Prove $\lim \limits_{h \to 0} \frac{a^h-1}{h} = \ln (a)$ without using L'hospital's.

Question

I'm a Calc 2 student and was curious as to why $\frac{d}{dx}a^x = a^x\ln a$. Using the limit definition you can arrive at $\frac{d}{dx}a^x = \lim \limits_{h \to 0} \frac{a^x(a^h-1)}{h}$ so the part of the limit involving $h$ must be $\ln (a)$. I was thinking it'd be cool to use the definition of $e^x = \sum_{n=1}^\infty \frac{x^n}{n!}$ by setting $y=\sum_{n=1}^\infty \frac{x^n}{n!}$ and solving for $x$ which I'm guessing yields some sort of sum representation for $\ln(x)$.

Given the above I have a couple questions.

• How can I turn a limit expression such as $\lim \limits_{h \to 0} \frac{(a^h-1)}{h}$ into a Riemann sum and vice versa?
• How would someone go about solving for $x$ in $y =\sum_{n=1}^\infty \frac{x^n}{n!}$

I know there are prob way easier ways to figure that out, but I'm curious as to whether this way of solving it works and how it pans out.

Thanks.

Answer 1

Since you used power series:

$$\frac{a^h-1}h=\frac{e^{h\log a}-1}h=\frac1h\left(\sum_{n=0}^\infty \frac{h^n\log^na}{n!}-1\right)=\frac1h\sum_{n=1}^\infty\frac{h^n\log^na}{n!}=$$

$$=\log a+\sum_{n=2}^\infty\frac{h^{n-1}\log^na}{n!}\xrightarrow[h\to0]{}\log a$$

Answer 2

I'll use the definition that $a^x$ (for positive $a$) is the unique continuous function satisfying $f(0) = 1, f(1) = a, f(x+y) = f(x) * f(y)$.

From this definition, we can prove that $(a^b)^x = a^{bx}$ as follows : We want to prove that $a^{bx}$ satisfies the above properties. Clearly , $a^{b0} = a^0 = 1$, and $a^{b1} = a^b$. $a^{b(x+y)} = a^{bx + by} = a^{bx} a^{by}$.

Now use any of the standard arguments to show that the above definition of $e^x$ is equivalent to any of the others (the above definition is characterization 5 in the article). By another standard argument, $\frac{d}{dx}e^x = e^x$.

By defining $ln(x)$ to be the inverse of $e^x$, we have $e^{ln(a)} = a$, so $a^x = (e^{ln(a)}) ^x = e^{x ln(a)}$. Now just use the chain rule to get $\frac{d}{dx} a^x = a^x ln(a)$.

Answer 3

Use $a^x=1+\frac{x\ln a }{1!}+\frac{(x\ln a) ^2}{2!}+O(x^3)$ Then $$L=\lim_{h \to 0} \frac{a^h-1}{h}=\lim_{h \to 0} \frac{h \ln a+(h \ln a)^2/2+O(h^3)}{h}=\ln a$$

Answer 4

I would argue that both current answers stray close to circular logic regarding the derivative of $a^x$. I offer this proof which instead defines $e^x$ and its derivatives and uses this to prove facts about $a^x$.

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Define

$$\ln(x)=\int_{1}^x \frac{1}{t}dt$$

Note that here we are not assuming that $\ln(x)=\log(x)$ although we will eventually show this to be the case.

$1)$ For any $r\in\mathbb{R}$ we have

$$\frac{d}{dx}\ln(x^r)=\frac{du}{dx}\frac{d}{du}\ln(u)$$

where $u=x^r$. Then

$$=rx^{r-1}\frac{1}{u}=\frac{rx^{r-1}}{x^r}=\frac{r}{x}$$

In a similar manner, we have

$$\frac{d}{dx} r\ln(x)=\frac{r}{x}$$

Since the derivatives are equal, we may conclude

$$\ln(x^r)=r\ln(x)+C$$

for some constant $C$. To find this, note that at $x=1$ we have

$$\ln(1)=\int_1^1\frac{1}{t}dt=0$$

Thus

$$0=\ln(1)=\ln(1^r)=r\ln(1)+C=C$$

We conclude that for all $r\in\mathbb{R}$ we have $\ln(x^r)=r\ln(x)$.

$2)$ Consider

$$\lim_{x\to 0}\frac{\ln(1+x)}{x}$$

We can bound the quantity by geometric arguements

$$\frac{1}{1+|x|}=\frac{1}{x}\cdot x\cdot \frac{1}{1+|x|}\leq \frac{1}{x}\int_1^{1+x}\frac{1}{t}dt\leq \frac{1}{x}\cdot x\cdot \frac{1}{1-|x|}=\frac{1}{1-|x|}$$

By the squeeze theorem we may conclude

$$\lim_{x\to 0}\frac{\ln(1+x)}{x}=1$$

$3)$ Define

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

Note that we are not assuming that $e=\sum_{n=0}^\infty\frac{1}{n!}$. We have

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

Since $e\geq 1$ and $\ln(x)$ is continuous for all positive real numbers, this becomes

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

Using the limit proved in $2)$ this becomes

$$=\lim_{n\to\infty}n\cdot \frac{1}{n}=\lim_{n\to\infty}1=1$$

$4)$ This then implies

$$\ln(e^x)=x\ln(e)=x\cdot 1=x$$

Since both $\ln(x)$ and $e^x$ are increasing functions on their domains, this is enough to conclude that these are inverse functions.

$5)$ We will now state the inverse function theorem:

Suppose that $f(x)$ and $g(x)$ are inverse functions. Then

$$\frac{d}{dx}f(x)=\frac{1}{g'(f(x))}$$

Using this with $\ln(x)$ and $e^x$ gives us

$$\frac{d}{dx}e^x=\frac{1}{\left.\frac{d}{dx}\ln(x)\right|_{e^x}}=e^x$$

as desired.

$6)$ With this, we can now construct the Taylor series for $e^x$ about $x=0$. By induction, we have that

$$e^x=\sum_{n=0}^\infty \frac{x^n}{n!}$$

and it is not difficult to show that the series converges absolutely with an infinite radius of convergence. However, we already know that $\ln(x)$ is the inverse function of this series. We conclude that $\log(x)=\ln(x)$.

$7)$ Finally, we have

$$\frac{d}{dx}a^x=\frac{d}{dx} e^{\ln(a^x)}=\frac{d}{dx} e^{x\ln(a)}=\ln(a)e^{x\ln(a)}=\ln(a)e^{\ln(a^x)}=\ln(a)a^x$$

However, using the limit definition we have

$$\frac{d}{dx}a^x=\lim_{h\to 0}\frac{a^{x+h}-a^x}{h}=a^x\lim_{h\to 0}\frac{a^{h}-1}{h}$$

We conclude

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