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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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    $\begingroup$ Well often this is taken as the definition of exp(x), so I suppose it depends on your definition. $\endgroup$
    – Three
    Commented Apr 11, 2013 at 22:43
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    $\begingroup$ @LordSoth Consider $x\mapsto 0$. $\endgroup$
    – Git Gud
    Commented Apr 11, 2013 at 22:46
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    $\begingroup$ maybe you want $\lim \left ( 1+ \frac{1}{n}\right )^n=e$ then it implies $\lim \left (1 +\frac{x}{n} \right )^n=e^x$? $\endgroup$
    – clark
    Commented Apr 11, 2013 at 22:47
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    $\begingroup$ @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 $\endgroup$
    – Three
    Commented Apr 11, 2013 at 22:56
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    $\begingroup$ 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. $\endgroup$
    – Pedro
    Commented Apr 11, 2013 at 23:56

14 Answers 14

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I would like to cite here an awesome German mathematician, Konrad Königsberger. He writes in his book Analysis I as follows:

Fundamental lemma. For every sequence of complex numbers $w_n$ with a limit $w$ it is true that $$\lim_{n \to \infty} \Bigl(1 + \frac{w_n}{n}\Bigr)^n = \sum_{k=0}^\infty \frac{w^k}{k!}.$$ Proof. For every $\varepsilon > 0$ and sufficiently large index $K$ we have the following estimations: $$\sum_{k=K}^\infty \frac{(|w|+1)^k}{k!} < \frac \varepsilon 3 \quad\mbox{and}\quad |w_n| \le |w|+1.$$Therefore if $n \ge K$ then $$\left|\Bigl(1 + \frac{w_n}{n}\Big)^n - \exp w \right| \le \sum_{k=0}^{K-1} \left|{n \choose k}\frac{w_n^k}{n^k} - \frac{w^k}{k!}\right| + \sum_{k=K}^n{n\choose k} \frac{|w_n|^k}{n^k} + \sum_{k=K}^\infty \frac{|w|^k}{k!}.$$ The third sum is smaller than $\varepsilon / 3$ based on our estimations. We can find an upper bound for the middle one using $${n \choose k} \frac 1 {n^k} = \frac{1}{k!} \prod_{i = 1}^{k-1} \Bigl(1 - \frac i n \Bigr) \le \frac 1 {k!}.$$ Combining this with $|w_n| \le |w| + 1$, $$\sum_{k=K}^n {n \choose k} \frac{|w_n|^k}{n^k} < \sum_{k=K}^n \frac{(|w|+1)^k}{k!} < \frac \varepsilon 3$$ Finally, the first sum converges to $0$ due to $w_n \to w$ and ${n \choose k} n^{-k} \to \frac 1 {k!}$. We can choose $N > K$ such that it's smaller than $\varepsilon / 3$ as soon as $n > N$.

Really brilliant.

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    $\begingroup$ I examined proof technique for $w_n = w$ (no sequence) then went full bore. Agree - brilliant! I can use it to show the exp power series takes addition to multiplication. Did not have to get in the weeds with rearrangements, absolute convergence/ commutativity, etc. $\endgroup$ Commented Jul 9, 2017 at 23:10
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    $\begingroup$ +1.... This appears to be the only complete and rigorous answer ( so far) that considers all complex $w$. And provides a reference (as the proposer requested). And actually has more than what was asked... A masterful exposition by K.K. $\endgroup$ Commented Aug 30, 2018 at 15:49
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    $\begingroup$ A simpler proof of this Fundamental Lemma, for the case $w_n$ are all real, can be found at Proposition 4.6.1 in Koski's Lecture Notes on Probability and Random Processes (math.kth.se/matstat/gru/sf2940/lectnotemat5.pdf). $\endgroup$ Commented Feb 16, 2020 at 16:37
  • $\begingroup$ ( on page 133 ) $\endgroup$ Commented Feb 16, 2020 at 16:44
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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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  • $\begingroup$ Side note: The difficult part here is deducing the first limit holds over real $n$, opposed to just natural $n$. $\endgroup$ Commented Aug 10, 2019 at 15:01
  • $\begingroup$ @SimplyBeautfulArt It's not that difficult. For anyone who wants to learn more about defining real exponents with limits, I recommend Paramanand Singh's blog post. $\endgroup$
    – Poder Rac
    Commented Aug 13, 2020 at 15:24
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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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  • $\begingroup$ 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... $\endgroup$
    – DonAntonio
    Commented Apr 11, 2013 at 23:36
  • $\begingroup$ 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) $\endgroup$
    – Andrew D
    Commented Apr 11, 2013 at 23:39
  • $\begingroup$ 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. $\endgroup$
    – DonAntonio
    Commented Apr 11, 2013 at 23:42
  • $\begingroup$ @DonAntonio The log's continuity assumption is just fine, though. Since $\exp$ is its inverse, it is continuous. $\endgroup$
    – Pedro
    Commented Apr 11, 2013 at 23:50
  • $\begingroup$ Yeah, thankfully that is covered by the inverse function theorem (which I've now linked/discussed above, along with some other things) $\endgroup$
    – Andrew D
    Commented Apr 11, 2013 at 23:51
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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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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 $$ \lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n=e^x $$ EDIT: This idea actually works for all reals—if we use $f(x)=\ln|x|$ instead, then we get eventually get: $$ e^x=\lim_{n\to\infty}\left|1+\frac{x}{n}\right|^{n}=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n $$ Where the last equality come from the fact that $n$ always eventually dominates $x$, so that the absolute value function becomes redundant.

