I knew that $e^x=\lim \limits_{n\to+\infty }{\left(1+\frac{x}{n}\right)^n}$. But I've never seen its proof. So I tried to prove it using $\exp(\ln x)=\ln(\exp(x))=x$. Here is what I've tried so far :

$$ \left(1+\frac{x}{n}\right) ^n=e^{n\ln(1+\frac{x}{n})}$$ $$\text{I'll now study just } {n\ln\left(1+\frac{x}{n}\right)}.$$$$ \text{If this function has the line }y=x \text{ as oblique asymptote, then the equality is proven.}$$

$$ n\ln\left(1+\frac{x}{n}\right) = n\ln\left(\frac{n+x}{n}\right)$$

$$=n[\ln(n+x)-ln(n)]$$ $$=n\left[\int_1^{n}\frac{dt}{t}+\int_{n}^{x}\frac{dt}{t}-\int_1^{n}\frac{dt}{t}\right]$$ $$=n[\ln(x)-\ln(n)]$$

But I just don't know how to show that this expression has an oblique asymptote $y=x$. I've thought that if there is an oblique asymptote as $n$ goes to infinity, than for a huge $n$, we have :

$$\ln\left(1+\frac{x}{n}\right)\approx \frac{x}{n}\approx0$$ Which looks correct but we could have any other function $f(x)$, $\ln\left(1+\frac{x}{n}\right)\approx\frac{f(x)}{n}\approx 0$. Which doesn't prove the oblique asymptote because $x$ is constant.

So how can prove $e^x=\lim \limits_{n\to +\infty } \left(1+\frac{x}{n}\right)^n$? And where did I go wrong?

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    $\begingroup$ What is your definition of $e^x$? $\endgroup$ – Ink Jun 14 '13 at 6:39
  • 3
    $\begingroup$ What is your definition of $\exp (x)$? $\endgroup$ – Git Gud Jun 14 '13 at 6:40
  • $\begingroup$ @moray95: Your integral should be $\int_1^{nx}\frac{dt}{t}+ \int_{nx}^{nx+1} \frac{dt}{t}- \int_1^n \frac{dt}{t}$. $\endgroup$ – Seirios Jun 14 '13 at 6:41
  • $\begingroup$ I'm using the definition $\exp(\ln(x))=\ln(\exp(x))=x$ $\endgroup$ – moray95 Jun 14 '13 at 6:50
  • 1
    $\begingroup$ Are you defining $\exp$ as the inverse of $\ln$, is that it? $\endgroup$ – Git Gud Jun 14 '13 at 6:53

11 Answers 11


I don't know if it helps you, it is just a suggestion, if you know the fundamental limite: $$\lim_{n\to \infty}\left(1+\frac{1}{n}\right)^n=e$$ Then you have for $$\lim_{n\to \infty}\left(1+\frac{x}{n}\right)^n$$ replacing $k=\frac{n}{x}$ we get $$\lim_{n\to \infty}\left(1+\frac{1}{k}\right)^{kx}= \left(\lim_{k\to \infty}\left(1+\frac{1}{k}\right)^{k}\right)^x =e^x$$

  • $\begingroup$ I've thought about something like that but then I'll need to prove first $\lim_{n\to \infty}(1+\frac{1}{n})^n=e$. $\endgroup$ – moray95 Jun 15 '13 at 10:23
  • $\begingroup$ @moray95 This is often taken as the definition of $e$. The only other definition I know of is the power series, which you can show is equal to this expression using binomial expansion. $\endgroup$ – Slade Jan 7 '14 at 20:44

Start with the functions $$ f_n(x) = \left(1 + \frac{x}{n}\right)^n. $$ Then $$ f'_n(x) = \left(1 + \frac{x}{n}\right)^{n-1} = \left(1 + \frac{x}{n}\right)^{-1}f_n(x) $$ If we take the limit and call $f(x) = \lim_{n\rightarrow\infty}f_n(x)$, then $$ f'(x) = f(x) $$ and $f(0) = 1$. This first-order ODE has the unique solution $f(x) = e^x$.

  • $\begingroup$ Nice answer! I like it :) $\endgroup$ – AIM_BLB Oct 20 '14 at 16:48

