Proof that $\lim_{x\to+\infty} (1+1/x)^x$ is finite I'm trying to understand more about the $e$ constant. I know it can be defined in many ways, like $\lim_{x\to+\infty} (1+1/x)^x$ or as the sum of all $(n!)^{-1}$ from $0$ to infinity, or as the only value for $a$ so that the derivative of $y = a^x$ is itself. But, however I define it, there's always something that's unclear to me. For example, how do you prove that $\lim_{x\to+\infty} (1+1/x)^x$ is finite and doesn't go bigger than a value, like for instance a logarithmic function does? And for the $y' = y$ definition, how do you know that the $y$ function goes up like an exponential function? Why isn't it another type of function? (of course the "sum" definition comes from the Taylor series of $y = e^x$ at $x = 0$, so that's a consequence of the "$y = y'$" definition)
Hope my question was clear.
 A: Take the limit along integer x. $$\left(1+\frac{1}{x}\right)^x=\sum_{h=0}^x{x \choose h}\frac{1}{x^h}=\sum_{h=0}^x\frac{x(x-1)\dotsb (x-h+1)}{h!x^h}=\sum_{h=0}^x \frac{1}{h!}\left(1-\frac{1}{x}\right)\left(1-\frac{2}{x}\right)\dotsb \left(1-\frac{h-1}{x}\right)$$
This is monotonically increasing as $x\to\infty$ and is also $\leq\sum_{h=0}^x \frac{1}{h!}$ which converges, so the limit also converges.
A: Your expression is less than $\left(1+\frac{1}{n}\right)^{n+1}$ where $n\leq x <n+1$, and this sequence is decreasing, and thus $e<(1+1)^2=4$.
A: Try expanding by using binomial expansion and apply the limit. Since x tends to infinity reciprocal of powers of x tends to 0 and u will get the series expansion for e. Also any function whose derivative is itself is of the form f(x)=ce^x where c is any real no. U can try proving it by using dy/dx=y and solving this D.E by using variable separable method for y u will get y=ce^x.
A: Many great answers were given while I was writing this up. Nevertheless, here's the longer approach anyway ;).

The limit
$$\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^x=e$$
relies upon the following theorems, properties, and a definition.

Theorem 1. Say functions $u=u(x)$, $v=v(x)$ follow the inequality $$v\geq u.$$ Then for $x\to\infty$ $$\lim v\geq\lim u.$$
Theorem 2. If a function $v(x)$ is increasing and bounded above by $M\in\Bbb{R}$, then
  $$\lim_{x\to \infty}v\leq M.$$
Property 1. $$\frac{1}{1\cdot2\cdot3}<\frac{1}{2^2},\ \ \ \frac{1}{1\cdot2\cdot3\cdot4}<\frac{1}{2^3},\ \dots, \ \ \frac{1}{n!}<\frac{1}{2^{n-1}}$$
Property 2. Sum $S$ of a geometric sequence $$S=a_1\frac{1-q^n}{1-q}$$


I will prove the result for $x=n\in\Bbb{N}^+$ but it can be generalised. Besides showing that the limit is finite, we shall also see that $2\leq e\leq3$.
Begin by writing the sequence in question:
$$\left(1+\frac{1}{n}\right)^n$$
where $n$ is increasing. By the binomial expansion theorem,
$$\left(1+\frac{1}{n}\right)^n=\sum_{k=0}^n\begin{pmatrix}n\\k\end{pmatrix}1^{n-k}\left(\frac{1}{n}\right)^k.$$
After a bunch of simplification,
$$\tag{*}\left(1+\frac{1}{n}\right)^n=1+1+\frac{1}{2}\left(1-\frac{1}{n}\right)+\ldots+\frac{1}{n!}\left(1-\frac{1}{n}\right)\cdot\left(1-\frac{2}{n}\right)\cdots\left(1-\frac{n-1}{n}\right).$$
Note that 
$$\tag{**}0<\frac{1}{n}\leq1\implies 0\leq1-\frac{1}{n}<1,\ \ \dots,$$
and
$$\tag{**}1-\frac{1}{n}<1-\frac{1}{n+1};\ \ \ 1-\frac{2}{n}<1-\frac{2}{n+1},$$
etc.
So from $(*)$ we already have two useful implications:
$$\left(1+\frac{1}{n}\right)^n \text{is increasing},\ \ \ 2\leq \left(1+\frac{1}{n}\right)^n.$$
Due to our inequalities $(**)$ above
$$\left(1+\frac{1}{n}\right)^n<1+1+\frac{1}{2!}+\dots+\frac{1}{n!}.$$
Via Property 1
$$\left(1+\frac{1}{n}\right)^n<1+\underbrace{1+\frac{1}{2}+\frac{1}{2^2}+\dots+\frac{1}{2^{n-1}}}_\text{geometric sequence}.$$
Property 2 applies. We see that $a_1=1$, $q=\frac{1}{2}.$
$$\left(1+\frac{1}{n}\right)^n<1+\frac{1-\left(\frac{1}{2}\right)^n}{1-\frac{1}{2}}=1+\left[2-\left(\frac{1}{2}\right)^{n-1}\right]<3$$
In summary the sequence is increasing and bounded. By theorems 1 and 2
$$2\leq \lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n \leq3.$$
And here's the definition.

Definiton. The number $2\leq e\leq3$ is defined to be the limit.$$\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n\equiv e\approx2.718$$

A: Bernoulli's Inequality says
$$
\begin{align}
\frac{\left(1+\frac1n\right)^n}{\left(1+\frac1{n-1}\right)^{n-1}}
&=\frac{n}{n-1}\left(1-\frac1{n^2}\right)^n\\
&\ge\frac{n}{n-1}\left(1-\frac1n\right)\\[9pt]
&=1
\end{align}
$$
Therefore, $a_n=\left(1+\frac1n\right)^n$ is an increasing sequence.

Bernoulli's Inequality also says
$$
\begin{align}
\frac{\left(1+\frac1{n-1}\right)^n}{\left(1+\frac1n\right)^{n+1}}
&=\frac{n-1}{n}\left(1+\frac1{n^2-1}\right)^{n+1}\\
&\ge\frac{n-1}{n}\left(1+\frac1{n-1}\right)\\[9pt]
&=1
\end{align}
$$
Therefore, $b_n=\left(1+\frac1n\right)^{n+1}$ is a decreasing sequence.

Note that for any $n$,
$$
a_n\le\left(1+\frac1n\right)a_n=b_n
$$
Therefore, for any $m$ and $n$,
$$
\overbrace{a_m\le a_{\max(m,n)}}^\text{$a_n$ is increasing}\le \overbrace{b_{\max(m,n)}\le b_n}^\text{$b_n$ is decreasing}
$$
Thus, $a_n$ is increasing and bounded above, while $b_n$ is decreasing and bounded below. Furthermore,
$$
\begin{align}
\lim_{n\to\infty}\frac{b_n}{a_n}
&=\lim_{n\to\infty}\left(1+\frac1n\right)\\
&=1
\end{align}
$$
Thus,
$$
\lim_{n\to\infty}a_n=\lim_{n\to\infty}b_n
$$
and both limits exist.
