Classifying the discontinuity of a rational-radical function

Multiply numerator and denominator by the conjugate $\sqrt{x+5}+3$: $e(x) = \frac{(\sqrt{x+5}-3)(\sqrt{x+5}+3)}{(x-4)(\sqrt{x+5}+3)} = \frac{(x+5)-9}{(x-4)(\sqrt{x+5}+3)} = \frac{x-4}{(x-4)(\sqrt{x+5}+3)}.$ For $x \neq 4$, the factor $(x-4)$ cancels, leaving $e(x) = \frac{1}{\sqrt{x+5}+3}.$

Now take the limit of the simplified expression as $x \to 4$: $\lim_{x\to 4} \frac{1}{\sqrt{x+5}+3} = \frac{1}{\sqrt{9}+3} = \frac{1}{6}.$ Since this limit exists and is finite but $e(4)$ is undefined, the graph of $e(x)$ has a hole at $x=4$ that could be patched by defining $e(4)=\frac{1}{6}$. This is a removable discontinuity.