How to calculate $\lim \limits_{h \to 0}{\frac{a^h-1}{h}}$?

Question

As the title says, I would like to prove for $f(x) = a^x$ there is always some constant c such that $f'=cf$. Is calculating the limit the right approach to solve this problem? Also, how to show there is only one solution when $c=1$? (the $e^x$)

Answer 1

Assuming $a>0$, you need to use the definition of real exponentiation: $$a^h=e^{h\ln a}=\sum_{n=0}^\infty\frac{(h\ln a)^n}{n!}$$

Then $$\lim_{h\to0}\frac{a^h-1}h=\lim_{h\to 0}\sum_{n=1}^\infty h^{n-1}\frac{(\ln a)^n}{n!}=\ln a$$

Answer 2

To answer the question in your text rather than in the title, if $f(x) = a^x$, then by definition $f(x) = e^{\ln(a)\cdot x}$ and $f'(x)$ exists for all $x$ by the chain rule and is equal to $\ln(a)\cdot e^x$, so you can take $c = \ln(a)$.

As for your follow-on question, $e^x$ is not the only only solution to $f(x) = f'(x)$. The general solution is $f(x) = de^x$ for some constant $d$. To see this, let $g(x) = f(x)e^{-x}$ and note that: $$ g'(x) = f'(x)e^x - f(x)e^x = 0 $$ so $g(x) = f(x)e^{-x} = \frac{f(x)}{e^x}$ is some constant $d$ implying that $f(x) = de^x$.