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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.

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How do you intuitively define $e^x$? – Thomas Andrews Apr 17 '13 at 23:51
You've got to define $e^x$ somewhere. Is it the limit definition? The 'it's its own derivative'? Is it 'inverse of the integral of $1/x$? – Henry Swanson Apr 17 '13 at 23:52
@HenrySwanson: I suppose, for this question, you get to choose your favourite... – Aryabhata Apr 18 '13 at 0:16
I had in mind that $e$ is the base of an exponential function equal to its own derivative. My own posted answer explains how we know intuitively that such a thing exists before we know the result to be "proved". – Michael Hardy Apr 18 '13 at 0:25
@ThomasAndrews, if I start walking at a $1$ meter mark, reach $e$ meters at $1$ second, $e^2$ at $2$ seconds, ..., $e^t$ at $t$ seconds, the distance travelled is $e^t$. So is the velocity. And the acceleration. And its variation... ...**ad infinitum*... That's the most intuitive explanation I'd find. – JMCF125 Mar 7 '14 at 19:03

15 Answers 15

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 $f(0) = 1$.

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+1: I was thinking along the same lines! – Aryabhata Apr 18 '13 at 0:13
@EuYu: I'm guessing it's a reference to proof by intimidation – Hurkyl Apr 18 '13 at 0:16
@EuYu, You cannot exchange limit and differentiation without proper justification. – Lord Soth Apr 18 '13 at 0:18
(Euler will be proud). :-) – Aryabhata Apr 18 '13 at 0:20
Maybe Euler is the original author of this one. – Michael Hardy Apr 18 '13 at 0:28

I think that the most intuitive proof is the most simple $$\left(1+\frac xn\right)^n=e^{n\log\left(1+\frac xn\right)}\sim_\infty e^{n\times \frac xn}=e^x$$

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+1 (as $\log(1+\epsilon) = \epsilon + O(\epsilon^2)$ for small $\epsilon$...) – Lord Soth Apr 18 '13 at 0:00
This is a rigorous proof :-). +1 though. – Aryabhata Apr 18 '13 at 0:12
I appreciated Aryabhat's comment, but I think this argument is better than it looks. But maybe not quite as good as I'd like. – Michael Hardy Apr 18 '13 at 0:13
Here is a formalization of this argument, if necessary: Since $\log(1+x) \leq x,\,\forall x\geq 0$, we obtain $(1+\frac{x}{n})^n \leq e^x$. Also, $\forall x>0, \forall\epsilon>0$, there is an $n_0\in\mathbb{N}$ s.t. $\log(1+\frac{x}{n}) \geq (1-\epsilon)\frac{x}{n},\,\forall n \geq n_0$. This gives us $\lim_{n\rightarrow\infty}(1+\frac{x}{n})^n \geq e^{x(1-\epsilon)}$. Since $\epsilon>0$ can be chosen arbitrarily,... – Lord Soth Apr 18 '13 at 0:28
Even simpler? $e^{x/n}\sim_\infty1+\frac{x}{n}$ and raise both sides to $n$? – alex.jordan Apr 18 '13 at 3:53

Another way of looking at it:

Let $$f_n(x) = \left(1+\frac{x}{n}\right)^n$$ and we are interested in $f(x) = \lim_{n \to \infty} f_n(x)$


$$f_n(x) f_n(y) = \left(1+\frac{x}{n}\right)^n\left(1+\frac{y}{n}\right)^n$$ $$ = \left(1+\frac{x+y +\frac{xy}{n}}{n}\right)^n = f_n\left(x+y +\frac{xy}{n}\right)$$

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.

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Nice. You could also found that $f(ax)=f(x)^a$ – leonbloy Apr 18 '13 at 15:37
@leonbloy: Right, that works too, but I guess an answer quite similar to that was added already.. – Aryabhata Apr 18 '13 at 15:48
It's not clear why $f_n(x+y+xy/n)\to f(x+y)$. You can't in general prove that $f_n(x_n)\to f(x)$ just because $x_n\to x$ and $f_n\to f$ pointwise. You need something specific. – Thomas Andrews Mar 7 '14 at 23:26
Being intuitive doesn't allow it to use flawed intuition. This intuition can give some very wrong results... – Thomas Andrews Mar 8 '14 at 3:02
@ThomasAndrews: Thanks for pointing out that this intuition can lead to wrong results. We need proof and cannot leave it as a bare statement, I agree with you. But, is your claim that this intuition cannot be turned into a proof for this problem? Note that the original problem was to find "intuitive proofs" that the function is exponential. I would say that this answer qualifies, irrespective of whether the intuition used leads to wrongs results in some cases. – Aryabhata Mar 8 '14 at 5:57

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


which is


and term by term as $n\to\infty$,


I'm not sure if this is what you are looking for, but it's certainly not a rigorous proof!

