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Let $$ e = \lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^{n} $$ where $ n \in \mathbb{R} $.

Let $$ \exp(x) = \lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^{n} $$ where where $ n \in \mathbb{R} $ and $ x \in \mathbb{R} $.

How can I prove that $$ e^x = \exp(x) $$ where $ e^x $ is $ e $ raised to the power of $ x $ and $ x \in \mathbb{R} $?

Lone Learner
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    For $x=0$ it is obvious. For $x\ne 0$, multiply and divide the exponent in $\left( 1+ \frac{x}{n}\right)^n$ by $x$ then use the change of variable $m=\frac{n}{x}$. – Giuseppe Negro Aug 23 '17 at 10:27
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    Of course you need a definition for $e^x$. For $x$ integer, or even rational, we get a definition from algebra. But what about general $x$? – GEdgar Aug 23 '17 at 12:39
  • Using $n\in\mathbb{Z} ^{+} $ is simpler as it easier to justify the existence of these limits in that case and my answer assumes this. If $n\in\mathbb{R} $ then the existence of these limits is a challenge but the current question is trivial. Also by convention the variable $n$ is $n\to\infty$ is considered to be a positive integer. – Paramanand Singh Aug 24 '17 at 07:43
  • Of course the issues about my comment that have been pointed out are serious. Those are the same ones as in this answer. – Giuseppe Negro Sep 10 '17 at 21:38

3 Answers3

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By definition, $$e = \lim_{n\rightarrow\infty}\left(1+\frac 1n\right)^n$$ Now replace $n$ by $\frac nx$, $$e = \lim_{\frac nx \rightarrow\infty}\left(1+\frac xn\right)^\frac nx$$ Now by power law for limits, we know that $$e = \Bigr[\lim_{\frac nx \rightarrow\infty}\left(1+\frac xn\right)^n\Bigr]^\frac1x$$

Also we know that when $\frac nx\rightarrow\infty$ and $x$ is an arbitary value, then $n\rightarrow\infty$. So replacing $\frac nx\rightarrow\infty$ by $n\rightarrow\infty$ and cancelling the power $\frac1x$ by raising both sides by the power $x$, we get $$e^x=\lim_{n\rightarrow\infty}\left(1+\frac xn\right)^n = \exp(x)$$ and the proof is complete.

Faiq Irfan
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  • If you want to understand it deeply, expand (1 +x/n)^n you will get something like: 1+x+ ∑ (x/n)^j ][1-1/n][1-2/n].../ j! . when n →∞ , 1/n →0 and you finally get: (1+x/n)^n = 1+x +x^2/2! +x^3/3! + . . . which is equal to e^x. – sirous Aug 23 '17 at 12:35
  • @sirous You are right. The proof can also be carried out by the use of Taylor series and binomial theorem. ;) But I find series methods a bit lengthy and awkward... :p – Faiq Irfan Aug 23 '17 at 13:01
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    This argument does not hold for $x \leq 0$. – bloomers Aug 23 '17 at 20:11
  • For x < 0, you can define a new series $b_n = \left(1 - \frac{x^2}{n^2} \right)^n$ and if we choose any $|x| < n$ it will also mean $x^2 < n^2$ and so $0 < \frac{x^2}{n^2} < 1$, and so we can apply bernoulli formula and get $1 + \frac{-x^2}{n} \leq \left(1 - \frac{x^2}{n^2}\right)^n \leq 1$ and because it holds true for any n, from sandwich theorem we get that $\lim b_n = 1$, and so $\lim \frac{b_n}{e_n(-x)} = e^x$ while $\frac{b_n}{e_n(-x)} = \frac{e_n(x)e_n(-x)}{e_n(x)} = e_n(x)$, and so $\lim e_n(x) = e^{x} \Box$ – Nadav96 Mar 20 '20 at 12:17
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In what follows we assume that $n$ is a positive integer.

For $x\in\mathbb{Q}$ the expression $e^{x} $ is defined via processes of algebra and it's existence is justified via processes of analysis. For irrational $x$ we need a specific definition of $e^{x} $ and there are multiple approaches. One approach is to define it via the limit $$\lim_{n\to\infty} \left(1+\frac{x}{n}\right)^{n}\tag{1}$$ in question. Another is to use continuous extension of $e^{x} $ for irrational $x$ (this one is more common and Rudin popularized it heavily).

