=\exp\left(n^2\log\left(1+ \frac{2}{n}\right)-2n\right).$$ Now, by using the expansion $$\log(1+t)=t-\frac{t^2}{2}+o(t^2)$$ at $$t=0$$, we have that $$\log\left(1+ \frac{2}{n}\right)=\frac{2}{n}-\frac{2}{n^2}+o(1/n^2).$$ Can you take it from here?

• Okay, I got exp(-2). Thanks! – Mathsaddict Jan 6 at 10:19
• Yes, that's it! – Robert Z Jan 6 at 10:34
• maybe $O(1/n^3)$? – John Joy Jan 6 at 11:08
• @JohnJoy Yes, here we can replace $o(1/n^2)$ with $O(1/n^3)$, but $o(1/n^2)$ suffices. – Robert Z Jan 6 at 13:31