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.
 A: 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,
$$
\lim_{n\to\infty }\left(1+\frac{1}{f(n)}\right)^{g(n)} 
= e^0 = 1.
$$
A: Note that
$$\lim_{n \to \infty}\frac{g(n)}{f(n)} = 0 \implies \bigg(1+\frac{1}{f(n)}\bigg)^{g(n)} = \Biggl[\bigg(1+\frac{1}{f(n)}\bigg)^{f(n)}\Biggl]^{\frac{g(n)}{f(n)}} = e^\frac{g(n)}{f(n)} \to  e^0 = 1$$
A: We can use that
$$ \left(1+\frac{1}{f(n)}\right)^{g(n)} =\left[\left(1+\frac{1}{f(n)}\right)^{f(n)}\right]^{\frac{g(n)}{f(n)}}$$
A: 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}
A: 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.