For rational values of $x$ the proof is simple. The harder approach is to prove that the limit $(1)$ above exists for all real $x$ and defines a function of $x$ commonly denoted by $\exp(x) $ and establish the property that $\exp(x+y) =\exp (x) \exp(y) $. Using this one can easily show via algebra that $\exp(x) =\{\exp(1)\} ^{x} =e^{x}$ for rational values of $x$.

The easier approach is as follows (this avoids the proof of existence of limit $(1)$ for all $x$). Let $x$ be a positive integer and then we can see that $$\lim_{n\to\infty} \left(1+\frac{x}{n}\right)^{n}=\lim_{n\to\infty}\prod_{i=1}^{x}\dfrac{\left(1+\dfrac{1}{n+i-1}\right)^{n+i-1}}{\left(1+\dfrac{1}{n+i-1}\right)^{i-1}}=\prod_{i=1}^{x}e=e^{x}$$ If $x=-y$ is a negative integer then we can see that $$\lim_{n\to\infty} \left(1-\frac{y}{n}\right)^{n}=\lim_{n\to\infty}\prod_{i=1}^{y}\dfrac{\left(1+\dfrac{1}{n-i}\right)^{-i}}{\left(1+\dfrac{1}{n-i}\right)^{n-i}}=\prod_{i=1}^{y}\frac{1}{e}=e^{-y}$$ (It is best to put some values of $x, y$ like $x=3,y=2$ and write the products in expanded form to convince of the validity of algebraic manipulation above. I have avoided writing expanded form to reduce labor of typing on smartphone). So we have proved the desired result for all integers $x$ (the case $x=0$ is trivial by the way).

Lets assume that $x=p/q$ where $p\in\mathbb{Z} $ and $q\in\mathbb{Z} ^{+} $. Then we can see that $$\lim_{n\to\infty} \left(1+\frac{x}{n}\right)^{n}=\lim_{n\to\infty}\sqrt[q]{\left(1+\frac{p}{qn}\right)^{qn}}=\sqrt[q]{e^{p}}=e^{x}$$

Paramanand Singh
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Here is a simple proof I could come up with. Please review this and let me know if this proof looks good.

$$ \exp(x) = \lim_{n \to \infty} \left( 1 + \frac{x}{n} \right)^n = \lim_{\frac{n}{x} \to \infty} \left( 1 + \frac{1}{\frac{n}{x}} \right)^{\frac{n}{x} \cdot x} = \lim_{t \to \infty} \left( 1 + \frac{1}{t} \right)^{tx} = e^x $$

Lone Learner
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  • This proof assumes that $n$ is real and that the expression $a^{b} $ is defined somehow (without the use of exponential and logarithmic functions) for all real numbers $b$. This approach to $a^{b} $ is difficult and your solution appears simple only because you assume all the usual properties of $a^{b} $ without proof. – Paramanand Singh Sep 06 '17 at 05:51
  • My comment to your question tells us that the question is trivial if $n\in\mathbb{R} $ and your current answer shows the manner in which the problem is trivial. – Paramanand Singh Sep 06 '17 at 05:53
  • @ParamanandSingh I don't follow. When would $n$ not be real? – bloomers Sep 07 '17 at 11:58
  • @bloomers: The difference between $n$ real and $n$ positive integer in the symbol $n\to\infty$ matters a lot. Thus if $n$ is integer then $\lim_{n\to\infty} \sin n\pi=0$ and same limit does not exist if $n$ is real variable. – Paramanand Singh Sep 07 '17 at 12:02
  • @ParamanandSingh So when you say, "This proof assumes that $n$ is real...", do you mean to say, "This proof assumes that $n$ can be a real non-integer value..."? – bloomers Sep 07 '17 at 12:07
  • @bloomers: yes that is my intended meaning and that makes a lot of difference here. In my answer below I have explicitly mentioned that $n$ is a positive integer. This makes the problem non-trivial and interesting. – Paramanand Singh Sep 07 '17 at 12:08
  • @ParamanandSingh Yes, it does. With the way you have it worded, I interpreted your meaning to be "This proof assumes that $n$ cannot be imaginary..." But I follow now. – bloomers Sep 07 '17 at 12:23
  • @bloomers: oh my bad! Will be more careful with my language so that there are less chances of misinterpretation. – Paramanand Singh Sep 07 '17 at 12:25