Proove the limit as x approaches 4 for f(x)=x^2-8x= -16
Definition of Precise Limits
The Attempt at a Solution
I know that I want x^2-8x+16 (after moving the -16 over per the limit definition) to look like |x-4|
Factoring gets me (x-4)(x-4)<e
Because I have two factors, I want to bound one of them so that -1<|x-4|<1.
Adding the four to both sides, I get 3<x<5, but when I add four back in (since both factors are the same), I'm back to -1<|x-4|<1. This leads me to believe that C=1, but I don't think I'm doing this part right. I know the answer isn't 'e'.
I'm lost at this point.