Prove $\lim_{x \to \infty} (1+1/x)^x = e$ using sequences I'm aware of the proof of $\lim_{n \to \infty }(1+\frac 1n)^n = e$ using monotone convergence theorem but I want to prove $\lim_{x \to \infty} (1+\frac 1x)^x = e$ using sequential criterion for limits . How I can write it rigorously ?
More precisely : For every $a_n \to \infty$ s.t $a_n \in D_f$ prove $(1+\frac 1{a_n})^{a_n} \to e$ $\ \ \ \ \ \ \ \ \ \ (\star)$
 A: Take any sequence of real numbers $(x_n)_{n\in\mathbb N}$ such that $x_n\to +\infty$.
$$\lim_{x\to +\infty}\left(\left(1+\frac{1}{x}\right)^x\right)=\lim_{n\to +\infty}\left(\left(1+\frac{1}{x_n}\right)^{x_n}\right)$$
For sufficiently large $x_n$ as $x_n\to +\infty$ as $n\to +\infty$, e.g. $x_n\ge 1$, we have
$$\left(1+\frac{1}{x_n}\right)^{x_n}\le \left(1+\frac{1}{\lfloor x_n\rfloor}\right)^{\lfloor x_n\rfloor +1}=$$
$$=\left(1+\frac{1}{\lfloor x_n\rfloor}\right)^{\lfloor x_n\rfloor}\left(1+\frac{1}{\lfloor x_n\rfloor}\right)\stackrel{n\to +\infty}\to e\cdot (1+0)=e$$
$$\left(1+\frac{1}{x_n}\right)^{x_n}\ge \left(1+\frac{1}{\lfloor x_n\rfloor+1}\right)^{\lfloor x_n\rfloor}=$$
$$=\frac{\left(1+\frac{1}{\lfloor x_n\rfloor + 1}\right)^{\lfloor x_n\rfloor +1}}{1+\frac{1}{\lfloor x_n\rfloor+1}}\stackrel{n\to +\infty}\to \frac{e}{1+0}=e$$
By Squeeze Theorem $$\lim_{x\to +\infty}\left(\left(1+\frac{1}{x}\right)^x\right)=e$$
$$\lim_{n\to -\infty}\left(\left(1+\frac{1}{n}\right)^{n}\right)\stackrel{m=-n}=\lim_{m\to +\infty}\left(\left(1-\frac{1}{m}\right)^{-m}\right)=$$
$$=\lim_{m\to +\infty}\left(\left(1+\frac{1}{m-1}\right)^{m-1}\cdot \left(1+\frac{1}{m-1}\right)\right)=$$
$$=e\cdot (1+0)=e$$
The proof that $$\lim_{x\to -\infty}\left(\left(1+\frac{1}{x}\right)^x\right)=e$$ is similar to what I've done before.
A: Consider the sequences 
$$a_{n}=\left(1+\frac{1}{n}\right)^{n}$$ 
and 
$$b_{n}=\left(1+\frac{1}{n}\right)^{n+1}$$
Using Bernoulli's inequality you can show that:


*

*$a_{n}$ is monotone increasing and upper bounded

*$b_{n}$ is monotone decreasing and lower bounded

*$\underset{n\to\infty }{\lim}{\ a_n}=e$ 

*$\underset{n\to\infty }{\lim}{\ b_n}=e$ 


Finally:
$$a_{n}\leq \left(1+\frac{1}{x}\right)^{x} \leq b_{n}$$
A: You can also prove a more general statement:

Suppose that $\underset{n\to \infty}{\lim}{\ (f_{n}-1)g_n}$ exists. If $\underset{n\to \infty}{\lim}{\ f_{n}}=1$ and $\underset{n\to \infty}{\lim}{\ g_{n}}=\infty$
$$\underset{n\to \infty}{\lim}{\ \displaystyle{f_{n}^{g_n}}}=e^{\ \underset{n\to \infty}{\lim}{\ (f_{n}-1)g_n}}$$

This is explained in B.P. Demidovich book.
Then, using this you can see that:

If $f$ and $g$ are functions such that
$$\underset{x\to a}{\lim}{\ f(x)}=1$$
$$\underset{x\to a}{\lim}{\ g(x)}=\infty$$
the limit bellow exists
$$\underset{x\to a}{\lim}{\ (f(x)-1)g(x)}$$
then
$$ \underset{x\to a}{\lim}{\ f(x)^{g(x)}}=e^{\ \underset{x\to a}{\lim}{\ (f(x)-1)g(x)}} $$

