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This question already has an answer here:

The limit I have to evaluate is this - $$\lim_{x \to \infty} \frac{e^x}{\left(1+\frac{1}{x}\right)^{x^2}}$$

I first checked if L'Hopital's rule applies here. The limit of both numerator and denominator is $\infty$. But differentiating the denominator yield a even more complicated expression.

I am not getting how to approach this question using some other method. Thank you.

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marked as duplicate by YuiTo Cheng, Thomas Shelby, cmk, José Carlos Santos limits Jun 29 at 17:15

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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let $$\begin{align}y &= \frac{e^x}{\left(1 + \frac 1 x\right)^{x^2}},\\\ \ln y &= x - x^2 \ln \left(1 + \frac 1x\right) \\ &= x - x^2\left(\frac 1x - \frac1 {2x^2} + \frac 1 {3x^3 }+\cdots\right)\\ &=\frac 12 - \frac 1 {3x} + \frac 1 {4x^2}+\cdots \rightarrow \frac 12 \end{align}$$

therefore $$\lim_{x \to \infty}\frac {e^x} {\left(1 + \frac 1 x\right)^{x^2}}= \sqrt e.$$

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  • $\begingroup$ Except that it's $\sqrt e~$. $\endgroup$ – Lucian Mar 21 '15 at 15:08
  • $\begingroup$ nice work @Lucian. $\endgroup$ – abel Mar 21 '15 at 15:09
  • $\begingroup$ There's a $\frac{1}{2}$ in the second term of expansion and $\frac{1}{3}$ in the third term and so on. Thanks! $\endgroup$ – Arpan Mar 21 '15 at 15:20
  • $\begingroup$ @ArpanBanerjee, thanks man. it is too early in the morning for maclaurin expansion. $\endgroup$ – abel Mar 21 '15 at 15:24
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    $\begingroup$ Once more, you win by a couple of minutes ! No mercy for the old man ! Cheers. $\endgroup$ – Claude Leibovici Mar 21 '15 at 15:26
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Hint

Define $$A= \frac{e^x}{\left(1+\frac{1}{x}\right)^{x^2}}$$ and take logarithms of both sides; so $$\log(A)=x -x^2\log(1+\frac 1 x)$$ Now, using the fact that, by Taylor, for small $y$, $\log(1+y)=y-\frac{y^2}{2}+\frac{y^3}{3}+O\left(y^4\right)$, replace $y$ by $\frac 1x$ to get $$\log(A)=x-x^2\Big(\frac{1}{x}-\frac{1}{2 x^2}+\frac{1}{3 x^3}+\cdots\Big)=\frac{1}{2}-\frac{1}{3 x}+\cdots$$

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  • $\begingroup$ Just a doubt, what does the $O$ mean? $\endgroup$ – Arpan Mar 21 '15 at 15:26
  • $\begingroup$ If you did not learn it yet, it is called the big $O$ notation which describes the limiting behavior of a function when the argument tends towards a particular value. In simpler words, for you case, say that it collects the other terms which are less and less significant. Don't worry, you will hear soon about it and you will have plenty of assignments with that. Cheers. $\endgroup$ – Claude Leibovici Mar 21 '15 at 15:32

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