This leaves the case where $x=0$, but that is a trivial matter.

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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:

For every $|x|<\frac12n$, $$ \exp\left(x-\frac{x^2}{n}\right)\leqslant\left(1+\frac{x}n\right)^n\leqslant\mathrm e^x. $$

Finally, $x^2/n\to 0$ when $n\to\infty$ and the exponential is continuous at $0$, hence we are done.

Facts/Definitions used:

  • The logarithm has derivative $t\mapsto1/t$.
  • The exponential is the inverse of the logarithm.
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  • $\begingroup$ We need to evangelize the use of $\leqslant$ and $\geqslant$ in MSE. $\endgroup$
    – Pedro
    Commented Aug 10, 2013 at 4:11
  • $\begingroup$ I used this in an application to lower bound $(1+x/n)^n$, thank you. $\endgroup$ Commented Aug 16, 2016 at 11:55
  • $\begingroup$ Didier, I like this approach. So (+1). I was wondering if you've seen a way to establish the same lower bound for $\left(1+\frac xn\right)^n$ by using the limit definition of the exponential function and without appealing to calculus. The upper bound is trivial for $x>-n$. $\endgroup$
    – Mark Viola
    Commented Jan 5, 2017 at 19:39
  • $\begingroup$ @Dr.MV This reduces to showing $1+t\geqslant \exp(t-t^2)$, that is, $\frac1{1+t}\leqslant\exp(-t+t^2)$. What you call the trivial upper bound yields $\frac1{1+t}=1-\frac{t}{1+t}\leqslant\exp\left(-\frac{t}{1+t}\right)$ hence if $\frac{t}{1+t}\geqslant t-t^2$, we are done. This is asking that $t\geqslant (t-t^2)(1+t)=t(1-t^2)$ hence indeed, we are done for every $t$ in $(0,1)$. This fails for $t$ negative but similar arguments might work. $\endgroup$
    – Did
    Commented Jan 8, 2017 at 9:17
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Let $t\in[1,1+\frac{1}{n}]$, and we have: $$\frac{1}{1+\frac{1}{n}}\leq \frac{1}{t}\leq 1$$ $$\int_1^{1+\frac{x}{n}}\frac{1}{1+\frac{1}{n}}dt\leq \int_1^{1+\frac{x}{n}}\frac{1}{t}\leq \int_1^{1+\frac{x}{n}} 1dt$$ $$\frac{x}{n+1}\leq \ln\Big(1+\frac{x}{n}\Big)\leq\frac{x}{n}$$ Since $e^x$ is increasing on $\mathbb{R}$, we have: $$e^{\frac{x}{n+1}}\leq 1+\frac{x}{n}\leq e^{\frac{x}{n}}$$ Since $x^n$ is increasing on $(0,\infty)$, on the RHS we have: $$\Big(1+\frac{x}{n}\Big)^n\leq e^x$$ On the LHS we have: $$e^x\leq \Big(1+\frac{x}{n}\Big)^{n+1} \ \ \Longrightarrow \ \ \frac{e^x}{1+\frac{x}{n}}\leq \Big(1+\frac{x}{n}\Big)^n$$ Then we have: $$\frac{e^x}{1+\frac{x}{n}}\leq \Big(1+\frac{1}{n}\Big)^n\leq e^x$$ As $n\to\infty$ we know $\Big\{\displaystyle\frac{e^x}{1+\frac{x}{n}}\Big\}\to e^x$. So, by Squeeze Theorem, we have proved the limit.

