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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.

  • 1
    I was wondering who was that sad man who downvoted both answers...2012-04-10

2 Answers 2

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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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    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.2012-04-11
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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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    I didn't downvote this, but perhaps this was downvoted because you don't deal with the case: $f(x) = 0$ infinitely often?2012-04-11