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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
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@LordSoth Consider $x\mapsto 0$. –  Git Gud Apr 11 '13 at 22:46
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@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
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@Three I suggest you read www-history.mcs.st-and.ac.uk/HistTopics/e.html –  Lord Soth Apr 11 '13 at 22:59
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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):

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

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)

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

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 (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)

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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

$ (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
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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
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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 :).

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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,

$$\lim_{u\rightarrow0^+}f(u)=\lim_{u\rightarrow0^+}{x/(1+ux)\over1}=x$$

Now exponentiate $f$:

$$e^{f(u)}=(1+ux)^{1/u}$$

By continuity of the exponential function, we have

$$\lim_{u\rightarrow0^+}(1+ux)^{1/u}=\lim_{u\rightarrow0^+}e^{f(u)}=e^{\lim_{u\rightarrow0^+}f(u)}=e^x$$

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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