Intuitive proofs that $\lim\limits_{n\to\infty}\left(1+\frac xn\right)^n=e^x$ At this link someone asked how to prove rigorously that
$$
\lim_{n\to\infty}\left(1+\frac xn\right)^n = e^x.
$$
What good intuitive arguments exist for this statement?  
Later edit: . . . where $e$ is defined as the base of an exponential function equal to its own derivative.
I will post my own answer, but that shouldn't deter anyone else from posting one as well.
 A: If $n$ is really large, then $\int_1^{1+x/n}\frac{n}{t}\,dt$ is approximately an $\frac{x}{n}\times n$ rectangle with area $x$.

So $$\begin{align}n\int_1^{1+x/n}\frac{1}{t}\,dt&\approx x\\
\implies e^{ \textstyle n\int_1^{1+x/n}\frac{1}{t}\,dt}&\approx e^x\\
\implies \left(e^{ \textstyle \int_1^{1+x/n}\frac{1}{t}\,dt}\right)^n&\approx e^x\\
\end{align}$$
Now $\int_1^{e^z}\frac{1}{t}\,dt$ is a linear function of $z$ with slope $1$, since its derivative works out to be $\frac{1}{e^z}e^z=1$ (FToC, Chain Rule, and the OPs definition of $e$). Further, its value at $z=0$ is clearly $0$. Therefore $\int_1^{e^z}\frac{1}{t}\,dt=z$.
So $$\begin{align}\left(1+\frac{x}{n}\right)^n&\approx e^x\\
\end{align}$$
And the approximation only gets better as $n$ gets larger. If you like, you can even follow the error which is, at the first step, approximately that little triangle of area $\frac{1}{2}\frac{x}{n}x$.
A: This is a rigourous argument, but I think it gets at an intuition.
At the heart of this argument is that $$f(x)=\frac{e^x}{1+x}$$ has the property that $f(0)=1$ and $f'(0)=0$, and therefore that $$\lim_{n\to\infty} f(x/n)^n = 1$$
To get there, we'll use a key fact.
The key fact about this sort of limit is that if $g(n)$ is a function and $\lim_{n\to\infty} ng(n) = 0$, then  $$\lim_{n\to\infty}(1+g(n))^n\to 1$$
I'll prove this key result later. It's essentially a nearly trivial result of the binomial theorem.
Now, if $f(0)=1$ and $f'(0)=0$ then $$\lim_{h\to 0}\frac{f(xh)-1}{h} = xf'(0)=0$$.
Letting $h=1/n$, this means that $\lim_{n\to\infty} n(f(x/n)-1) = 0$. Letting $g(n)=f(x/n)-1$, then, the "key fact" shows that $$\lim_{n\to\infty}f(x/n)^n = 1$$
Now, given $f_1,f_2$ two functions differentiable at $0$ with $f_1(0)=f_2(0)\neq 0$ and $f_1'(0)=f_2'(0)$, we can define $f(x)=\frac{f_1(x)}{f_2(x)}$, and see that $f(0)=1$ and $f'(0)=0$. This shows that:
$$\lim_{n\to\infty} \left(\frac{f_1(x/n)}{f_2(x/n)}\right)^n=\lim_{n\to\infty} f(x/n)^n=1$$
Then let $f_1(x)=e^x$ and $f_2(x)=1+x$ to get your limit.
Essentially, the fact that the derivative of $e^z$ at $0$ is $1$ means that $e^z$ is "close enough" to $1+z$ when $z$ is small to allow us to use our "key fact."

Back to proving our "key fact."
Lemma: If $ng(n)\to 0$ as $n\to\infty$, then $\lim_{n\to\infty}(1+g(n))^n=1.$
Proof: We use a binomial theorem argument. When $|ng(n)|<1$ we have:
$$\begin{align}\left|(1+g(n))^n - 1\right| &\leq \sum_{k=1}^n \binom{n}{k}\left|g(n)\right|^k\\
&\leq \sum_{k=1}^n  n^k|g(n)|^k \leq \sum_{k=1}^\infty (n|g(n)|)^k\\&=\frac{n|g(n)|}{1-|ng(n)|}
\end{align}$$
So $(1+g(n))^n\to 1$ since $\frac{ng(n)}{1-ng(n)}\to 0$.

