I need some basic introduction to limits So, I know you can obviously cut out a value if it is multiplying and dividing something at the same time, right? Like:
$$\frac{4h-2xh-h^2}{h} = \frac{h(4-2x-h)}{h} = 4-2x-h$$
But then I saw this same equation on an explample about finding the differential of a function
$$f(x)=4x-x^2$$
Which is given by
$$\frac{(4(x+h)-(x+h)^2)-(4x-x^2)}{h}=\frac{4x+4h-x^2-2xh-h^2-4x+x^2}{h}$$
$$=\frac{4h-2xh-h^2}{h} = \frac{h(4-2x-h)}{h} = 4-2x$$
For some reason the other $h$ on the numerator got cut out too, and I know it has something to do with an implied limit as $h\to0$ on this formula for finding the differential, but I don't have a clue on what is that mean.
 A: You are right in the last equation $\frac{4h-2xh-h^2}{h} = \frac{h(4-2x-h)}{h} = 4-2x$, when it must be $\frac{h(4-2x-h)}{h} \rightarrow 4-2x$ as $h \rightarrow 0$.
In the beginning they used the definition of the derivative:
$$
f'(x) = \lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}.
$$
They calculated $\frac{f(x+h)-f(x)}{h}$ first, and after calculating they took the limit.
A: The derivative is that expression as $h$ approaches zero.  That's how the other $h$ got "cut out" (set to $0$) at the very end:
$$\lim_{h \to 0} (4 - 2x - h) = 4 - 2x - 0 = 4 - 2x.$$
A: If what you've written is a verbatim quote, then the source is incorrect. They are being sloppy with their limits.
If $f(x)=4x-x^2,$ then for any non-zero $h,$ we can say that $$\frac{f(x+h)-f(x)}{h}=4-2x-h,$$ as you determined. What this means is that $\frac{f(x+h)-f(x)}h$ and $4-2x-h$ are exactly the same for any non-zero $h.$ As we let $h$ get closer and closer to $0$, then, both expressions tend toward $4-2x,$ so we can say that $$\lim_{h\to 0}\frac{f(x+h)-f(x)}h=\lim_{h\to 0}[4-2x-h]=4-2x.$$ That would be a correct statement, and is what they are trying to say.
