# 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.

• 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
• @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. – Michael Hardy Nov 10 '17 at 22:12

## 14 Answers

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 $f(0) = 1$.

• What's the point of "intimidate" on the first equality? – EuYu Apr 18 '13 at 0:15
• @EuYu: I'm guessing it's a reference to proof by intimidation – user14972 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 $$\left(1+\frac xn\right)^n=e^{n\log\left(1+\frac xn\right)}\sim_\infty e^{n\times \frac xn}=e^x$$

• +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 $f(x) = \lim_{n \to \infty} f_n(x)$

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.

• 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

$$1+\binom{n}{1}\frac{x}{n}+\binom{n}{2}\frac{x^2}{n^2}+\binom{n}{3}\frac{x^3}{n^3}+\cdots$$

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!

• 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

$f(x) = e^x$ 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

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

• I remember attending a course in quantum physics, some 10 years ago, in which the teacher did this, but without justifying the convergence (I am a mathematician). I liked it a lot, because it applies equally well to the case of $a$ being an operator. But I couldn’t make it rigorous. So I like this answer very much, for bringing back a nice memory WITH a rigorous proof! – Giuseppe Negro Feb 13 at 0:32

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

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

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.

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.

• 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

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}

• 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

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

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

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

• 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-Álvarez 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-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}

• 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 :) – Mr Pie May 13 '18 at 8:40
• @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. – Paramanand Singh May 13 '18 at 13:40
• Thank you for letting me know :) – Mr Pie 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 $$P_n(0) = 1$$ : $$\textstyle P_n(x) = \ \ \color{red}{\sum_{k=0}^n \frac {x^k}{k!} \ \ \to \ \ e^x} \qquad\qquad \text{uniformly on } [-A,A]$$