Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I was wondering if it is possible to get a link to a rigorous proof that $$\displaystyle \lim_{n\to\infty} \left(1+\frac {x}{n}\right)^n=\exp x$$

share|improve this question
Well often this is taken as the definition of exp(x), so I suppose it depends on your definition. –  Three Apr 11 '13 at 22:43
@LordSoth Consider $x\mapsto 0$. –  Git Gud Apr 11 '13 at 22:46
@LordSoth, actually that's false. $\exp(x)$ was originally discovered by a Bernoulli as the limit of compound interest -- in fact, exactly as the OP has written it. Only later was the calculus studied: en.wikipedia.org/wiki/Exponential_function –  Three Apr 11 '13 at 22:56
@Three I suggest you read www-history.mcs.st-and.ac.uk/HistTopics/e.html –  Lord Soth Apr 11 '13 at 22:59
How do you define $\exp$? This is really a matter of definition. What tools do you have available? Can you use continuity of $\exp$? Can you use $\log$? &c... Whenever you make this kind of questions, you must state what definitions and available tools are, always. Else we're just guessing what you want. –  Pedro Tamaroff Apr 11 '13 at 23:56

7 Answers 7

From the very definition (one of many, I know):


we can try the following, depending on what you have read so far in this subject:

(1) Deduce that

$$e=\lim_{n\to\infty}\left(1+\frac{1}{f(n)}\right)^{f(n)}\;,\;\;\text{as long as}\;\;f(n)\xrightarrow[n\to\infty]{}\infty$$

and then from here ($\,x\neq0\,$ , but this is only a light technicality)


2) For $\,x>0\,$ , substitute $\,mx=n\,$ . Note that $\,n\to\infty\implies m\to\infty\,$ , and

$$\left(1+\frac{x}{n}\right)^n=\left(\left(1+\frac{1}{m}\right)^m\right)^x\xrightarrow[n\to\infty\iff m\to\infty]{}e^x$$

I'll leave it to you to work out the case $\,x<0\,$ (hint: arithmetic of limits and "going" to denominators)

share|improve this answer

Firstly, let us give a definition to the exponential function, so we know the function has various properties:

$$ \exp(x) := \sum_{n=0}^{\infty} \frac{x^n}{n!}$$

so that we can prove that (as exp is a power series) :

  • The exponential function has radius of convergence $\infty$, and is thus defined on all of $\mathbb R$
  • As a power series is infinitely differentiable inside its circle of convergence, the exponential function is infinitely differentiable on all of $\mathbb R$
  • We can then prove that the function is strictly increasing, and thus by the inverse function theorem (http://en.wikipedia.org/wiki/Inverse_function_theorem) we can define what we know as the "log" function

Knowing all of this, here is hopefully a sufficiently rigorous proof (at least for positive a):

As $\log(x)$ is continuous and differentiable on $(0,\infty)$, we have that $\log(1+x)$ is continuous and differentiable on $[0,\frac{a}{n}]$, so by the mean value theorem we know there exists a $c \in [0,\frac{a}{n}]$ with

$$f'(c) = \frac {\log(1+ \frac{a}{n} ) - \log(1)} {\frac {a}{n} - 0 } $$ $$ \Longrightarrow \log[{(1+\frac{a}{n})^n}] = \frac{a}{1+c}$$ $$ \Longrightarrow (1+\frac{a}{n})^n = \exp({\frac{a}{1+c}})$$

for some $c \in [0,\frac{a}{n}]$ . As we then want to take the limit as $n \rightarrow \infty$, we get that:

  • As $c \in [0,\frac{a}{n}]$ and $\frac{a}{n} \rightarrow 0$ as $n \rightarrow \infty$, by the squeeze theorem we get that $ c \rightarrow 0$ as $n \rightarrow \infty$
  • As $ c \rightarrow 0$ as $n \rightarrow \infty$, $\frac{a}{1+c} \rightarrow a$ as $n \rightarrow \infty$
  • As the exponential function is continuous on $\mathbb R$, the limit can pass inside the function, so we get that as $\frac{a}{1+c} \rightarrow a$ as $n \rightarrow \infty$

$$ \exp(\frac{a}{1+c}) \rightarrow \exp(a) $$ as $n \rightarrow \infty$. Thus we can conclude that

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

(Of course, this is ignoring that one needs to prove that $\exp(a)=e^a$, but this is hardly vital for this question)

share|improve this answer
If we're just about to define the exponential function (or at least show that it equals something), it seems to me the assumption of its continuity is highly suspicious... –  DonAntonio Apr 11 '13 at 23:36
This is true - although I can't see how this proof is nothing more than showing that the various definitions of the exponential function are equilivant, and thus I would presume continuity would have been proved before trying to prove statements such as this one (for example, in our lectures we defined it in terms of a power series, which means that we can prove it is continuous fairly straightforwardly) –  Andrew D Apr 11 '13 at 23:39
I agree with that, @Andrew D, but then perhaps mentioning some other definition from which continuity follows and then use that in it...perhaps too long a detour for a beginner, but absolutely possible indeed. –  DonAntonio Apr 11 '13 at 23:42
@DonAntonio The log's continuity assumption is just fine, though. Since $\exp$ is its inverse, it is continuous. –  Pedro Tamaroff Apr 11 '13 at 23:50
Yeah, thankfully that is covered by the inverse function theorem (which I've now linked/discussed above, along with some other things) –  Andrew D Apr 11 '13 at 23:51

