How to get an original function from the limit definition of a derivative? Say I have
$$\lim_{h\to0} \frac{e^h-1}{h}$$
If $$\frac{d}{dx}(e^x)|_{x=0} =
\lim_{h\to0} \frac{e^{0+h}-e^0}{h},$$
how would I “back engineer” the derivative limit definition to satisfy the expression meaning expression-I equals expression-II. But we’re only given expression-II the limit expression so how do we find the “mystery” expression-I from expression-II.
I basically want to remove the guess work required to satisfy the two expressions, and if there’s even a general “algorithm” to follow?
 A: 
The derivative of a function $f$ at a  point $a$ is defined as
$$f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}.$$

Setting $f(x)=e^x$ and $a=0$ this yields
$$\frac{d}{dx}e^x\mid_0 = \lim_{h\to 0}\frac{e^{0+h}-e^{0}}{h}=\lim_{h\to 0}\frac{e^{h}-1}{h}.$$
This would be the solution to your problem.
If I understood you correctly you want to have a step by step derivation, from the limit to the derivative. I don't think that's the aim of the book, it's about observing that these two expressions are equal and not about calculations.
A: If I'm understanding your question correctly, it boils down to (1) identifying that a given limit is the derivative of a function $f(x)$ at some point $x=a$, namely
$$f'(a)=\lim_{h\to0}\frac{f(a+h)-f(h)}h$$
and (2) precisely picking out what the function $f(x)$ is supposed to be.
In the given example,
$$\lim_{h\to0}\frac{e^h-1}h=\lim_{h\to0}\frac{e^{\color{red}{0+h}}-e^{\color{red}0}}h=\lim_{h\to0}\frac{f(0+h)-f(0)}h\implies f(x)=e^x\text{ and }a=0$$
and hence this limit is exactly the derivative of $e^x$ at $x=0$. If you know that $(e^x)'=e^x$, then the limit is simply $e$.
So the strategy is to recast the given limit into an expression involving the forward difference $f(a+h)-f(a)$.
Another example:
$$\lim_{h\to0}\frac{(1+h)^{\frac13}-1}h=\lim_{h\to0}\frac{(\color{red}{1+h})^{\frac13}-\color{red}{1}^{\frac13}}h\implies f(x)=x^{\frac13}\text{ and }a=1$$
Then this limit would be $\frac13$, or the derivative $\left(x^{\frac13}\right)'=\frac1{3x^{\frac23}}$ at $x=1$.
