Intuitive proofs that $\lim\limits_{n\to\infty}\left(1+\frac xn\right)^n=e^x$

Question

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.

Answer 1

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 $f(0) = 1$.

Answer 2

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

Answer 3

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.

Answer 4

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!

Answer 5

$f(x) = e^x$ 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

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

Answer 6

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

Answer 7

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."

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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 $\frac{ng(n)}{1-ng(n)}\to 0$.

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

Answer 8

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

Answer 9

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

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.

Answer 10

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.

Answer 11

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

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

Answer 12

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

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

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

Answer 13

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

Answer 14

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}