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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    $\begingroup$ How do you intuitively define $e^x$? $\endgroup$ – Thomas Andrews Apr 17 '13 at 23:51
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    $\begingroup$ 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 $1/x$? $\endgroup$ – Henry Swanson Apr 17 '13 at 23:52
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    $\begingroup$ @HenrySwanson: I suppose, for this question, you get to choose your favourite... $\endgroup$ – Aryabhata Apr 18 '13 at 0:16
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    $\begingroup$ 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". $\endgroup$ – Michael Hardy Apr 18 '13 at 0:25
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    $\begingroup$ @GuyFsone : Maybe you need to read more carefully. Look at the two words that are set in bold type in this question. This is not a duplicate of the question to which you link. $\endgroup$ – Michael Hardy Nov 10 '17 at 22:12

14 Answers 14


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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    $\begingroup$ What's the point of "intimidate" on the first equality? $\endgroup$ – EuYu Apr 18 '13 at 0:15
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    $\begingroup$ @EuYu: I'm guessing it's a reference to proof by intimidation $\endgroup$ – Hurkyl Apr 18 '13 at 0:16
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    $\begingroup$ @EuYu, You cannot exchange limit and differentiation without proper justification. $\endgroup$ – Lord Soth Apr 18 '13 at 0:18
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    $\begingroup$ (Euler will be proud). :-) $\endgroup$ – Aryabhata Apr 18 '13 at 0:20
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    $\begingroup$ Maybe Euler is the original author of this one. $\endgroup$ – 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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    $\begingroup$ +1 (as $\log(1+\epsilon) = \epsilon + O(\epsilon^2)$ for small $\epsilon$...) $\endgroup$ – Lord Soth Apr 18 '13 at 0:00
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    $\begingroup$ This is a rigorous proof :-). +1 though. $\endgroup$ – Aryabhata Apr 18 '13 at 0:12
  • $\begingroup$ 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. $\endgroup$ – Michael Hardy Apr 18 '13 at 0:13
  • $\begingroup$ 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,... $\endgroup$ – Lord Soth Apr 18 '13 at 0:28
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    $\begingroup$ Even simpler? $e^{x/n}\sim_\infty1+\frac{x}{n}$ and raise both sides to $n$? $\endgroup$ – 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.

  • $\begingroup$ Nice. You could also found that $f(ax)=f(x)^a$ $\endgroup$ – leonbloy Apr 18 '13 at 15:37
  • $\begingroup$ @leonbloy: Right, that works too, but I guess an answer quite similar to that was added already.. $\endgroup$ – Aryabhata Apr 18 '13 at 15:48
  • $\begingroup$ 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. $\endgroup$ – Thomas Andrews Mar 7 '14 at 23:26
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    $\begingroup$ Being intuitive doesn't allow it to use flawed intuition. This intuition can give some very wrong results... $\endgroup$ – Thomas Andrews Mar 8 '14 at 3:02
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    $\begingroup$ @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. $\endgroup$ – 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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    $\begingroup$ I'm not sure it counts as "intuitive" either :) $\endgroup$ – Thomas Andrews Apr 17 '13 at 23:56
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    $\begingroup$ 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". $\endgroup$ – EuYu Apr 17 '13 at 23:57
  • $\begingroup$ @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. $\endgroup$ – Marc van Leeuwen Apr 18 '13 at 17:33
  • $\begingroup$ @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. $\endgroup$ – EuYu Apr 18 '13 at 17:50
  • $\begingroup$ @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 $\lim\{(1+x/n)^n\}$. $\endgroup$ – 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 $\frac{1}{2}\frac{x}{n}x$.


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 $x$, $$\cos \frac{x}{n} +i\sin \frac{x}{n} = 1+\frac{ix}{n}+O(1/n^2)$$


$f(x) = e^x$ 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

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


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


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

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.


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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    $\begingroup$ But this assumes the special case in which $x=1$, which is nearly as substantial as the generalization that this argument establishes. $\endgroup$ – Michael Hardy Apr 18 '13 at 0:12
  • $\begingroup$ @MichaelHardy I think this goes back to how you actually define $e$. What would you take as the most intuitive definition? $\endgroup$ – EuYu Apr 18 '13 at 0:16
  • $\begingroup$ @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?) $\endgroup$ – Glen O Apr 18 '13 at 0:18
  • $\begingroup$ 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 $x=1$. $\endgroup$ – Michael Hardy Apr 18 '13 at 0:24
  • $\begingroup$ 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. $\endgroup$ – 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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    $\begingroup$ 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. $\endgroup$ – 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 $m=\frac{n}{x}$


  • $\begingroup$ 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. $\endgroup$ – Mariano Suárez-Álvarez Mar 8 '14 at 6:10
  • $\begingroup$ 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 $\endgroup$ – Alan Mar 11 '14 at 5:16
  • $\begingroup$ 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. $\endgroup$ – Mariano Suárez-Álvarez 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 $$\lim_{h \to 0}\frac{g(1 + h)}{h} = 1\tag{1}$$

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}

  • $\begingroup$ Here, you have two brackets in the last few equations $((\ldots ))$, but in the following link, you use brackets and then braces, namely, $\{(\ldots )\}$ like Ramanujan. math.stackexchange.com/questions/540816/… I am finding it difficult to decide which one I will use. Which one is best, in your honest opinion, and why? And also, under which circumstance is the latter notation used? Thanks :) $\endgroup$ – Feeds May 13 '18 at 8:40
  • $\begingroup$ @user477343: well, it does not matter which one you use. The idea is that the innermost brackets/braces are handled first and then the next innermost. The typical practice is to nest them like $[\{(\dots) \}]$ but this is not necessary and you may use the same symbol $($ or $\{$ or $[$ everywhere. $\endgroup$ – Paramanand Singh May 13 '18 at 13:40
  • $\begingroup$ Thank you for letting me know :) $\endgroup$ – Feeds May 13 '18 at 14:12

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 $P_n(0) = 1$ : $$\textstyle P_n(x) = \ \ \color{red}{\sum_{k=0}^n \frac {x^k}{k!} \ \ \to \ \ e^x} \qquad\qquad \text{uniformly on } [-A,A]$$

protected by Alex M. Nov 11 '16 at 13:55

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