The reason I say the above is a "key fact" is that if instead we define $e^x$ as $\lim(1+x/n)^n$, we can then use the "key fact" to show that $e^{x+y}=e^xe^y$, which follows since $$\frac{(1+x/n)(1+y/n)}{1+(x+y)/n} = 1+O(1/n^2)$$
We can also use it to show that $e^{ix}=\cos x+i\sin x$ by having approximations $\cos \frac x n = 1+O(1/n^2)$ and $\sin \frac{x}{n}=\frac{x}{n}+O(1/n^2)$.
We can prove those approximations for $\sin x$ and $\cos x$ essentially geometrically as follows.
We have that $\sqrt{2-2\cos \theta}$ is the length of the chord from $1+0i$ to $\cos \theta+i\sin \theta$, and thus that length is less than the length of the circle arc, $\theta$, so $0\leq 2-2\cos\theta \leq \theta^2$, or $|\cos \theta -1|=O(\theta^2)$.
We can also show geometrically that $x\cos x\leq \sin x \leq x$, so $$0\leq x-\sin x\leq x(1-\cos x)=xO(x^2)=O(x^3)$$
That $\sin x\leq x$ can be seen because $\sin x$ is the shortest distance from $\cos x+i\sin x$ to the real line, while $x$ is the length of the circle arc from the same point to the real line.
The other inequality is a little harder. We can find a path of length $2\tan x$ between $cos 2x + i\sin 2x$ and $1+0i$ that is strictly outside the circle except at the endpoints, thus showing that $2\tan x \geq 2x$ or $\sin x\geq x\cos x$.
With these two approximations for the trigonometric functions, we get, for fixed $x$, $$\cos \frac{x}{n} +i\sin \frac{x}{n} = 1+\frac{ix}{n}+O(1/n^2)$$
A: How about,
\begin{align}
\frac{d}{dx}\left(\lim_{n\rightarrow\infty} \left(1+\frac{x}{n}\right)^{n} \right) & \overset{\text{intimidate}}{=} \lim_{n\rightarrow\infty} \frac{d}{dx}\left(1+\frac{x}{n}\right)^{n} \\
& = \lim_{n\rightarrow\infty} \left(1+\frac{x}{n}\right)^{n-1} \\
& = \lim_{n\rightarrow\infty} \frac{\left(1+\frac{x}{n}\right)^{n}}{\left(1+\frac{x}{n}\right)} \\
& = \frac{\lim_{n\rightarrow\infty} \left(1+\frac{x}{n}\right)^{n}}{\lim_{n\rightarrow\infty} \left(1+\frac{x}{n}\right)} \\
& = \lim_{n\rightarrow\infty} \left(1+\frac{x}{n}\right)^{n}
\end{align}
We now solve the differential equation $f'(x) = f(x)$ with condition $f(0) = 1$. 
A: Let $f(u)=b^u$ be a typical exponential function.  It's easy to show from the definition of differentiation that $f'(u) = (b^u\cdot\text{constant})$, and by doing some intermediate-value stuff, that for some $b$, the constant is $1$.  And it's quite easy to show that $b$ is then between $2$ and $4$.  It can be narrowed down more by crude brute force but the arithmetic gets messy.
Let $n$ be an infinitely large integer.
Every time we add $x/n$ to the argument to $f$, we multiply by the same amount $m$.  But if $f$ grows at a rate equal to its present size (i.e. $f'=f$) then when we multiply by $m$, what we must add to the value of $f(u)$ is $f(u)$ times what we added to its argument, namely $f(u)\cdot x/n$.  Therefore
$$
f\left(u+\frac xn\right) = f(u)\left(1+\frac xn\right).
$$
Repeating this $n$ times, we have
$$
f(u+x)=f(u)\left(1+\frac xn\right)^n.
$$
But $f(u+x)=b^x f(u)$, and hence $\displaystyle\left(1+\frac xn\right)^n=b^x$, where $b$ is the base for which $f'=f$.
