What is wrong with this argument that differentiation using the First Principle leads to division by $0$:
$$
f'(x)=\lim_\limits{h \to 0} \frac{f(x+h)-f(x)}{h} \\
$$
Using the quotient limit law:
$$
\lim_\limits{h \to 0} \frac{f(x+h)-f(x)}{h}=\frac{\lim_\limits{h \to 0}f(x+h)-f(x)}{\lim_\limits{h \to 0}h}
$$
$$
\lim_\limits{h \to 0}h = 0
$$
Therefore, there the top half of the fraction is divided by $0$. Here is my reasoning for why $\lim_\limits{h \to 0}h = 0$:
As $h$ approaches $0$, its value becomes smaller (and will become smaller than any number strictly greater than $0$). For example, you cannot evaluate the limit as equalling $0.001$, because at some point $h$ will be lower than this. $0$ is the largest number that does not have this problem. Therefore, the limit expression is equal to $0$.
Thank you for reading.
Best Answer
The part which is wrong in your reasoning is this: $$\lim_\limits{h \to 0} \frac{f(x+h)-f(x)}{h}=\frac{\lim_\limits{h \to 0}f(x+h)-f(x)}{\lim_\limits{h \to 0}h}$$
The formula $\lim \frac{f}{g}=\frac{\lim f}{\lim g}$ can only be used when $\lim(g)\neq 0$. If $\lim g \neq 0$ the RHS of your formula, does not make any sense.
To understand what happens here, just look at $$\lim_{h \to 0} \frac{h}{h}$$
$\frac{h}{h}=1$ and $\lim_1=0$. But you cannot write $$\lim_{h \to 0} \frac{h}{h}=\frac{\lim_{h \to 0} h}{\lim_{h \to 0} h}$$ since the RHS makes no sense.