TITLE: Prove that $\lim\limits_{n\rightarrow\infty} \left(1+\frac{1}{a_{n}} \right)^{a_{n}}=e$ if $\lim\limits_{n\rightarrow\infty} a_{n}=\infty$
QUESTION [7 upvotes]: What would be the nicest proof of the following theorem:

If $\lim\limits_{n \rightarrow \infty} a_{n} = \infty$, then $\lim\limits_{n \rightarrow \infty} \left(1 + \frac{1}{a_{n}} \right) ^ {a_{n} } = e$.
If $\lim\limits_{n \rightarrow \infty}b_{n} = 0$, then $\lim\limits_{n \rightarrow \infty} \left(1 + b_{n} \right) ^ {\frac {1} {b_{n}} } = e$.

I somehow failed to find a proof here on the website and in the literature.

REPLY [4 votes]: Given that $\lim_{n\to\infty }b_n =  \lim_{n\to\infty }\frac{1}{a_{n}}=0$
we have,
$$\lim_{n \to\infty} \left(1 + \frac{1}{a_{n}} \right) ^ {a_{n} } = \lim_{n \to\infty} \exp\left(\frac{\ln\left(1 + \frac{1}{a_{n}} \right)}{\frac1{a_{n}}} \right)=  \lim_{h \to0} \exp\left(\frac{\ln\left(1 +h \right)}{h} \right)=e$$
similarly
$$\lim_{n \to\infty} \left(1 +b_n \right) ^ { \frac{1}{b_{n}}} = \lim_{n \to\infty} \exp\left(\frac{\ln\left(1 + b_n\right)}{b_n} \right)=  \lim_{h \to0} \exp\left(\frac{\ln\left(1 +h \right)}{h} \right)=e$$