Let's start with the "where did I go wrong" part of the question. Where you wrote


you should have written


Note, the correct upper limit in the middle integration is $n+x$, not just $x$. Otherwise you were on the right track. The corrected integral leaves us with

$$n\ln\left(1+\frac{x}{n}\right)=n\int_n^{n+x}{dt\over t}$$

Now as soon at $n\gt x$, we have

$${1\over n+|x|}\le{1\over t}\le{1\over n-|x|}$$

on the interval $t\in[n-|x|,n+|x|]$, which certainly includes the interval between $n$ and $n+x$. If you are careful with the minus signs, you can conclude that

$${nx\over n+x}\le n\int_n^{n+x}{dt\over t}\le{nx\over n-x}$$

and it now follows easily that



If you're allowed to use Taylor (power) expansions this is pretty simple:

$$n\log\left(1+\frac xn\right)=n\sum_{k=1}^\infty (-1)^{k+1}\frac{x^k}{k\,n^k}=n\left(\frac xn+\mathcal O\left(\frac1{n^2}\right)\right)=$$

$$=x+\mathcal O\left(\frac1n\right)\xrightarrow[n\to\infty]{}x$$

  • 2
    $\begingroup$ Great response but I'd like to prove it without the power series... $\endgroup$ – moray95 Jun 14 '13 at 15:51
  • 1
    $\begingroup$ Ok...did you read my comment under your question? Because you haven't yet corrected your work there... $\endgroup$ – DonAntonio Jun 14 '13 at 15:53
  • $\begingroup$ Edited my question with your remark but still, I still have the same thing at the end... $\endgroup$ – moray95 Jun 14 '13 at 17:07

If you already know the exponential function, then you probably also know the inequality $e^x\ge 1+x$. Then also


Now $e^x=(e^{\frac xn})^n$, so for $n>|x|$

$$ \left(1+\tfrac xn\right)^n \le e^x \le \left(1+\tfrac{x}{n}\right)^n\frac1{\left(1-\tfrac{x^2}{n^2}\right)^n}\le\left(1+\tfrac{x}{n}\right)^n\frac1{1-\tfrac{x^2}{n}} $$

The last inequality is by the Bernoulli inequality.


First, let's prove that : $$(1)\space\space e = \lim_{n \to \infty} \bigg (1+\frac{1}{n} \bigg)^n$$

For that, we need to know : $$(2)\space\space\lim_{x \to 0} \frac{ln(1 + x)}{x}$$

Let's say : $\forall x > -1,\space f(x) = ln(1 + x)$

$f(0) = ln(1) = 0$, so : $$\frac{ln(1+x)}{x} = \frac{f(x) - f(0)}{x} = \frac{f(0+x) - f(0)}{(0 + x) - 0}$$

By using the definition of the derivative, $$\lim_{x \to 0} \frac{ln(1 + x)}{x} = \lim_{x \to 0} \frac{f(0+x) - f(0)}{(0 + x) - 0} = f'(0)$$

$$\forall x > -1,\space f'(x) = \frac{1}{1+x}\space And\space f'(0) = 1$$

Finally, $$\lim_{x \to 0} \frac{ln(1 + x)}{x} = 1$$

Now (2) is computed, let's prove (1) :

Let's say : $\forall x \neq 0, \space y=\frac{1}{x}$

So : $$x = \frac{1}{y}$$ $x \to 0 \implies y \to \infty$ $$\lim_{x \to 0} \frac{ln(1 + x)}{x} = \lim_{y \to \infty} \frac{ln(1 + \frac{1}{y})}{\frac{1}{y}} = \lim_{y \to \infty} y \space ln\bigg(1 + \frac{1}{y}\bigg)=\lim_{y \to \infty} ln\bigg(1+\frac{1}{y}\bigg)^y$$

$$\lim_{y \to \infty} ln\bigg(1+\frac{1}{y}\bigg)^y = \lim_{x \to 0} \frac{ln(1 + x)}{x} = 1$$

$$(3)\space\space\lim_{y \to \infty} ln\bigg(1+\frac{1}{y}\bigg)^y = 1$$

Let's say exp the inverse function of ln : $$exp(1) = \lim_{y \to \infty} \bigg(1+\frac{1}{y}\bigg)^y$$

Let's say $e = exp(1)$, and replace y by n, we finally prove : $$(1)\space\space e = \lim_{n \to \infty} \bigg (1+\frac{1}{n} \bigg)^n$$

(1) could also be written : $$(1)\space\space e = \lim_{f(n) \to \infty} \bigg (1+\frac{1}{f(n)} \bigg)^{f(n)}$$ Let's say : $\forall x \neq 0$, $f(n) = \frac{n}{x}$

Let's separate the proof in 3 cases : (a) $x>0$, (b) $x<0$ and (c) $x=0$

Case (a) : $x>0$ so $f(n) \to \infty \implies n \to \infty$ $$e = \lim_{n \to \infty} \bigg (1+\frac{1}{\frac{n}{x}} \bigg)^{\frac{n}{x}} = \lim_{n \to \infty} \bigg (1+\frac{x}{n} \bigg)^{\frac{n}{x}}$$

By raising at power x : $$\forall x>0, \space e^x = \lim_{n \to \infty} \bigg (1+\frac{x}{n} \bigg)^{n}$$

