Natural Logarithm expressed as Limit

Question

In the process of trying to prove the derivative of $f(x)=a^x$ (for $a\in\mathbb{R}$) using the definition of the derivative, one arrives at the following equation:

\begin{align} \frac{df}{dx} = \frac{d}{dx}\left[a^x\right] = \lim_{\Delta x \rightarrow 0}\left[\frac{a^{(x+\Delta x)}-a^x}{\Delta x}\right] = \lim_{\Delta x \rightarrow 0}\left[\frac{(a^{\Delta x} - 1)a^x}{\Delta x}\right] \end{align}

At this point, I wish to show that

\begin{align} \lim_{\Delta x \rightarrow 0}\left[\frac{a^{\Delta x} - 1}{\Delta x}\right] = \ln(a). \end{align}

How can one show that this is true in this context WITHOUT Taylor Series and WITHOUT the knowledge of the derivative of $e^x$? (i.e. from first principles in the context of the proof?) L'Hopital's rule seems to be ineffective here since it would involve assuming what we are trying to prove.

Answer 1

In order to show that this limit is $\ln(a)$ you have to bring in the definition of the natural logarithm. And it is not good enough to say that $x = \ln(a) \Leftrightarrow a = e^x$ because that begs the question of how to define $e$.

One typical way to define the natural logarithm is as the integral of $1/x$; but that immediately lets you show that the derivative of $e^x$ is $e^x$ so it may not satisfy your desire.

So I think the best you can do without injecting knowledge of natural logarithms or of $e$, is to prove that the limit you present exists for all positive $a$ (though it might -- does -- depend on $a$) so that the derivative of $a^x$ will be $C(a) a^x$.

Answer 2

Hint you can prove it using basic limit theorem (exponential limit) ...

$\left(1+\frac{1}{x}\right)^x=e.$ By assuming $ a^x-1=y$ then you may see as $x\to 0, y\to 0$ then $\log(y+1)/\log a=x$. then substituting and using the exponential limit you can prove it

Answer 3

In order to show that this is in fact $\ln(a)$, you need to have in mind a particular definition of the function $\ln(x)$. There are several ways that we may fundamentally define $\ln(x)$ without assuming knowledge of the derivative of $a^x$:

One may define $\ln(x)$ to be the unique antiderivative of $y=1/x$ on the interval $(0,\infty)$ satisfying the initial condition $\ln(1)=0$. That is, we have $\ln(x)=\int_{t=1}^x(t^{-1}dt)$.

Or one can first define the natural base $e$ as the number $\lim_{n\to\infty}(1+1/n)^n$ (or equivalently the number $\sum_{n=0}^{\infty}(1/n!)$). Then, we may define the natural logarithm as the logarithm to base $e$, i.e. $\ln(x)=\log_e(x)$.

Or one can first define the exponential function $\exp(x)$ as the unique function that is its own derivative and satisfies the initial condition $\exp(0)=1$. Then, we may define the natural logarithm as the inverse function of $\exp$.

Using any one of these definitions, it is possible to prove that $\ln(a) = \lim_{x\to 0}(\frac{a^x-1}{x})$.

For instance, if we use the definition that $\ln(x)=\int_{t=1}^x(t^{-1} dt)$, then we derive the above limit as follows:

We first note the general rule for antiderivatives:

$\int_{t=b}^{t=a}(t^n dt)=\frac{a^{n+1}-b^{n+1}}{n+1}$.

One might at first think we can directly apply this to the case $n=-1$. However, we see we get the indeterminate expression $0/0$ if directly plug in $n=-1$ into the above formula. What do we do?

Well, although we cannot plug in $n=-1$ directly, we know the formula holds for all $n$ arbitrarily close to $-1$, so it should also work in the limit $n\to -1$. (This is a step that actually needs to be justified, but using tools from real analysis, this is not hard to do).

Thus, we have $\int_{t=b}^{t=a}(t^{-1} dt)=\lim_{n\to -1}(\frac{a^{n+1}-b^{n+1}}{n+1})$.

Applying this to the case $b=1$, we get that

$\int_{t=1}^{t=a}(t^{-1} dt) = \lim_{n\to -1}(\frac{a^{n+1}-1^{n+1}}{n+1}) = \lim_{n\to -1}(\frac{a^{n+1}-1}{n+1})$.

However, this above integral is by definition equal to $\ln(a)$, so we conclude that

$\ln(a)=\lim_{n\to -1}(\frac{a^{n+1}-1}{n+1})=\lim_{x\to 0}(\frac{a^x-1}{x})$,

which is what we wanted to prove.

This is just one of several ways we can prove your wanted limit from "fundamental principles". I hope this helps.

Answer 4

How about something like this...

You've shown that the derivative of $b^x$ is proportional to $b^x$, i.e., $b^x$ is an eigenfunction of the derivative operator with eigenvalue $\lambda_b$ given by:

$$\lambda_b = \lim_{\Delta x\rightarrow\infty}\frac{b^{\Delta x} - 1}{\Delta x}$$

Now, suppose that there is some $b$ such $\lambda_b = 1$. Denote that particular value of $b$ (if it exists) as $e$ so that (by stipulation):

$$\lambda_e = \lim_{\Delta x\rightarrow\infty}\frac{e^{\Delta x} - 1}{\Delta x} = 1$$

Now, using the fundamental relationship $b^{log_ba} = a$, write $b^x$ as $(e^{\log_eb})^x = e^{x\log_eb}$ and find the derivative:

$$\frac{\mathrm{d}}{\mathrm{d}x}b^x = \frac{\mathrm{d}}{\mathrm{d}x}e^{x\log_eb} = \lambda_ee^{x\log_eb}\log_eb = b^x\log_eb$$

But you've already shown that:

$$\frac{\mathrm{d}}{\mathrm{d}x}b^x = b^x\lim_{\Delta x\rightarrow\infty}\frac{b^{\Delta x} - 1}{\Delta x}$$

Thus:

$$\Rightarrow \lim_{\Delta x\rightarrow\infty}\frac{b^{\Delta x} - 1}{\Delta x} = \log_eb $$

Note: In the above, I've simply stipulated that there is some special $b$ such that $\lambda_b = 1$, I've denoted that special value of $b$ (if it exists) with the letter $e$, and I then used the relationship $b^{log_ba} = a$ to equate $\lambda_b$ with $\log_eb$. That is, I don't think that I've snuck in knowledge of the derivative of $e^x$ or the natural log here.