# Finding exponential function limit definition from definition of e

I am trying to prove, from the following limit definition of $$e$$ $$e=\lim_{n\to+\infty} \left(1+\frac{1}{n}\right)^n$$ the following definition for the exponential function: $$e^x =\lim_{n\to+\infty} \left(1+\frac{x}{n}\right)^n$$

I tried to follow the answer of this post, but there is something I don't understand: Prove $e^x$ limit definition from limit definition of $e$.

Here is a screenshot:

To substitute $$n$$ for $$u$$ in the limit "index", we need to make sure that $$u$$ goes to $$+\infty$$ as $$n$$ goes to $$+\infty$$. Therefore, we should have: $$\lim_{n\to+\infty} u = \lim_{n\to+\infty} nx \stackrel{?}{=}+\infty$$ Which is, in some way, $$(+∞)*x$$. That is fine if $$x$$ is positive, but what if $$x$$ were negative? Wouldn't that limit then be equal to $$(-∞)$$, which would invalidate that "re-indexing"?

Is the proof missing something, or am I wrong?

Thanks in advance! (The reason I am writing this here and not commenting is that the site settings forbid me to, because of my limited activity.)

• You should note that $(1+x/n)^n$ tends to $e^x$ even if $n\to-\infty$ Commented Aug 22, 2021 at 1:51
• I dislike that argument for another reason. It is an (unstated) rule that $\lim_{n\to\infty} f(n)$ is a limit of a sequence, $f(1),f(2),f(3),…$ But $x$ is not necessarily an integer when computing $e^x.$ Replace $xn$ with $m$ and then take $m\to\infty,$ because $m$ is not necessarily an integer. Commented Aug 22, 2021 at 1:53
• @ParamanandSingh, thank you. Now that you mention it, it seems completely reasonable. Commented Aug 22, 2021 at 2:26
• @ThomasAndrews, I understand the convention and the confusion it causes in that context. However, this is only for my personal understanding, but if I were to redact a formal proof of it, I'll remember that. Thanks! Commented Aug 22, 2021 at 2:26

I will assume exponentiation is continuous. Let $$f(x)=\lim_{n\to\infty}(1+\frac xn)^n$$. For $$x>0$$, I will show $$e^x=f(x)$$. [The case when $$x<0$$ is similar.]

Lemma $$f(x)$$ is continuous.

Proof I will prove it is continuous at $$x=x_0$$, for arbitrary $$x_0\in\mathbb R$$. For all $$\epsilon>0$$, let $$\delta=\epsilon/f(|2x_0|)$$. Let $$x\in\mathbb R$$ be such that $$|x-x_0|<\delta$$. Then,

\begin{align} |f(x)-f(x_0)|&=\lim_{n\to\infty}\left|\big(1+\frac xn\big)^n-\big(1+\frac {x_0}n\big)^n\right|\\ &=\lim_{n\to\infty}\left|\sum_{i=1}^n{n\choose i}\frac{x^i-x_0^i}{n^i}\right|\\ &<\delta\lim_{n\to\infty}\sum_{i=1}^n{n\choose i}\frac{i|2x_0|^{i-1}}{n^i}. \end{align} Here, the sum in the limit is $$\sum_{i=1}^n{n\choose i}\frac{i|2x_0|^{i-1}}{n^i}=\sum_{i=1}^n{n-1\choose i-1}\frac{|2x_0|^{i-1}}{n^{i-1}}\to f(|2x_0|) \ (n\to\infty).$$ Thus, we obtain $$|f(x)-f(x_0)|. QED

Thus, since both $$e^x$$ and $$f(x)$$ are continuous, it suffices to check they are equal for rational numbers $$x=p/q$$ with $$p,q$$ positive integers.

It is well-known that taking a subsequence does not change the limit of a convergent sequence. Thus,

\begin{align} e^{p/q}&=\lim_{n\to\infty}\big(1+\frac1n\big)^{pn/q}\\ &=\lim_{n\to\infty}\big(1+\frac1{qn}\big)^{pn}, \end{align} and \begin{align} f(p/q)&=\lim_{n\to\infty}\big(1+\frac{p}{qn}\big)^n\\ &=\lim_{n\to\infty}\big(1+\frac1{qn}\big)^{pn}, \end{align} equal each other.

If you don't have any restrictions on how to prove, then you can use logarithm method-

Let $$f(n) = (1 + \frac{1}{n})^{n}$$. Then,

$$\lim\limits_{n \to \infty} f(n) = \lim\limits_{n \to \infty} (1 + \frac{1}{n})^{n} = \Large e^{\big[\large \lim\limits_{n \to \infty} \large\ln \big(1 + \frac{1}{n}\big)^{n}\big]}$$

So, instead of finding $$\lim\limits_{n \to \infty} f(n)$$, try to find $$\lim\limits_{n \to \infty} \ln(f(n))$$.

So, $$\lim\limits_{n \to \infty} \ln(f(n)) = \lim\limits_{n \to \infty} \ln \big(1 + \frac{1}{n}\big)^{n}$$

$$= \lim\limits_{n \to \infty} n\ln \big(1 + \frac{1}{n}\big) = \lim\limits_{n \to \infty} \frac{\ln \big(1 + \frac{1}{n}\big)}{1/n}$$

Now this is $$\frac{0}{0}$$ form, so using L'Hôpital's rule,

$$\lim\limits_{n \to \infty} \ln(f(n)) = \lim\limits_{n \to \infty} \frac{1}{1+1/n}*\big(\frac{-1}{n^{2}}\big)*\left(\Large\frac{1}{\frac{\Large -1}{\Large n^{2}}}\right)$$

$$= \lim\limits_{n \to \infty} \frac{1}{1+1/n} = 1$$

So, $$\lim\limits_{n \to \infty} f(n) = e^{1} = e$$

Same can be used for proving $$\lim\limits_{n \to \infty} (1 + \frac{x}{n})^{n} = e^{x}$$