I was looking for the proof of $\frac{\mathrm{d}}{\mathrm{d}x}a^x$ as the $\lim_{h\to0}\frac{a^{x+h}-a^{x}}{h}$ which is $a^x\lim_{h\to0}\frac{a^{h}-1}{h}$. To solve that, I need to know the $\lim_{h\to0}\frac{a^{h}-1}{h}$.

In terms of $t$

I defined $\frac{1}{t}=a^h-1=e^{h\ln(a)}-1$, therefore $e^{h\ln(a)} = 1+\frac{1}{t}$ and $h=\frac{\ln\left (1+\frac{1}{t} \right )}{\ln(a)}$.
When $h\to0$, $t\to\infty$.

$$\begin{aligned}\lim_{h\to0}\frac{a^{h}-1}{h}&=\lim_{t\to\infty}\frac{\frac{1}{t}}{\frac{\ln\left (1+\frac{1}{t} \right )}{\ln(a)}}\\ &=\lim_{t\to\infty}\frac{\ln(a)}{t\cdot \ln\left (1+\frac{1}{t} \right )}\\ &=\lim_{t\to\infty}\frac{\ln(a)}{\ln\left (\left (1+\frac{1}{t} \right )^t \right )}\\ &=\frac{\ln(a)}{\ln\left ( \lim_{t\to\infty}\left (1+\frac{1}{t} \right )^t \right )}\end{aligned}$$

In terms of $n$

Now, I need to solve the limit $\lim_{t\to\infty}\left (1+\frac{1}{t} \right )^t$. I defined $t=\frac{1}{n}$.
When $t\to\infty$, $n\to0$.

$$\begin{aligned}\lim_{t\to\infty}\left (1+\frac{x}{t} \right )^t&=\lim_{n\to0}\left (1+\frac{n}{\frac{1}{x}} \right )^\frac{1}{n}\\ &=\lim_{n\to0}\left (\frac{\frac{1}{x}+n}{\frac{1}{x}} \right )^\frac{1}{n}\\ &=e^{\lim_{n\to0}\ln\left (\frac{\frac{1}{x}+n}{\frac{1}{x}} \right )^\frac{1}{n}}\\ &=e^{\lim_{n\to0}\frac{\ln\left (\frac{\frac{1}{x}+n}{\frac{1}{x}} \right )}{n}}\\ &=e^{\lim_{n\to0}\frac{\ln\left ( \frac{1}{x}+n \right )-\ln\left ( \frac{1}{x} \right )}{n}}\\ &=e^{\frac{\mathrm{d}}{\mathrm{d}\frac{1}{x}}\ln\left ( \frac{1}{x} \right )}\end{aligned}$$

Now I tried to solve $\frac{\mathrm{d}}{\mathrm{d}x}\ln(x)=\frac{1}{\frac{\mathrm{d}}{\mathrm{d}\ln(x)}x}=\frac{1}{\frac{\mathrm{d}}{\mathrm{d}\ln(x)}e^{\ln(x)}}$.
And I get back to $\frac{\mathrm{d}}{\mathrm{d}x}a^x$, when $a = e$.
Is there a way to solve one of the limits or the derivatives in another way that does not create a loop?
Note that I can't use L'Hopital rule, since I did now proof the derivatives.

  • $\begingroup$ $\lim_{t\to\infty}\left (1+\frac{1}{t} \right )^t$ is frequently the definition of $e.$ If it isn't what definition do you have? Maybe we can show $\lim_{t\to\infty}\left (1+\frac{1}{t} \right )^t = e$ by that definition. $\endgroup$ – Doug M Sep 28 '17 at 20:11
  • $\begingroup$ $f'(x)=f(x)$ is frequently the definiiton of $e^x$. $\endgroup$ – Hagen von Eitzen Sep 28 '17 at 20:12
  • $\begingroup$ @Ziv I've edited the formatting of your answer and added subheadings to make it easier to read. Please let me know if I've changed the meaning of anything. $\endgroup$ – Jam Sep 28 '17 at 20:39
  • $\begingroup$ The answer to your question crucially depends on how you define the symbol $a^{x}$ for all real values of $x$. It is an act of sheer intellectual dishonesty that most introductory calculus textbooks avoid proper development of exponential and logarithmic functions. A plethora of such questions as the current one are a testimony to the above fact. $\endgroup$ – Paramanand Singh Sep 30 '17 at 5:17

I thought it might be useful to present an approach that uses only elementary analysis (i.e., pre-calculus tools only) to evaluate the limit of interest. We will use only (1) a standard inequality that can be obtained from elementary analysis and (2) the squeeze theorem. We now proceed.

Note that we have


Now, in THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the exponential function satisfies the inequalities

$$\bbox[5px,border:2px solid #C0A000]{1+x\le e^x\le \frac{1}{1-x}}\tag1$$

for $x<1$.

Therefore, using $(1)$ with $x=h\log(a)$, we have for $h\log(a)<1$

$$h\log(a)\le e^{h\log(a)}-1\le \frac{h\log(a)}{1-h\log(a)}\tag 2$$

Dividing by $h$ for $h>0$ in $(2)$ yields

$$\log(a)\le \frac{e^{h\log(a)}-1}{h}\le \frac{\log(a)}{1-h\log(a)} \tag 3$$

while dividing by $h$ for $h<0$ in $(2)$ yields

$$ \frac{\log(a)}{1-h\log(a)} \le \frac{e^{h\log(a)}-1}{h}\le\log(a)\tag 4$$

Applying the squeeze theorem to $(3)$ and $(4)$ shows that the limits from the left and right are equal. Hence, we assert that

$$\bbox[5px,border:2px solid #C0A000]{\lim_{h\to 0}\frac{e^{h\log(a)}-1}{h}=\log(a)}$$

as expected!

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@Mark Viola's answer covers a general proof of $\lim_{h\to0}\frac{a^h-1}{h}$. There are other proofs of $(a^x)'$ on the internet, which shouldn't be hard to find.

However, if you wanted to adapt your proof from when you want to prove $\lim(1+1/t)^t=e$, you can use the steps proof here: http://www.maths.manchester.ac.uk/~mprest/elimit.pdf .

You could also adapt your proof from the later steps, with $n$. We have that $\frac{\mathrm{d}}{\mathrm{d}\left(1/x\right)}\ln(1/x)=\frac{\mathrm{d}}{\mathrm{d}(1/x)}(x)\cdot\frac{\mathrm{d}}{\mathrm{d}x}\ln(1/x)$ by the chain rule, which in turn equals $\frac{1}{\left(\frac{\mathrm{d}}{\mathrm{d}x}(1/x)\right)}\cdot\frac{\mathrm{d}}{\mathrm{d}x}\ln(1/x)$ since $\frac{\mathrm{d}x}{\mathrm{d}y}=\frac{1}{\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)}$. So finally, we can use the chain rule again to show that this derivative is $\frac{1}{-x^{-2}}\cdot\left((-x^{-2})\frac1{(1/x)}\right)=x$. So then $\lim(1+\frac{x}{t})^t=e^x$. You can then, of course, substitute $\lim(1+\frac{1}{t})^t=e$ into the earlier part of your proof. Obviously, this isn't the most efficient way to proceed with the proof but it goes to show that you weren't that far off from finishing, and as long as your steps are logical, which they were, you should be able to get to the right answer.

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