Value of $\lim_{h\rightarrow 0}\frac{a^h-1}{h}$

Question

I'm given a question by my friend. State the value of $\lim_{h\rightarrow 0}\frac{a^h-1}{h}$ for $a=2$.

Then I've used l'Hopital's rule to solve it and I got the answer of $\ln 2$.

But my friend said need to be solved by first principles method. I wonder how. Can anyone show me? Thanks.

Have a look at: How to prove that $\lim\limits_{h \to 0} \frac{a^h - 1}{h} = \ln a$. These two questions are about derivatives, but they are basically just a reformulation of the same question: Proof of the fact that $\ln(a) = f '(0)$ for $f(x) = a^x$? and Show that $d/dx (a^x) = a^x\ln a$. — Martin Sleziak, Feb 03 '16 at 13:03
http://math.stackexchange.com/questions/1828962/is-it-valid-to-write-1-lim-x-rightarrow-0-fracex-1x-frac-lim-x — , Jul 04 '16 at 00:26

Hosein Rahnama · Answer 1 · 2016-02-03T11:49:54.777

5

Hints

Use the definition of the derivative of $f(x)=a^x$ at $x=0$ to construct your limit.
Next, use the differentiation formula $(a^x)'=a^x \ln a$.

edited Feb 03 '16 at 11:49

answered Feb 02 '16 at 12:06

Hosein Rahnama

14,882

score 0 · Answer 2 · answered Feb 03 '16 at 12:58

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

Now identify the last limit as the definition of the derivative of $e^x$ at $x=0$.

Value of $\lim_{h\rightarrow 0}\frac{a^h-1}{h}$

2 Answers2