# Proving that $\lim_{n\to\infty} \left(1+\frac{1}{f(n)}\right)^{g(n)} = 1$

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.

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

We can use that $$\left(1+\frac{1}{f(n)}\right)^{g(n)} =\left[\left(1+\frac{1}{f(n)}\right)^{f(n)}\right]^{\frac{g(n)}{f(n)}}$$

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}

• 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 $[0,\infty]$). – Clement C. Dec 1 '18 at 18:37
• @ClementC. Yes, I agree that one should be careful here. – José Carlos Santos Dec 1 '18 at 18:48
• 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. – user1717828 Dec 2 '18 at 1:37
• @user1717828 Yes.This happens around here. Get used to it. – José Carlos Santos Dec 2 '18 at 11:09
• @user1717828 what can you do... – Clement C. Dec 4 '18 at 22:17

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

• THIS is the rigorous way (+1) – Robert Z Dec 2 '18 at 17:10

Note that

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

• The equality with the exponential is just... false? (and even in terms of limits, there would be assumptions hidden) – Clement C. Dec 1 '18 at 18:35

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.

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