I'm going to assume you meant to say
[tex]\displaystyle \lim_{x\to-4} \frac{x+4}{x^2-16}[/tex]
Factorize the denominator to reveal a removable discontinuity in the limand:
[tex]\dfrac{x+4}{x^2-16} = \dfrac{x+4}{(x+4)(x-4)}[/tex]
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
[tex]\displaystyle \lim_{x\to-4} \frac{x+4}{x^2-16} = \lim_{x\to-4} \frac1{x-4} = \frac1{-4-4} = \boxed{-\frac18}[/tex]
