Question about the derivative definition The derivative at a point $x$ is defined as:
$\lim\limits_{h\to0} \frac{f(x+h) - f(x)}h$
But if $h\to0$, wouldn't that mean:
$\frac{f(x+0) - f(x)}0 = \frac0{0}$ which is undefined?

Think of how fast some terms of $f(x+h)-f(x)$ vanish compared to $h$ (no matter how small $h>0$ is), while others are independent of it. 
In particular, $f(x+h)-f(x)$ consists of the derivative times $h$ (hence independent of $h$), plus a sublinear term $u(h)$ that tends faster to 0 than $h$, i.e. $\lim_{h \to 0} \frac{u(h)}{h} = 0$.
To illustrate this, take $f(x) = x^3$. Then $$f(x+h)-f(x) = x^3 + 3xh^2 + 3x^2h + h^3 - x^3 = 3x^2 h + h^2(3x + h)$$
The derivative term is $3x^2$, while $u(h) = h^2(3x+h)$ satisfying $\lim_{h \to 0} \frac{u(h)}{h} = 0$. Thus, $$\lim_{h \to 0}\frac{f(x+h)-f(x)}{h} = \lim_{h \to 0} 3x^2\frac{h}{h} + \lim_{h \to 0} \frac{u(h)}{h} = 3x^2 + 0$$
