How to evaluate a limit with high school math? So my textbook's explanation of the derivative of e is very sketchy. They used lots of approximations and plugging things into the calculator. Basically I want to know how you can work out as h approaches 0
$$
\lim_{h\to0}\frac {10^{x+h}-10^x }h
$$
 A: The limit is equivalent to $$10^x \lim_{h \to 0} \frac{10^h-1}{h}.$$ Now the troubles begin, since this limit can be understood only after you tell us how you define $10^h$ for real $h$.
In higher mathematics, the quickest way is probably to define $$e^x=\sum_{n=0}^\infty \frac{x^n}{n!}$$ and use some theorem about swapping limits and series. Then $10^h = e^{10 \log h}$.
In high-school mathematics we usually teach that this limit is fundamental, without any further detail. This is the reason why your textbook is rather vague and gives only numerical hints.
A: We have
$$
f(x) = 10^x
$$
the limit can be rewritten as
$$
10^x \lim_{h \rightarrow 0} \frac{10^h-1}{h}
$$
where
$$
\lim_{h \rightarrow 0} \frac{10^h-1}{h} = f'(0)
$$
so
$$
f'(x) = f'(0) \cdot a^x
$$
by rewriting $a^x$:
$$
a=e^{ln(a)} \Rightarrow a^x = e^{x ln(a)}
$$
and using the chain rule:
$$
f'(x) = e^{x ln(a)} \cdot ln(a) = a^x ln(a)
$$
EDIT:
I realized I was cheating a bit above since finding out $f'(0)$ requires that we know $f'(x)$ and I get this without using limits..
 so here's another way of looking at it:
Let
$$
f(x) = a^x = e^{x \ ln(a)}
$$
then we have
$$
f'(x) = \lim_{h \rightarrow 0} \frac{e^{(x + h) \ ln(a)} - e^{x \ ln(a)}}{h} = 
e^{x \ ln(a)} \ \lim_{h \rightarrow 0} \frac{e^{h \ ln(a)} - 1}{h}
$$
substitute $t = h \ ln(a)$ and we get
$$
e^{x \ ln(a)} \lim_{t \rightarrow 0} \frac{e^t - 1}{\frac{t}{ln(a)}} = 
ln(a) \ e^{x \ ln(a)} \lim_{t \rightarrow 0} \frac{e^t - 1}{t}
$$
now
$$
\lim_{t \rightarrow 0} \frac{e^t - 1}{t} = 1
$$
if we can accept that this is a standard limit for now. We get the result
$$
f'(x) = ln(a) \ e^{x \ ln(a)} = a^x \ ln(a)
$$
