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Is it true that : $\lim_{ x\to\infty } \left( 1+\frac{f \left( x \right) }{x} \right) ^x = \exp \left( \lim_{ x\to\infty } f \left( x \right) \right)$ ?

Assumption is that limit of $\lim_{ x\to\infty } f(x)$ exists.

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    $\begingroup$ Yes. Proof: what can you say about $\log(1+f(x)/x)$ when $x\to\infty$? $\endgroup$ – Did Apr 10 '12 at 21:00
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    $\begingroup$ I was wondering who was that sad man who downvoted both answers... $\endgroup$ – Norbert Apr 10 '12 at 22:45
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\begin{eqnarray} \lim_{x\to\infty}\left(1+\frac{f(x)}x\right)^x &=&\lim_{x\to\infty}\exp \left(x\log \left(1+\frac{f(x)}x \right) \right)=\exp \left(\lim_{x\to\infty}x\log \left(1+\frac{f(x)}x\right)\right)\\ \ \\ &=&\exp \left(\lim_{x\to\infty}x \left(\frac{f(x)}x+o\left(\frac1{x^2}\right)\right)\right) =\exp \left(\lim_{x\to\infty}f(x)+o\left(\frac1{x}\right)\right)\\ \ \\ &=&\exp \left(\lim_{x\to\infty}f(x)\right). \end{eqnarray}

The limit exchange in the second equality is justified by the fact that the exponential is continuous and that the expression inside converges.

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  • $\begingroup$ Not sure why this was downvoted. This is essentially correct, but I guess some people don't like seeing the second equality before seeing a proof of the convergence of the expression inside. $\endgroup$ – Aryabhata Apr 11 '12 at 5:27
  • $\begingroup$ Could be. Actually, it's probably much better to address the limit inside first. What I don't like is when people downvote without saying why. $\endgroup$ – Martin Argerami Apr 11 '12 at 5:40
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    $\begingroup$ Can't help it. People aren't obligated to explain their upvotes or downvotes. In this case, the person downvoting probably lost an opportunity to learn something from your answer. $\endgroup$ – Aryabhata Apr 11 '12 at 5:45
  • $\begingroup$ One more possible reason for the downvote: You haven't mentioned that $\frac{f(x)}{x} \to 0$, in order to justify the third equality. $\endgroup$ – Aryabhata Apr 11 '12 at 7:21
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Since there exist $\lim\limits_{x\to 0}f(x)=l$ we have that $f$ is bounded in the neighborhood of $\infty$. So $\lim\limits_{x\to\infty}\frac{f(x)}{x}=0$ and we can write $$ \lim\limits_{x\to\infty}\left(1+\frac{f(x)}{x}\right)^x= \lim\limits_{x\to\infty}\left(\left(1+\frac{f(x)}{x}\right)^{\frac{x}{f(x)}}\right)^{f(x)}= \lim\limits_{x\to\infty}\left(\left(1+\frac{f(x)}{x}\right)^{\frac{x}{f(x)}}\right)^{\lim\limits_{x\to\infty}f(x)}= e^l $$

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  • $\begingroup$ Is the second limit in second equality justified by the fact that (1+f(x)/x)^(x/f(x)) converges ? $\endgroup$ – Qbik Apr 10 '12 at 21:16
  • $\begingroup$ I didn't downvote this, but perhaps this was downvoted because you don't deal with the case: $f(x) = 0$ infinitely often? $\endgroup$ – Aryabhata Apr 11 '12 at 5:22

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