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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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    $\begingroup$ This is not enough; there are infinitely many terms, so you need to show that you can swap two limits here. $\endgroup$ Commented Apr 11, 2013 at 23:17
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    $\begingroup$ What you want to do is work with $\limsup$ and $\liminf$ here, and show $e^x\leq\liminf $ and $e^x\geq \limsup$ $\endgroup$
    – Pedro
    Commented Apr 11, 2013 at 23:53
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    $\begingroup$ How would you show that you can swap the two limits? $\endgroup$ Commented Mar 26, 2017 at 22:54
  • $\begingroup$ This equality is also only valid if $n$ is an integer (the index on the sum should go to infinity in full generality, and would then only be valid for $x/n < 1$) $\endgroup$ Commented Jun 28, 2020 at 3:06
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Using a single interval upper and lower Riemann sum for $\ln x:=\int_1^x\dfrac 1t\operatorname dt$, we get $\dfrac x{n+x}\le\ln(1+\frac xn)\le\dfrac xn$.

So, $e^{\frac x{n+x}}\le 1+\dfrac xn\le e^{\frac xn}$.

Now $e^{\frac {nx}{n+x}}\le (1+\dfrac xn)^n\le e^ x$.

Let $n\to\infty $.

I think I saw this in Best and Penner's Calculus.

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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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  • $\begingroup$ 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. $\endgroup$
    – DonAntonio
    Commented Apr 11, 2013 at 23:40
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    $\begingroup$ 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. $\endgroup$
    – user67133
    Commented Apr 11, 2013 at 23:45
  • $\begingroup$ 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. $\endgroup$
    – DonAntonio
    Commented Apr 11, 2013 at 23:47
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    $\begingroup$ I cannot but totally agree. Thank you for your suggestions, I am going to edit the post to make it clearer! $\endgroup$
    – user67133
    Commented Apr 12, 2013 at 0:07
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There is at most one function $g$ on $\mathbb{R}$ such that $$g'(x)=g(x)\text{ for all } x\text{ in }\mathbb{R}\quad\text{and}\quad g(0)=1\,.$$ If you let $f_n(x)=(1+x/n)^n$ and you can demonstrate that it compactly converges to some function $f$, you can demonstrate that $f'(x)=f(x)$ and $f(0)=1$. Likewise, if you take $f_n(x)=\sum_{k=0}^n x^k/k!$ and demonstrate this sequence converges compactly, you can show that this limit satisfies the same conditions. Thus it doesn't matter what your definition is. The uniqueness criterion is what you should probably have in mind when you think of "the exponential".

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Start with the binomial theorem: $$\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n=\lim_{n\to\infty}\sum_{m=0}^n\frac{n!}{m!(n-m)!}\frac{x^m}{n^m}=\lim_{n\to\infty}\sum_{m=0}^n\frac{n!}{n^m(n-m)!}\frac{x^m}{m!}.$$ Define $$f(\mu,\nu)=\sum_{m=0}^{\nu}\frac{\mu!}{\mu^m(\mu-m)!}\frac{x^m}{m!}.$$ Thus you need to prove $$\lim_{n\to\infty}f(n,n)=\lim_{n\to\infty}\sum_{m=0}^n\frac{x^m}{m!},$$ which amounts to proving that $$\lim_{n\to\infty}f(n,n)=\lim_{\nu\to\infty}\left[\lim_{\mu\to\infty}f(\mu,\nu)\right]$$ and $$\forall{m\in\mathbb{N}},\lim_{\mu\to\infty}\frac{\mu!}{\mu^m(\mu-m)!}=1.$$ The latter can be proven rather easily using the principle of induction. The former follows immediately from the Tannery's theorem. With this done, we have that $$\lim_{n\to\infty}\sum_{m=0}^n\frac{n!}{n^m(n-m)!}\frac{x^m}{m!}=\lim_{\nu\to\infty}\left(\sum_{m=0}^{\nu}\left[\lim_{\mu\to\infty}\frac{\mu!}{\mu^m(\mu-m)!}\right]\frac{x^m}{m!}\right)=\lim_{\nu\to\infty}\sum_{m=0}^{\nu}\frac{x^m}{m!}:=\exp(x).$$ Even if the definition of $\exp$ you are using is not via its Maclaurin series, any other definition can be easily proven from Maclaurin series rather easily.

