lim h→ 0 of [((h−9)^2−9^2)/h]=?

Question

I do not know how to factorize this in the slightest

Answer 1

lim h→ 0 of [((h−9)^2−9^2)/h] = -18

Answer 2

It's been a while since I've done these, but I think you can work it out as follows:

((h−9)^2−9^2)/h = (h^2 - 18h + 9^2 - 9^2)/h = h - 18

Therefore lim h→0 (h - 18) = -18.

Answer 3

to factorize this u can use the identity :
(a^2 - b^2 ) = (a-b)(a+b) for any a and b

hence for your problem
=[{(h-9)-9} { (h-9)+9}/h
=(h-18)(h)/h
=(h-18) for h not equal to 0
as in our case h -->0 hence h close to 0 but not exactly 0.
= -18