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I'm not sure it counts as "intuitive" either :) – Thomas Andrews Apr 17 '13 at 23:56
This is very close to the standard argument connecting the limit to the power series. I think it can be "rigorized" quite easily. On the other hand, I wouldn't count the power series characterization of $e^x$ as "intuitive". – EuYu Apr 17 '13 at 23:57
@EuYu: It is very easy to see that the power series for $e^x$ is the unique one with constant term $1$ and equal to its own formal derivative. This is much easier than to show that the exponential function exists and is the unique function with (certain conditions and) value $1$ at $0$ and equal to its own derivative. – Marc van Leeuwen Apr 18 '13 at 17:33
@MarcvanLeeuwen I don't know... To me intuitive means that I should be able to explain it to a high school student and for them to understand it heuristically. If someone has never seen the concept of a power series before then I imagine it comes as a surprise that functions can be represented by an infinite series. I guess my main complaint is about power series being unintuitive unless you know about them beforehand. – EuYu Apr 18 '13 at 17:50
@EuYu Yeah, this was the first answer posted, and I wasn't sure yet what OP was asking for. Just as a counter point, in my own sequence of learning these things, power series came earlier than special sequence limits like $\lim\{(1+x/n)^n\}$. – alex.jordan Apr 18 '13 at 21:54

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 $x$.

The integral of 1/t from 1 to 1+x/n

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 $\frac{1}{2}\frac{x}{n}x$.

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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." If $ng(n)\to 0$ as $n\to\infty$, 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 $x$, $$\cos \frac{x}{n} +i\sin \frac{x}{n} = 1+\frac{ix}{n}+O(1/n^2)$$

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$f(x) = e^x$ 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

$$e^a = \lim_{n \to \infty} (1+\frac{a}{n})^n$$

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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 $$ b^x = \left(1+\frac xn\right)^n. $$

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$$\lim_{n \to \infty}\left ( 1 + \frac{x}{n} \right )^n \stackrel{(1)}{=} \lim_{n \to \infty} \exp \left ( \log \left ( 1 + \frac{x}{n} \right )^n \right) \stackrel{(2)}{=} \exp \left (\lim_{n \to \infty} \log \left ( 1 + \frac{x}{n} \right )^n \right)$$ $$ \stackrel{(3)}{=} \exp \left (\lim_{n \to \infty} n \log \left ( 1 + \frac{x}{n} \right ) \right) = \exp \left (\lim_{n \to \infty} \frac{ \log \left ( 1 + \frac{x}{n} \right )}{\frac{1}{n}} \right) = \exp \left (\lim_{t \to 0^+} \frac{ \log \left ( 1 + xt \right )}{t} \right)$$ $$\stackrel{(4)}{=} \exp \left (\lim_{t \to 0^+} \frac{x}{1 + xt} \right) = \exp(x),$$

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.

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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 $(\ln f(x))'=1$.

$$\begin{align} \left(\ln \lim_{n\to\infty} \left(1+\frac1n\right)^{nx}\right)'&=\left(\lim_{n\to\infty} \ln \left(1+\frac1n\right)^{nx}\right)'\\ &=\left(\lim_{n\to\infty} nx\ln \left(1+\frac1n\right)\right)'\\ &=\left(x\lim_{m\downarrow0} \frac{\ln (1+m)}{m}\right)'\\ &=\lim_{m\downarrow0} \frac{\ln (1+m)}{m}\\ &=\lim_{m\downarrow0} \frac{\ln (1+m)-\ln 1}{m}\\ &=\left[(\ln k)'\right]_{k=1}\\ &= \left[\frac1k\right]_{k=1}\\ &= 1 \end{align}$$

