I assume the limit you're asked to copmute is supposed to be
[tex]\displaystyle \lim_{x\to5} \frac{\sqrt{x+4}-3}{x-5}[/tex]
Rationalize the numerator by multiplying it by its conjugate:
[tex]\dfrac{\sqrt{x+4}-3}{x-5} \times \dfrac{\sqrt{x+4}+3}{\sqrt{x+4}+3} = \dfrac{\left(\sqrt{x+4}\right)^2-3^2}{(x-5) \left(\sqrt{x+4}+3\right)} = \dfrac{x-5}{(x-5) \left(\sqrt{x+4}+3\right)}[/tex]
Since x is approaching 5, we have x ≠ 5 and can cancel the factors of x - 5 to remove the discontinuity.
Then
[tex]\displaystyle \lim_{x\to5} \frac{\sqrt{x+4}-3}{x-5} = \lim_{x\to5} \frac1{\sqrt{x+4}+3} = \frac1{\sqrt9+3} = \boxed{\frac16}[/tex]
because the simplified limand is continuous at x = 5.
