What is wrong with this limit evaluation $\lim_{h\to0} \frac{a^h-1}{h}=\ln(a)$ and how to arrive to the correct one?

Question

I know this is wrong but I cannot see why. I also cannot get to the write answer even though I did this in the past.

$$\lim_{h\to0} \frac{a^h-1}{h}=\ln(a)$$

What I did was

$$\lim_{h\to0} \frac{a^h-1}{h}\left(\frac{h}{h}\right)$$ $$\lim_{h\to0} \frac{ha^h-h}{h^2}$$ Then I used l'Hopital's rule twice: $$\lim_{h\to0} \frac{h^2(h-1)a^{h-2}}{2}=0$$

So what am I doing wrong and how to do it right?

EDTI: Ok I see my mistake.

Now how do you actually compute this limit? BTW the goal is to actually derive $a^h$ with respect to h from first principles so I cannot simply use $a^h=a^h \ln(h)$

Answer 1

You are taking the derivative incorrectly. $$\frac{d}{dh}(ha^h-h)=h(a^h)'+(h)'a^h - (h)' = ha^h \ln a + a^h -1$$

There was no benefit from multiplying by $h/h$.

To do the limit from the OP one usually needs to prove somehow the lemma that $$\lim_{h\to 0} \frac{e^h-1}{h}=1$$

You can apply this to your problem via the exponent property that $a^x=e^{x\ln a}$.

Answer 2

Your first derivative is wrong; you treated $a$ as a variable and $h$ as a constant when calculating $\frac{d}{dh}(a^h)$. Your second derivative is wrong for the same reason.

Answer 3

You're chasing your tail if you intend to evaluate such limit using L'Hôpital, since you're trying to compute the derivative of $f(t)=a^t$ at $t=0$. You would have to agree on a definition of $a^t$ (for example, that $a^t=e^{t\log a}$), and then use that $e^t$ is its own derivative plus the chain rule.