This works since $n$ is infinite and $x$ is finite.  If we let $x$ grow to the point where it rivals $n$ in size, then obviously all this won't work.  Hence the convergence is not uniform.
Later edit: Let's try to be a bit neater:
$$
f\left(u+\frac xn\right) = f(u+du)=f(u)+f'(u)\,du = f(u)+f(u)\,du = f(u)(1+du) = f(u)\left(1+\frac xn\right).
$$
Iterating $n$ times, we have
$$
f(u+x)=f(u)\left(1+\frac xn\right)^n.
$$
Since $f(u+x)=b^x f(u)$, we have
$$
b^x = \left(1+\frac xn\right)^n.
$$
A: $$\lim_{n \to \infty}\left ( 1 + \frac{x}{n} \right )^n \stackrel{(1)}{=} \lim_{n \to \infty} \exp \left ( \log \left ( 1 + \frac{x}{n} \right )^n \right) \stackrel{(2)}{=} \exp \left (\lim_{n \to \infty}  \log \left ( 1 + \frac{x}{n} \right )^n \right)$$
$$ \stackrel{(3)}{=} \exp \left (\lim_{n \to \infty} n \log \left ( 1 + \frac{x}{n} \right ) \right) = \exp \left (\lim_{n \to \infty} \frac{ \log \left ( 1 + \frac{x}{n} \right )}{\frac{1}{n}} \right) =  \exp \left (\lim_{t \to 0^+} \frac{ \log \left ( 1 + xt \right )}{t} \right)$$
$$\stackrel{(4)}{=} \exp \left (\lim_{t \to 0^+} \frac{x}{1 + xt} \right) = \exp(x),$$
where $(1)$ is by definition of $\log$ as inverse of $\exp$, $(2)$ is by continuity of $\exp$ (and assumption that inside limit exists), $(3)$ is by properties of $\log$ (and hence properties of $\exp$), and $(4)$ is by l'Hospital's rule.
A: This is how it was taught to me in high school, which should hopefully be an indication of its simplicity:
$$
\begin{align}
f(x)&=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n\\
\ln(f(x))&=\ln\left(\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n\right)\\
&=\lim_{n\to\infty}\left(\ln\left(1+\frac{x}{n}\right)^n\right)\\
&=\lim_{n\to\infty}\left(n\ln\left(1+\frac{x}{n}\right)\right)\\
&=\lim_{n\to\infty}\left(\frac{\ln\left(1+\frac{x}{n}\right)}{\frac{1}{n}}\right)\\
&=\lim_{n\to\infty}\left(\frac{\left(\frac{-x}{n^2}\right)\left(\frac{1}{1+\frac{x}{n}}\right)}{\frac{-1}{n^2}}\right)\\
&=\lim_{n\to\infty}\left(\frac{x}{1+\frac{x}{n}}\right)\\
&=x\\
f(x)&=e^x
\end{align}
$$
The third line is perhaps not really rigorous, and the sixth line uses L'Hopital's rule, in case it wasn't clear.
A: I think that the most intuitive proof is the most simple
$$\left(1+\frac xn\right)^n=e^{n\log\left(1+\frac xn\right)}\sim_\infty e^{n\times \frac xn}=e^x$$
A: In my opinion, the most intuitive proof is the one that doesn't require any extraneous methods like using logarithms, power series, etc. So let's do it using only limits and basic algebra:
For $x=0$, the result is obvious. For $x>0$, let $k=\frac{n}x$, so you have
$$
\lim_{n\to \infty} \left(1+\frac{x}n\right)^n = \lim_{k\to \infty} \left(1+\frac1k\right)^{kx} = \left(\lim_{k\to\infty} \left(1+\frac1k\right)^k\right)^x = e^x
$$
For $x<0$, we need to proceed a little differently, as the limit goes in the "wrong" direction. We have $k=\frac{n}x$, which gives
$$
\lim_{n\to \infty} \left(1+\frac{x}n\right)^n = \lim_{k\to -\infty} \left(1+\frac1k\right)^{kx}=\left(\lim_{k\to-\infty} \left(1+\frac1k\right)^k\right)^x
$$
So we need to confirm that the limit in the brackets is still $e$ (it is, but we want to use the "normal" definition). So,
$$\begin{align}
\lim_{k\to-\infty} \left(1+\frac1k\right)^k&=\lim_{m\to\infty} \left(1-\frac1m\right)^{-m}\\
&= \lim_{m\to\infty}\left(\frac{m-1}{m}\right)^{-m}\\
&= \lim_{p\to\infty}\left(\frac{p+1}{p}\right)^{p+1}\\
&= \lim_{p\to\infty}\left(1+\frac1p\right)^p\\
&= e
\end{align}$$
Therefore, our limit is again $e^x$.
