lim h -> 0 ((-3 + h)^2 - 9) / h

Question

how do you get -1 instead of zero?

Answer 1

Recall that f'(x) = lim_h→0 [(f(x+h) - f(x))/h]

So, if we let f(x) = x², then the given limit is equal to f'(-3) = 2(-3) = -6

Answer 2

As I suggested on your previous post

Expand (-3 + h)² - 9 then divide by h and substitute 0 for h.

You should get -6 rather than -1