\(lim\left(x->4\right)\dfrac{\sqrt{x+5}-3}{x-4}=lim\left(x->4\right)\dfrac{x+5-9}{\left(x-4\right)\left(\sqrt{x+5}+3\right)}=lim\left(x->4\right)\dfrac{x-4}{\left(x-4\right)\left(\sqrt{x+5}+3\right)}=lim\left(x->4\right)\dfrac{1}{\sqrt{x+5}+3}\\ =\dfrac{1}{\sqrt{4+5}+3}=\dfrac{1}{3+3}=\dfrac{1}{6}\)