EDIT: With the added condition that $e$ is defined as the base for an exponential function equal to its own derivative, this requires a little more work. It is clear from the above that the limit takes the form $a^x$. Now we need only show that $(a^x)'=a^x$. This is actually remarkably simple, using the derivative rules. We have $f(x)=a^x$. Therefore, we wish to show that $(\ln f(x))'=1$.
$$\begin{align}
\left(\ln \lim_{n\to\infty} \left(1+\frac1n\right)^{nx}\right)'&=\left(\lim_{n\to\infty} \ln \left(1+\frac1n\right)^{nx}\right)'\\
&=\left(\lim_{n\to\infty} nx\ln \left(1+\frac1n\right)\right)'\\
&=\left(x\lim_{m\downarrow0} \frac{\ln (1+m)}{m}\right)'\\
&=\lim_{m\downarrow0} \frac{\ln (1+m)}{m}\\
&=\lim_{m\downarrow0} \frac{\ln (1+m)-\ln 1}{m}\\
&=\left[(\ln k)'\right]_{k=1}\\
&= \left[\frac1k\right]_{k=1}\\
&= 1
\end{align}$$
A: $$ e^x=\lim_{m\rightarrow \infty}\left(1+\frac{1}{m}\right)^{mx} $$
Let $mx=n$, so $m=\frac{n}{x}$
$$e^x=\lim_{n\rightarrow\infty}\left(1+\frac{x}{n}\right)^n$$
A: Another way of looking at it:
Let
$$f_n(x) = \left(1+\frac{x}{n}\right)^n$$ and we are interested in $f(x) = \lim_{n \to \infty} f_n(x)$
Now,
$$f_n(x) f_n(y) = \left(1+\frac{x}{n}\right)^n\left(1+\frac{y}{n}\right)^n$$
$$ = \left(1+\frac{x+y +\frac{xy}{n}}{n}\right)^n = f_n\left(x+y +\frac{xy}{n}\right)$$
Thus as $n \to \infty$, we probably have that
$$f(x)f(y) = f(x+y)$$
and so we can expect $f(x)$ to be exponential.
A: According to this question $e$ is defined by $e = f(1)$ where $f(x)$ is a function satisfying $f'(x) = f(x), f(0) = 1$ for all $x$. As I have proved elsewhere on this site that under this condition $f(x)$ has an inverse $g(x)$ with $g'(x) = 1/x$ and $g(x) = \int_{1}^{x}(1/t)\, dt$. Also from the fact that $g(1) = 0, g'(1) = 1$ it follows that $$\lim_{h \to 0}\frac{g(1 + h)}{h} = 1\tag{1}$$
Next it can be easily proved that $$g(xy) = g(x) + g(y)\tag{2}$$ for positive $x, y$ and this leads to $$f(x + y) = f(x)f(y)\tag{3}$$ for all $x, y$. Hence if $a > 0$ and $n$ is positive integer then it follows that $$a^{n} = f(ng(a))\tag{4}$$ It is now easy to show that $$\lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^{n} = f(x)$$ We can proceed as follows
\begin{align}
L &= \lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^{n}\notag\\
&= \lim_{n \to \infty}f\left(ng\left(1 + \dfrac{x}{n}\right)\right)\text{ (using equation (4))}\notag\\
&= f\left(\lim_{n \to \infty}ng\left(1 + \dfrac{x}{n}\right)\right)\text{ (by continuity of }f)\notag\\
&= f\left(\lim_{h \to 0}\frac{x}{h}\cdot g(1 + h)\right)\text{ (by putting }h = x/n)\notag\\
&= f(x)\text{ (using equation (1))}\notag
\end{align}
A: Starting from the definition $\ln(x) = \int_1^x \frac{dt}{t}$ and $e^{\ln x} = x$ for any $x > 0$.