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But this assumes the special case in which $x=1$, which is nearly as substantial as the generalization that this argument establishes. – Michael Hardy Apr 18 '13 at 0:12
@MichaelHardy I think this goes back to how you actually define $e$. What would you take as the most intuitive definition? – EuYu Apr 18 '13 at 0:16
@MichaelHardy: It doesn't assume that $x=1$ at all. What gave you that idea? (Did you notice that the $e$ obtained in the final sequence of equalities is the term inside the bracket of the previous expression?) – Glen O Apr 18 '13 at 0:18
I didn't say you assumed $x=1$; I said you reduced the general case where $x$ could be anything to the particular case where $x=1$. – Michael Hardy Apr 18 '13 at 0:24
Oh, I see. Well, I've always understood that the limit is one of the fundamental definitions of $e$. At the very minimum, it proves that the general limit is a number of the form $a^n$. Note that the specific definition of $e$ now found in the question was what you edited in, so I was answering the question as it stood. I'll edit in the best argument I can see for the final result. – Glen O Apr 18 '13 at 1:32

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.

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I'm not altogether comfortable with L'Hopital's rule in this context, since I wanted it to be things comprehensible to those who've only had precalculus courses. – Michael Hardy Apr 24 '13 at 14:35

$$ e^x=\lim_{m\rightarrow \infty}\left(1+\frac{1}{m}\right)^{mx} $$

Let $mx=n$, so $m=\frac{n}{x}$


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This cheats badly, unless you come up a meaning for powers with real exponent without using the exponential. Notice that $e^x$ is not defined as a power. – Mariano Suárez-Alvarez Mar 8 '14 at 6:10
My bad, no recuerdo haber leido la parte de que $e^x$ se definia como $\frac{d}{dx} e^x = e^x $. Aunque a pesar de eso no deja de ser "intuituva" :p – Alan Mar 11 '14 at 5:16
You need to have some definition of powers with real (non-rational) exponents for your argument to make sense. They can be defined in several ways, and you could sidestep the circularity using a definition which does not neet exponentials. – Mariano Suárez-Alvarez Mar 11 '14 at 17:20

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 $\lim_{h \to 0}\dfrac{g(1 + h)}{h} = 1$.

Next it can be easily proved that $g(xy) = g(x) + g(y), f(x + y) = f(x)f(y)$ and that if $a > 0$ and $n$ is positive integer then $a^{n} = f(ng(a))$. Its now a simple matter to show that $\lim_{n \to \infty}\left(1 + \dfrac{x}{n}\right)^{n} = f(x)$. We can proceed as follows:

$$ \begin{aligned}L &= \lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^{n}\\ &= \lim_{n \to \infty}f\left(ng\left(1 + \dfrac{x}{n}\right)\right)\\ &= f\left(\lim_{n \to \infty}ng\left(1 + \dfrac{x}{n}\right)\right)\text{ (by continuity of }f)\\ &= f\left(\lim_{h \to 0}\frac{x}{h}\cdot g(1 + h)\right)\text{ (by putting }h = x/n)\\ &= f(x)\end{aligned}$$

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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 $P_n(0) = 1$ : $$\textstyle P_n(x) = \ \ \color{red}{\left(1+\frac{x}{n}\right)^{n} \ \ \to \ \ e^x} \qquad\qquad \scriptstyle\text{uniformly on } [-A,A]$$ $$ $$
  • 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 $P_n(0) = 1$ : $$\textstyle P_n(x) = \ \ \color{red}{\sum_{k=0}^n \frac {x^k}{k!} \ \ \to \ \ e^x} \qquad\qquad \scriptstyle\text{uniformly on } [-A,A]$$
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First we prove that $b^{x'}=ab^x$, starting with the Taylor formula: $$f(x+\varepsilon)=f(x)+\varepsilon f'(x)$$ $$b^{x'}=\frac{b^{x+\varepsilon} -b^x}{\varepsilon}$$ $$b^{x'}=b^x\frac{b^\varepsilon-1}{\varepsilon}=ab^x$$ As $\varepsilon\rightarrow0$ the equation better resembles the continuous case, with $b$ as an arbitrary constant. If we set the whole fraction (i.e. $a$) to unity, $b^x$ equals its own derivative then (with $b=e$ and $\varepsilon=1/n$): $$\frac{e^{1/n}-1}{1/n}=1$$ $$e=(1+\frac{1}{n})^n$$ Note that $e$ is defined by setting $b^{x'}=b^{x}$: if $a$ is something other than one that will be the numerator of the fraction in the brackets. Also, note that although the Taylor formula can apply to finite cases people tend to consider the continuous case where $\varepsilon\rightarrow0$, and therefore $n\rightarrow\infty$. The immediate questions to arise from this are does $e$ have a value and how do we calculate it?

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