show that $\lim_{ h\to 0} (a^h-1)/h=\ln(a)$

Question

I would like to show:

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

So I've got two questions:

1) Find the value of $\lim_{h\to 0}(a^h-1)/h$. You don't know that the answer is $\ln(a)$, you can't guess, so you have to derive it without using it. You can only use the $\lim_{n\to\infty}(1+1/n)^n$ definition for $e$. The definition for $\ln(a)=x$ is $e^x=a$.

2) Show that $\lim_{h\to 0}(a^h-1)/h$ and $\ln(a)$ is the same by whatever means.

Answer 1

Hint: Substituting $$a^h-1=m$$ so $$h=\frac{\ln(m+1)}{\ln(a)}$$ and we get $$\frac{a^h-1}{h}=\frac{\ln(a)}{\ln(m+1)^{1/m}}$$ Since $h$ tends to zero, $m$ tends to zero.

Answer 2

Use L'Hôpital's Rule:

The prerequisites are met since $$\lim_{h\to 0}(a^h-1)=\lim_{h\to 0}(h)=0$$ So we apply: $$\frac{d}{dh}(a^h-1)=a^h\ln(a)$$ $$\frac{d}{dh}(h)=1$$

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