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?
On
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.
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 $x=1$.