Okay the title could use some work ^^
My question - How come when we evaluate $\lim_{h\to 0} \frac {(1+h)^2-1^2}{h}$ we actually evaluate the $-1^2$ as $-(1^2)$ instead of $(-1)^2$. If the answer is "because the neg. sign is not inside of the parentheses, then how are we supposed to always know if we should incl. the plus or minus sign into the parentheses when we convert from say $f'(1)\lim_{h\to 0} \frac {f(1+h)-f(1)}{h}$ to $\lim_{h\to 0} \frac {(1+h)^2-1^2}{h}$?
Thanks!
On
Edit: while I was answering you edited the question to remove the first part.
I think there's something you haven't told us before the first question. I suspect it's that $f(x) = x^2$. Then the equivalence of the two expressions is just what you get when you substitute $1+h$ and $1$ for $x$.
For the second, you need to know the convention that $-A^2$ is $-(A^2)$ and not $(-A)^2$.
Your problems here suggest that you should do a thorough review of your algebra before you tackle calculus.
Just by definition. Obviously, you mean $$f(x) := x^2.$$