Let $P_n(x)$ be a sequence of polynomials such that $\frac{P'_n(x)}{P_n(x)} \to 1$ uniformly on any interval $[-A,A]$ and $P_n(0) = 1$. Then $P_n(x) \to e^x$ uniformly on any interval $[-A,A]$.
proof : note that $(\ln P_n(x))' = \frac{P_n'(x)}{P_n(x)}$ so that $\ln P_n(x) = \ln P_n(0) + \int_0^x \frac{P'_n(t)}{P_n(t)} dt \to x$ uniformly, hence $P_n(x) \to e^{x}$ uniformly.


*

*Consider $P_n(x) = (1+\frac{x}{n})^{n}$. $ \ $ Thus $P_n'(x) =  (1+\frac{x}{n})^{n-1}$ so that $\frac{P_n'(x)}{P_n(x)} = \frac{1}{1 +\frac{x}{n}} \to 1$ uniformly on any interval $[-A,A]$. And since $P_n(0) = 1$ : $$\textstyle P_n(x) = \ \ \color{red}{\left(1+\frac{x}{n}\right)^{n} \ \ \to \ \ e^x} \qquad\qquad \text{uniformly on } [-A,A]$$
$$ $$

*Consider $P_n(x) = \sum_{k=0}^n \frac {x^k}{k!}$.  $ \ $  Thus  $P_n'(x) = \sum_{k=0}^{n-1} \frac {x^k}{k!} = P_n(x) - \frac{x^n}{n!}$ so that $\frac{P_n'(x)}{P_n(x)} = \frac{P_n(x)}{P_n(x)-\frac{x^n}{n!}} \to 1$ uniformly on any interval $[-A,A]$. And since $P_n(0) = 1$ :
$$\textstyle  P_n(x) = \ \  \color{red}{\sum_{k=0}^n \frac {x^k}{k!} \ \ \to \ \ e^x} \qquad\qquad \text{uniformly on } [-A,A]$$
A: If you have access to the power series of $e^x$ and the binomial theorem, then you can see it because the left side is
$$1+\binom{n}{1}\frac{x}{n}+\binom{n}{2}\frac{x^2}{n^2}+\binom{n}{3}\frac{x^3}{n^3}+\cdots$$
which is 
$$1+\frac{n}{n}x+\frac{1}{2!}\frac{n(n-1)}{n^2}x^2+\frac{1}{3!}\frac{n(n-1)(n-2)}{n^3}x^3+\cdots$$
and term by term as $n\to\infty$,
$$1+x+\frac{1}{2!}x^2+\frac{1}{3!}x^3+\cdots$$
I'm not sure if this is what you are looking for, but it's certainly not a rigorous proof!
A: $f(x) = e^x$ is the only solution to the differential equation $\dfrac{dy}{dx} = y$ with $f(0)=1$.
To approximate $f(a)$, we can use Euler's method on the interval $[0,a]$ with $n$ subintervals.
$$f(0) = 1, f'(0)=1 \implies f(\frac{a}{n}) \approx 1+\frac{a}{n}$$
$$f(\frac{a}{n}) \approx 1+\frac{a}{n}, f'(\frac{a}{n}) \approx 1+\frac{a}{n} \implies f(\frac{2a}{n}) \approx 1+\frac{a}{n} + \frac{a}{n}(1+\frac{a}{n}) =  (1+\frac{a}{n})^2$$
$$ \vdots $$
$$f(a) \approx (1+\frac{a}{n})^n$$
Since Euler's method actually converges in the limit, we have 
$$e^a = \lim_{n \to \infty} (1+\frac{a}{n})^n$$
