Question

X +4 2 - 16 lim X→-4 X

Answer 1

I'm going to assume you meant to say

$\displaystyle \lim_{x\to-4} \frac{x+4}{x^2-16}$

Factorize the denominator to reveal a removable discontinuity in the limand:

$\dfrac{x+4}{x^2-16} = \dfrac{x+4}{(x+4)(x-4)}$

In the limit, we're considering values of x *near* -4, and not x = -4 itself. So we can cancel the factors of x + 4 and we're left with

$\displaystyle \lim_{x\to-4} \frac{x+4}{x^2-16} = \lim_{x\to-4} \frac1{x-4} = \frac1{-4-4} = \boxed{-\frac18}$