Case (b) : $x<0$ so $f(n) \to \infty \implies n \to -\infty$ $$e = \lim_{n \to -\infty} \bigg (1+\frac{1}{\frac{n}{x}} \bigg)^{\frac{n}{x}} = \lim_{n \to -\infty} \bigg (1+\frac{x}{n} \bigg)^{\frac{n}{x}}$$

Let's say $m = -n$

$n \to -\infty \implies m \to \infty$

$$e = \lim_{m \to \infty} \bigg (1+-\frac{1}{\frac{m}{x}} \bigg)^{-\frac{m}{x}} = \lim_{m \to \infty} \bigg (1-\frac{x}{m} \bigg)^{\frac{m}{-x}}$$

By raising at power -x : $$e^{-x} = \lim_{m \to \infty} \bigg (1-\frac{x}{m} \bigg)^{m}$$

By replacing -x by x and m by n :

$$\forall x < 0, \space e^{x} = \lim_{n \to \infty} \bigg (1+\frac{x}{n} \bigg)^{n}$$

(1) and (2) implies :

$$\forall x \neq 0, \space e^{x} = \lim_{n \to \infty} \bigg (1+\frac{x}{n} \bigg)^{n}$$ Case (c) : $e \neq 0 \implies e^0 = 1$

Let's see if the formula valid for $x \neq 0$ is also valid for $x=0$ :

$$\space e^{0} = \lim_{n \to \infty} \bigg (1+\frac{0}{n} \bigg)^{n} = \lim_{n \to \infty} 1^n = 1$$

We just proved:

$$\forall x \in \mathbb{R}, \space e^{x} = \lim_{n \to \infty} \bigg (1+\frac{x}{n} \bigg)^{n}$$


Here's a way to do it with the integral: \begin{align*} n\log{\left(1+{x\over n}\right)} & = n\int_1^{1+{x/ n}}{dt\over t} \end{align*} and by the simplest conceivable estimate \begin{align*} {x\over 1+x/n} = n{x/n\over1+{x/ n}}\leq n\int_1^{1+{x/ n}}{dt\over t} \leq n{x\over n} = x. \end{align*} (The inequalities hold when $x$ is negative too, provided the expressions are defined.) Now make $n\to\infty$ and apply the squeeze theorem.


$f(x) = e^x$ 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.

$f(0) = 1, f'(0)=1 \implies f(\frac{a}{n}) \approx 1+\frac{a}{n}$

$f(\frac{a}{n}) \approx 1+\frac{a}{n}, f'(\frac{a}{n}) \approx 1+\frac{a}{n} \implies f(\frac{2a}{n}) \approx 1+\frac{a}{n} + \frac{a}{n}(1+\frac{a}{n}) = (1+\frac{a}{n})^2$




$f(a) \approx (1+\frac{a}{n})^n$

Since Euler's method actually converges in the limit, we have

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

  • $\begingroup$ p.s. If you think long and hard about this same proof, only applied to the line segment from $0$ to $i$ in the complex plane, you will understand why $e^{i\pi}=-1$ $\endgroup$ – Steven Gubkin Jan 7 '14 at 19:54

Let's say you define $e^x$ by the derivative $\frac{d}{dx} e^x=e^x$. Then, we try the same to what you have here. $\frac{d}{dx}(1+\frac{x}{n})^n=(1+\frac{x}{n})^{n-1}$. And, as n approaches infinity, n and n-1 essentially boils down to the same thing. Then, $\frac{d}{dx}(1+\frac{x}{n})^n=(1+\frac{x}{n})^n$.


Since $X=e^{\log X}$, we can write $$\lim_{n \to \infty} \bigg(1+\frac{x}{n}\bigg )^n=e^{\log\lim_{n \to \infty} \bigg(1+\frac{x}{n}\bigg )^n} =e^{\lim_{n \to \infty} n\log \bigg(1+\frac{x}{n}\bigg )}$$

Notice that the exponent on $e$ is an indeterminate form of type $\infty\cdot 0$, so rewrite the limit in the exponent as $\frac{\log \big(1+\frac{x}{n}\big)}{\frac{1}{n}}$ and use L'Hopital's Rule to show the the limit in the exponent of $e$ is indeed $x$.


As shown by @math_man, the limit is indeed the exponential of a numerical constant, with

$$\lim \limits_{n\to+\infty }{\left(1+\frac{x}{n}\right)^n}=\lim \limits_{nx\to+\infty }{\left(1+\frac{x}{nx}\right)^{nx}}=\left(\lim \limits_{n\to+\infty }{\left(1+\frac1n\right)^n}\right)^x=e^x.$$

We can compute the constant by the Binomial theorem,

$$\left(1+\frac1n\right)^n=1+\frac nn+\frac{n(n-1)}{2n^2}+\frac{n(n-1)(n-2)}{3!n^3}\cdots+\frac1{n^n},$$ which becomes $$e=1+1+\frac12+\frac1{3!}+\frac1{4!}\cdots$$at infinity.


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