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Not a rigorous but still a proof.

$\displaystyle \lim_{n \to \infty} \left(1+ \frac{x}{n} \right)^n = [1^ \infty]$ - uncertainty

Let $\displaystyle \lim_{x \to x_o}u(x)=1$ and $\displaystyle \lim_{x\to x_o}V(x)=\infty$, then

$\displaystyle \lim_{x \to x_o}u^V=\lim_{x \to x_o}\Big[ \Big(1+(u-1) \Big)^{\frac{1}{u-1}}\space \Big]^{(u-1)V}=e^{\displaystyle\lim_{x\to x_o}(u-1)V}$

Now let's apply what we got above to our problem. Here $ \{ u(n) = 1+\frac{x}{n}\} \to 1 $ and $\{V(n)=n\} \to \infty$ as $n\to \infty$.

$\displaystyle \lim_{n \to \infty} \left(1+ \frac{x}{n} \right)^n = [1^ \infty] = \exp \left[ \lim_{n\to \infty}\left(1+\frac{x}{n}-1 \right)n \right] = \exp \left[ \lim_{n\to \infty} \left( \frac{x}{n} \cdot n \right) \right] = \exp \left[ \lim_{n\to \infty}x \right] = \exp (x) = e^x$

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I have read through many answers but this is the only approach that is easy for me to understand:

before proofing that $\displaystyle \lim_{n\to\infty} \left(1+\frac {x}{n}\right)^n=\exp x$ ,

  1. let's first proof that $\frac{\mathrm{d} }{\mathrm{d} x} \ln(x) = \frac{1}{x}$:

$$y=\ln x$$ $$e^y=x$$

find derivative using implicit differentiation $$(e^y) \frac{dy}{dx}=\frac{d(x)}{dx}$$ $$e^y \frac{dy}{dx}=1$$ $$ \frac{dy}{dx}=\frac{1}{e^y}$$ replace $e^y$ with $x$, and $y$ with $ln(x)$: $$ \frac{\mathrm{d} }{\mathrm{d} x} \ln(x) = \frac{1}{x}$$

according to the definition of the derivative, it means that $$ \frac{\mathrm{d} }{\mathrm{d} x} \ln(x) = \lim_{h \to 0} \frac{ln(x + h) - ln(x)}{h} = \frac{1}{x} $$


  1. now proof that $\displaystyle \lim_{n\to\infty} \left(1+\frac {x}{n}\right)^n=\exp x$

$$\displaystyle \lim_{n\to\infty} \left(1+\frac {x}{n}\right)^n = \lim_{n\to\infty} \left(\exp\left(\ln\left(1+\frac {x}{n}\right)^n\right)\right) = \\ \lim_{n\to\infty} \left(\exp\left(n\ln\left(1+\frac {x}{n}\right)\right)\right) = \\ \lim_{n\to\infty} \left(\exp\left(\frac{x \cdot \ln\left(1+\frac {x}{n}\right)}{x/n}\right)\right)$$

let $h = \frac{x}{n}$. As $n\to\infty$, $h\to 0$. Replace $\frac{x}{n}$ with $h$:

$$\lim_{h\to 0} \left(\exp\left(\frac{x \cdot \ln\left(1+h\right)}{h}\right)\right) = \lim_{h\to 0} \left(\exp\left(x \left[\frac{\ln\left(1+h\right)}{h}\right] \right)\right) = \\ \exp\left(x \cdot \lim_{h\to 0}\left(\frac{\ln\left(1+h\right) - 0}{h}\right) \right) = $$ $$ \exp\left(x \cdot \lim_{h\to 0}\left(\frac{\ln\left(1+h\right) - \ln\left(1\right)}{h}\right) \right) \tag{1}$$

from the previous proof we can assume that: $$ \frac{\mathrm{d} }{\mathrm{d} x} \ln(x)|_{x=1} = \lim_{h \to 0} \frac{\ln(1 + h) - \ln(1)}{h} = \frac{1}{1} = 1 $$

After replacing the limit equation in $(1)$ we get: $$\lim_{h\to 0} \left(\exp\left(x \cdot \left(\frac{\ln\left(1+h\right)}{h}\right) \right)\right) = \exp \left(x \cdot \lim_{h\to 0}\left(\frac{\ln\left(1+h\right) - \ln\left(1\right)}{h}\right) \right) = \exp \left(x \cdot 1 \right) = \exp \left(x \right) $$

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