assuming you mean
[tex] \lim_{x \to 4} \frac{\sqrt{x+5}-3}{4-x}[/tex]
that means as x approaches 4
if we sub 0 for x we get
0/0
and intermitent form
use l'hopital's rule
so
take the derivitive of the top and bottom seperatly
l'hopitals rule is something like
if [tex] \lim_{x \to n} \frac{f(x)}{g(x)}[/tex] results in 0/0 or -∞/∞ or∞/∞ then keep doing it until f(n)/g(n) gives a form that isn't intermitent
so
take derivitive of top and bottom
[tex]\dfrac{\frac{1}{2\sqrt{x+5}}}{-1}[/tex]
now, if we subsitute 4 for x we get
[tex]\dfrac{\frac{1}{2\sqrt{4+5}}}{-1}[/tex]=
[tex]\dfrac{\frac{1}{2\sqrt{9}}}{-1}[/tex]=
[tex]\dfrac{\frac{1}{2(3)}}{-1}[/tex]=
[tex]\dfrac{\frac{1}{6}}{-1}[/tex]=
[tex]\dfrac{1}{-6}=\dfrac{-1}{6}[/tex]
