Answer
Domain at $$[-5,4) \cup (4, \infty)$$
Undefined point, or hole, at $$(4,\frac{1}{6})$$
Work Step by Step
First off, the general function for this equation is $$f(x)=\frac{\sqrt{x+5}-3}{x-4}$$
Since the limit is when "x is approaching 4", which can be written as $$\underset{x\rightarrow 4}{\lim}$$ that would mean that there will be an undefined point at $(4,y)$ since $4-4=0$ and anything with a denominator of $0$ is undefined.
So, since the function at the given limit is going to be undefined, then the first we can do is rationalize the numerator, which means you can multiply by $\sqrt{x+5}+3$, which is the conjugate of $\sqrt{x+5}-3$, which can be written as the following:
$$\begin{matrix}
\frac{\sqrt{x+5}-3}{x-4}\\\\
=\frac{\left(\sqrt{x+5}-3\right)\left(\sqrt{x+5}+3\right)}{\left(x-4\right)\left(\sqrt{x+5}+3\right)}\\\\
=\frac{\left(\sqrt{x+5}\right)^2-3^2}{\left(x-4\right)\left(\sqrt{x+5}+3\right)}\\\\
=\frac{\left(x+5\right)-9}{\left(x-4\right)\left(\sqrt{x+5}+3\right)}\\\\
=\frac{x+5-9}{\left(x-4\right)\left(\sqrt{x+5}+3\right)}\\\\
=\frac{x+(5-9)}{\left(x-4\right)\left(\sqrt{x+5}+3\right)}\\\\
=\frac{x+(-4)}{\left(x-4\right)\left(\sqrt{x+5}+3\right)}\\\\
=\frac{x-4}{\left(x-4\right)\left(\sqrt{x+5}+3\right)}\\\\
=\frac{1}{\left(1\right)\left(\sqrt{x+5}+3\right)}\\\\
=\frac{1}{\sqrt{x+5}+3}\\\\
\end{matrix}$$
Now, onto the plug and play with the limitation of $$\frac{1}{\sqrt{x+5}+3}$$
Since the limit is when "x is approaching 4", which also means $x=4$, this means the following:
$$\begin{matrix}
_{x\rightarrow 4}^{\lim}\textrm{f(x)}&=\frac{1}{\sqrt{x+5}+3}\\\\
&=\frac{1}{\sqrt{(4)+5}+3}\\\\
&=\frac{1}{\sqrt{9}+3}\\\\
&=\frac{1}{3+3}\\\\
&=\frac{1}{6}
\end{matrix}$$
Since the limit has an undefined point at $(4,y)$ and $_{x\rightarrow 4}^{\lim}\textrm{f(x)}=\frac{1}{6}$, which means that the (official) undefined point, or hole, at $$(4,\frac{1}{6})$$ with the graph plotted down below.
So, since the domain is how the $x$-values range from left to right and since $(4,\frac{1}{6})$ being the undefined point, that would means that the domain is from both $5\:to\:4$ and $4\:to\: \infty$, which means it can be written as the following:
$$[-5,4) \cup (4, \infty)$$
