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

Question

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?

Robert Z · Accepted Answer · 2019-01-06T10:15:44.297

3

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?

edited Jan 06 '19 at 10:15

answered Jan 06 '19 at 10:07

Robert Z

145,942

2

Okay, I got exp(-2). Thanks! – Mathaddict Jan 06 '19 at 10:19
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Yes, that's it! – Robert Z Jan 06 '19 at 10:34
maybe $O(1/n^3)$? – John Joy Jan 06 '19 at 11:08
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@JohnJoy Yes, here we can replace $o(1/n^2)$ with $O(1/n^3)$, but $o(1/n^2)$ suffices. – Robert Z Jan 06 '19 at 13:31

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

1 Answers1