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

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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: – Three Apr 11 '13 at 22:56
@Three I suggest you read – 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

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)

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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 ( 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)

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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.

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

Another answer, assuming $x>0$:

Let $f(x)=\ln(x)$. Then we know that $f'(x)=1/x$. Also, by the definition of derivative, we can write $$ \begin{align} f'(x)&=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\\ &=\lim_{h\to 0}\frac{\ln(x+h)-\ln(x)}{h}\\ &=\lim_{h\to 0}\frac{1}{h}\ln\frac{x+h}{x}\\ &=\lim_{h \to 0}\ln\left(\frac{x+h}{x}\right)^\frac{1}{h}\\ &=\lim_{h\to 0}\ln\left(1+\frac{h}{x}\right)^\frac{1}{h} \end{align} $$ Then, using the fact that $\ln(x)$ is a continuous function for all $x$ in its domain, we can exchange the $\lim$ and $\ln$: $$ f'(x)=\ln\lim_{h\to 0}\left(1+\frac{h}{x}\right)^\frac{1}{h} $$ Now, let $m=1/h$. Then $m\to\infty$ as $h\to 0^+$, and $$ f'(x)=\ln\lim_{m\to\infty}\left(1+\frac{1}{mx}\right)^m $$ Now, assuming $x>0$, define $n=mx^2$, and so $n\to\infty$ as $m\to\infty$. Then we can write $$ f'(x)=\ln\lim_{n\to\infty}\left[\left(1+\frac{x}{n}\right)^n\right]^{1/x^2} $$ and from before, we still have $f'(x)=1/x$, so $$ \ln\lim_{n\to\infty}\left[\left(1+\frac{x}{n}\right)^n\right]^{1/x^2}=\frac{1}{x} $$ Exponentiating both sides, we find $$ \lim_{n\to\infty}\left[\left(1+\frac{x}{n}\right)^n\right]^{1/x^2}=e^{1/x} $$ Finally, raising both sides to the $x^2$, we find $$ \lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n=e^x $$

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$ (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 $\exp(x)$

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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 $\log(1+x)$.

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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. – user67133 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! – user67133 Apr 12 '13 at 0:07

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:

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

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Just intuition.

Step 1 : Expand ( 1 + x/n )^n using binomial theorem.

Result :

( 1 + x/n )^n = nC0 * 1 + nC1*(x/n) + nC2*(x/n)^2 + nC3*(x/n)^3 .......

          = 1 + n*(x/n) + (n*(n-1)/2!)*(x/n)^2 + (n*(n-1)*(n-2)/3!)*(x/n)^3 ...

Step 2 : Apply limits i.e. lim n→∞ (n*(n-1)*(n-2)...(n-k)) / (n ^ k ) = 1.

Step 3 : ( 1 + x/n )^n = 1 + x + (x^2)/2! + (x^3)/3! ......

Step 4 : We know e^x = 1 + x + (x^2)/2! + (x^3)/3!... (Series expansion)

Step 5 : Therefore , lim n→∞ ( 1 + x/n )^n = e^x

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This isn't rigorous enough. Just because every term in the sum converges to $x^n/n!$ doesn't necessarily mean that the whole sum converges to $\sum x^n/n!$. To deduce that you would have to argue that the sum converges uniformly on some interval. – Wojowu Nov 15 '15 at 19:23
Why ? Can u give a counter example – Saurabh Nov 17 '15 at 20:35
Suppose you have defined $a_n=\delta_{n0}+\delta_{n1}+...+\delta_{ni}+...$, where $\delta_{ij}$ is $1$ if $i=j$, $0$ otherwise. Then clearly each term $\delta_{ni}$ (for fixed $i$) is $0$ for all $n>i$, so $\delta_{ni}$ converges to $0$. But $a_n$ doesn't converge to $0+0+...+0+...=0$, because each $a_n=1$ (check it yourself). – Wojowu Nov 17 '15 at 21:18

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