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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?

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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?

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

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