Consider the functions $u$ and $v$ defined for every $|t|\lt\frac12$ by $$ u(t)=t-\log(1+t),\qquad v(t)=t-t^2-\log(1+t). $$ The derivative of $u$ is $u'(t)=\frac{t}{1+t}$, which has the sign of $t$, hence $u(t)\geqslant0$. The derivative of $v$ is $v'(t)=1-2t-\frac{1}{1+t}$, which has the sign of $(1+t)(1-2t)-1=-t(1+2t)$ which has the sign of $-t$ on the domain $|t|\lt\frac12$ hence $v(t)\leqslant0$. Thus, for every $|t|\lt\frac12$, $$ t-t^2\leqslant\log (1+t)\leqslant t. $$ The function $z\mapsto\exp(nz)$ is nondecreasing on the same domain hence $$ \exp\left(nt-nt^2\right)\leqslant(1+t)^n\leqslant\exp\left(nt\right). $$ In particular, using this for $t=x/n$, one gets that, for every $n\gt2|x|$, $$ \exp\left(x-\frac{x^2}{n}\right)\leqslant\left(1+\frac{x}n\right)^n\leqslant\mathrm e^x. $$ Finally, $x^2/n\to 0$ and the exponential is continuous, hence we are done.

Facts/Definitions used: the logarithm has derivative $t\mapsto1/t$, and the exponential is the inverse of the logarithm.

share|improve this answer
We need to evangelize the use of $\leqslant$ and $\geqslant$ in MSE. –  Pedro Tamaroff Aug 10 '13 at 4:11

$ (1+x/n)^n = \sum_{k=0}^n \binom{n}{k}\frac{x^k}{n^k} $

Now just prove that $\binom{n}{k}\frac{x^k}{n^k}$ approaches $\frac{x^k}{k!}$ as n approaches infinity, and you will have proven that your limit matches the Taylor series for $\exp(x)$

share|improve this answer
This is not enough; there are infinitely many terms, so you need to show that you can swap two limits here. –  Qiaochu Yuan Apr 11 '13 at 23:17
What you want to do is work with $\limsup$ and $\liminf$ here, and show $e^x\leq\liminf $ and $e^x\geq \limsup$ –  Pedro Tamaroff Apr 11 '13 at 23:53

This one of the ways in which it is defined. The equivalence of the definitions can be proved easily, I guess. If for example you take the exponential function to be the inverse of the logarithm:

$\log(\lim_n(1 + \frac{x}{n})^n) = \lim_n n \log(1 + \frac{x}{n}) = \lim_n n \cdot[\frac{x}{n} - \frac{x^2}{2n^2} + \dots] = x$

EDIT: The logarithm is defined as usual: $\log x = \int_1^x \frac{dt}{t}$. The first identity follows from the continuity of the logarithm, the second it's just an application of one of the property of the logarithm ($\log a^b = b \log a $), while to obtain the third it sufficies to have the Taylor expansion of $\log(1+x)$.

share|improve this answer
The very first equality requires, me believes, a justification that I cannot see as very easy unless we already assume quite a bit (say, continuity...). After that things get even tougher as we need power series and then also, apparently, differentiation. –  DonAntonio Apr 11 '13 at 23:40
The logarithm is defined as $\int_1^x \frac{dt}{t}$, therefore, if we have integration we can also have continuity and differentiation, I suppose. –  user01123581321345589144... Apr 11 '13 at 23:45
Perhaps so and also perhaps mentioning this could clear things out a little, since we don't know, apparently, what the OP's background is. –  DonAntonio Apr 11 '13 at 23:47
I cannot but totally agree. Thank you for your suggestions, I am going to edit the post to make it clearer! –  user01123581321345589144... Apr 12 '13 at 0:07

Thank you for your answers! I have found one solution, but, written in French. exponential. In particular, II-1, 2, 3. Solutions are here: solution It may be almost understood, since this is math :).

share|improve this answer
Did you bother to accept an answer? –  user93957 Dec 7 '13 at 14:56

For any fixed value of $x$, define

$$f(u)= {\ln(1+ux)\over u}$$

By L'Hopital's Rule,


Now exponentiate $f$:


By continuity of the exponential function, we have


All these limits have been shown to exist for the (positive) real variable $u$ tending to $0$, hence they must exist, and be the same, for the sequence of reciprocals of integers, $u=1/n$, as $n$ tends to infinity, and the result follows:

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

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.