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I want to prove that $$\lim_{n\to\infty} \left(1+\frac{1}{f(n)}\right)^{g(n)} = 1$$ if $f(n)$ grows faster than $g(n)$ for $n\to\infty$ and $\lim_{n\to\infty} f(n) = +\infty = \lim_{n\to\infty}g(n)$.

It is quite easy to see that if $f = g$ the limit is $e$, but I can't find a good strategy to solve this problem.

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  • $\begingroup$ What is $\lim_{n\to\infty}f(n)$ and $\lim_{n\to\infty}g(n)$? $\endgroup$ – Thomas Shelby Dec 1 '18 at 18:24
  • $\begingroup$ @ThomasShelby they're both $+\infty$; in a special case, I need $n^n$ and $n!$, but a more general case is more interesting. $\endgroup$ – Riccardo Cazzin Dec 1 '18 at 18:26
  • $\begingroup$ Yeah, if $f(n)\to\infty$ and $g(n)/f(n)\to 1,$ your result is true. $\endgroup$ – Thomas Andrews Dec 1 '18 at 18:26
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We can use that $$ \left(1+\frac{1}{f(n)}\right)^{g(n)} =\left[\left(1+\frac{1}{f(n)}\right)^{f(n)}\right]^{\frac{g(n)}{f(n)}}$$

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That depends upon how you defined to grow faster than. But if implies that $\lim_{n\to\infty}\frac{g(n)}{f(n)}=0$,then\begin{align}\lim_{n\to\infty}\left(1+\frac1{f(n)}\right)^{g(n)}&=\lim_{n\to\infty}\left(\left(1+\frac1{f(n)}\right)^{f(n)}\right)^{\frac{g(n)}{f(n)}}\\&=e^0\\&=1.\end{align}

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    $\begingroup$ The second-to-last step does require a bit of justification though, since $\lim_n{a_n}^{b_n} = (\lim_n a_n)^{\lim_n b_n}$ is not a trivial fact (and is false in general, for instance if limits are in $[0,\infty]$). $\endgroup$ – Clement C. Dec 1 '18 at 18:37
  • $\begingroup$ @ClementC. Yes, I agree that one should be careful here. $\endgroup$ – José Carlos Santos Dec 1 '18 at 18:48
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    $\begingroup$ I like how the comment above says that one of the steps need more explanation, but the accepted answer is just the result with zero explanation. $\endgroup$ – user1717828 Dec 2 '18 at 1:37
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    $\begingroup$ @user1717828 Yes.This happens around here. Get used to it. $\endgroup$ – José Carlos Santos Dec 2 '18 at 11:09
  • $\begingroup$ @user1717828 what can you do... $\endgroup$ – Clement C. Dec 4 '18 at 22:17
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You have $$ \left(1+\frac{1}{f(n)}\right)^{g(n)} = \exp\left(g(n) \ln \left(1+\frac{1}{f(n)}\right) \right) $$ Since $\lim_{n\to\infty} f(n) = \infty$, we have $$ g(n) \ln \left(1+\frac{1}{f(n)}\right) = g(n)\cdot \left(\frac{1}{f(n)} + o\left(\frac{1}{f(n)}\right)\right) = \frac{g(n)}{f(n)} + o\!\left(\frac{g(n)}{f(n)}\right) $$ and by your assumption that $f$ "grows faster than $g$", this converges to $\ell=0$ (the result holds as long as $\lim_{n\to\infty} \frac{g(n)}{f(n)}$ exists, not necessarily $0$).

Then, $$ \lim_{n\to\infty }\left(1+\frac{1}{f(n)}\right)^{g(n)} = e^0 = 1. $$

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    $\begingroup$ THIS is the rigorous way (+1) $\endgroup$ – Robert Z Dec 2 '18 at 17:10
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Note that

$$\lim_{n \to \infty}\frac{g(n)}{f(n)} = 0 \implies \bigg(1+\frac{1}{f(n)}\bigg)^{g(n)} = \Biggl[\bigg(1+\frac{1}{f(n)}\bigg)^{f(n)}\Biggl]^{\frac{g(n)}{f(n)}} = e^\frac{g(n)}{f(n)} \to e^0 = 1$$

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  • $\begingroup$ The equality with the exponential is just... false? (and even in terms of limits, there would be assumptions hidden) $\endgroup$ – Clement C. Dec 1 '18 at 18:35
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Taking logarithms it suffices to prove that the limit of the logarithm is zero. To this end note that $$ g(n)\log\left(1+\frac{1}{f(n)}\right)=\frac{g(n)}{f(n)}\frac{\log\left(1+\frac{1}{f(n)}\right)}{1/f(n)}\to0 $$ since $$ \frac{g(n)}{f(n)}\to0 $$ as $f$ grows faster than $g$ and for the second term use the fact $1/f(n)\to0$ and $$ \lim_{x\to0}\frac{\log (1+x)-0}{x-0}=1 $$ by definition of the derivative.

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  • $\begingroup$ So the result is 0*1 ? $\endgroup$ – Samy Bencherif Dec 2 '18 at 6:42
  • $\begingroup$ No to the limit is $\exp(0\times 1)$ $\endgroup$ – Foobaz John Dec 2 '18 at 15:57

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