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

Question

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.

Answer 1

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

Answer 2

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}

Answer 3

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

Answer 4

Note that

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

Answer 5

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.