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

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

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?

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

2 Answers 2

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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 :

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

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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 $x = \frac{2}{n}$)

$$\log\left(1 + \frac{2}{n}\right) = \frac{2}{n} - \frac{4}{2n^2} + \ldots .$$

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    $\begingroup$ You could have added this as a comment to the other answer rather than posting it as a separate one. $\endgroup$
    – Gary
    Feb 12, 2021 at 14:34
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    $\begingroup$ I think it's better, because in chat we can't see the preview of answer. $\endgroup$ Feb 12, 2021 at 14:35

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