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!

• Maybe it would be goo to define $f(x) := x^2$. Then everything should be clear. – TheGeekGreek Apr 8 '18 at 15:33
• What is $f(x)$? How do you define $f'(x)$? – Simply Beautiful Art Apr 8 '18 at 15:33
• @oijioj Is this a joke? – TheGeekGreek Apr 8 '18 at 15:35
• For the $-1^2$ question, recall your order of operations. PEMDAS tells us we need to first do parentheses (e.g. $(1+h)$), then exponents (e.g. $(1+h)^2$ and $1^2$), then multiplying/division, and finally adding/subtracting. – Simply Beautiful Art Apr 8 '18 at 15:36
• By your reasoning, $x-1^2$ should equal $x(-1)^2 = x$. There is a grammer for equations and its rules say that you treat $x^2-1^2$ as $x^2-(1)^2$. In real life, a mathematician will break a rule and do things differently if it simplifies what he or she is trying to achieve. So you can do it your way if you want. Except that it is considered polite to point out what you are doing and why. – steven gregory Jul 3 '18 at 8:43

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 $(-A)^2$.
Just by definition. Obviously, you mean $$f(x) := x^2.$$