Prove that $\lim\limits_{n\rightarrow\infty} \left(1+\frac{1}{a_{n}} \right)^{a_{n}}=e$ if $\lim\limits_{n\rightarrow\infty} a_{n}=\infty$ 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.
 A: 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
$$\lim_{n \to\infty} \left(1 +b_n \right) ^ { \frac{1}{b_{n}}} = \lim_{n \to\infty} \exp\left(\frac{\ln\left(1 + b_n\right)}{b_n} \right)=  \lim_{h \to0} \exp\left(\frac{\ln\left(1 +h \right)}{h} \right)=e$$ 
A: Let's do the 1st one ...
One of the logarithmic inequalities says: $$\frac{x}{1+x} < \ln (1 + x) < x, \forall x > -1$$
Because $\lim\limits_{n\rightarrow \infty} a_n \rightarrow \infty$, then $a_n > 0$ and $\frac{1}{a_n} > 0$ from some $n$ onwards. So
$$0<\frac{1}{a_n+1}=\frac{\frac{1}{a_n}}{1+\frac{1}{a_n}} < \ln\left(1 + \frac{1}{a_n}\right) < \frac{1}{a_n}$$
Given $e^x$ is ascending:
$$1<e^{\frac{1}{a_n+1}} <  1 + \frac{1}{a_n} < e^{\frac{1}{a_n}}$$
and from some $n$ onwards
$$e^{\frac{a_n}{a_n+1}} <  \left(1 + \frac{1}{a_n}\right)^{a_n} < e$$
Squeeze theorem finishes the proof.
A: For $|t|<1$, note that
$t-{1 \over 2} t^2 \le \log(1+t) \le t$ and so
$|\log(1+t)-t| \le {1 \over 2} t^2$.
Hence for $|x|>1$ we have
$|\log(1+{1 \over x})-{1 \over x}| \le {1 \over 2} ({1\over x})^2$ and
so
$|x\log(1+{1 \over x})-1| \le {1 \over 2} {1\over |x|}$.
Hence $\lim_{x \to \infty} x\log(1+{1 \over x}) =\lim_{x \to \infty} \log(1+{1 \over x})^x =  1$ from which it
follows that $\lim_{x \to \infty} (1+{1 \over x})^x = e$.
A: Hint:
$$\left(1 + \frac{1}{\lfloor a_n \rfloor+1} \right)^{\lfloor a_n \rfloor} \leqslant \left(1 + \frac{1}{a_n} \right)^{a_n} \leqslant \left(1 + \frac{1}{\lfloor a_n \rfloor} \right)^{\lfloor a_n \rfloor+1}, $$
and 
$$\left(1 + \frac{1}{n+1} \right)^n, \left( 1 + \frac{1}{n} \right)^{n+1} \to e$$
A: One does L'Hospital to show that $\lim_{x\rightarrow\infty}\left(1+\dfrac{1}{x}\right)^{x}=e$, then one does sequential characterisation of limit. Similar reasoning applied to $b_{n}\rightarrow 0$ case.
A: For any $a_n$ exists $x\in \mathbb N$ such that $x\le a_n\le x+1$ and
$$\left(1 + \frac{1}{x +1} \right)^{x} \leq \left(1 + \frac{1}{a_n} \right)^{a_n} \leq \left(1 + \frac{1}{x} \right)^{x+1}$$
and
$$\left(1 + \frac{1}{x +1} \right)^{x} =\frac{\left(1 + \frac{1}{x +1} \right)^{x+1}}{1 + \frac{1}{x +1} } \to \frac e 1=e$$
$$\left(1 + \frac{1}{x} \right)^{x+1}=\left(1 + \frac{1}{x} \right)^{x}\left(1 + \frac{1}{x} \right)\to e \cdot 1=e$$
then we can conclude by squeeze theorem.
