How can we prove that there is a number $a$ such that $\lim_{h\to 0}\frac{a^{h} -1}{h}=1$?

Question

One definition of $e$ that I am fond of is that it is the number $a$ such that $$ \lim_{h\to 0}\frac{a^{h} -1}{h}=1 $$ The reason for this is that it cuts to the heart of the special property of all exponential functions. If we have $f(x)=a^x$, then \begin{align} f'(x)&=\lim_{\Delta x \to 0}\frac{a^{x+\Delta x} -a^x}{\Delta x} \\ &=\lim_{\Delta x \to 0}a^x\frac{a^{\Delta x} - 1}{\Delta x} \\ &=a^x\lim_{\Delta x \to 0}\frac{a^{\Delta x} - 1}{\Delta x} \end{align} However, presumably there is a caveat to this approach. We need to show that a number like $e$ exists in the first place! In other words, we need to show that $g(a)=\lim_{\Delta x \to 0}\frac{a^{\Delta x} -1}{\Delta x}$ takes the value of $1$ for some value of $a$. How might we do this?

Answer 1

Independently, we can prove that there exists a number $e$ such that $e = \lim_{n \to \infty} (1+1/n)^n$ along with the property that for all $n \in \mathbb{N}$,

$$\left(1 + \frac{1}{n} \right)^n < e < \left(1 + \frac{1}{n} \right)^{n+1}$$

It follows that

$$1 < n(e^{1/n}-1) < n\left[\left(1 + \frac{1}{n} \right)^{1/n}\left( 1+ \frac{1}{n}\right)-1 \right] \leqslant 1 + \frac{1}{n} + \frac{1}{n^2},$$

where the right-hand inequality is obtained using Bernoulli's inequality $(1 + 1/n)^{1/n} \leqslant 1 + 1/n^2$.

By the squeeze theorem we get

$$\tag{*}\lim_{n \to \infty}\frac{e^{1/n}-1}{\frac{1}{n}} = 1$$

From here it is not difficult to show that

$$\tag{**} \lim_{h \to 0+}\frac{e^{h}-1}{h} = 1$$

Taking $n = \lfloor1/h\rfloor$ when $h > 0$, we have $n \leqslant 1/h < n+1$ and

$$\frac{n}{n+1}(n+1)(e^{1/n+1} - 1)= n(e^{1/n+1} - 1 ) \leqslant \frac{e^h -1 }{h} \leqslant (n+1)(e^{1/n} -1) = \frac{n+1}{n}n(e^{1/n}-1)$$

Since $n \to \infty$ if and only if $h \to 0$ we obtain (**) by the squeeze theorem using the previous result (*).

With some more work we can show that the limit $1$ is also attained as $h \to 0-$.

Answer 2

Let $f(x)$ be given by the

$$f(x)=\int_1^x \frac1t\,dt$$

It is easy to show that $f(x)$ is continuous and increasing for $x>0$ with $f(1)=0$ and $\lim_{x\to\infty}f(x)=\infty$. Then, by the intermediate value theorem, there exists a number $a>1$ such that

$$f(a)=\int_1^a \frac1t\,dt=1\tag1$$

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We enforce the substitution $t\mapsto (1+ht)^{1/h}$ in $(1)$ to obtain

$$1=\int_0^{(a^h-1)/h}\frac1{1+ht}\,dt\tag2$$

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For $h>0$, $\frac{a^h-1}{h}>0$ and $1\le 1+ht\le a^h$ when $t\in [0,(a^h-1)/h]$. Therefore, from $(2)$ we find that

$$\left(\frac{a^h-1}{h}\right)a^{-h}\le 1\le \left(\frac{a^h-1}{h}\right)\tag3$$

Rearranging $(3)$, we have the bounds

$$1\le \left(\frac{a^h-1}{h}\right)\le a^h$$

whence application of the squeeze theorem yields

$$\lim_{h\to 0^+}\left(\frac{a^h-1}{h}\right)=1$$

We leave it as an exercise for the reader to show that the left-side limit is also $1$ from which we conclude

$$\lim_{h\to 0}\left(\frac{a^h-1}{h}\right)=1$$

for some $a>1$ such that $1=\int_1^a \frac1t\,dt$.

Answer 3

Let me suggest one more, if it can be so called, functional, approach. Let's consider function $f$ with following 3 property:

1. $f(x_1+x_2)=f(x_1)f(x_2)$ for $\forall x_1,x_2 \in \mathbb{R}$
2. $f(0)=1,f(1)=a$ where $a>0$
3. $f$ is continuous in $x=0$

It can be proved, that exists unique continuous on all $\mathbb{R}$ function, which satisfies brought properties and it we define as $a^x$.

Can be proved, that exists limit $$\lim\limits_{x \to \infty}\left( 1+\frac{1}{x} \right)^x= \lim\limits_{x \to 0}\left( 1+ x\right)^{\frac{1}{x}}$$ and we denote it by $e$. For $e$, at last, we obtain $$\lim\limits_{x \to 0}\frac{e^x-1}{x}=1$$ If this approach is satisfactory and you are interested in the details, then write which part you would like to see more fully written.