First, and by definition, when dealing with
$$\lim_{x\to x_0}f(x)$$
we must assume $\,f\,$ is defined in some neighborhood of $\,x_0\,$ except , perhaps, on $\,x_0\,$ itself, and from here that in the process of taking the limit we have the right and the duty to assume $\,x\,$ approaches $\,x_0\,$ in any possible way but it is never equal to it.
Thus, and since in our case we always have $\,x\ne x_0=5\,$ during the limit process , we can algebraically cancel for the whole process.
$$\frac{x^2-25}{x-5}=\frac{(x+5)\color{red}{(x-5)}}{\color{red}{x-5}}=x+5\xrightarrow[x\to 5]{}10$$
The above process shows that the original function behaves exactly as the straight line $\,y=x+5\,$ except at the point $\,x=5\,$ , where there exists "a hole", as you mention.
A precondition in l'Hospital's Rule is that in order for it to apply, the limit
$$
\lim_{x\to\infty}\frac{f'(x)}{g'(x)}
$$
must exist (but is allowed to be $\pm \infty$). In this case, the limit does not exist, so it does not apply.
Best Answer
Here are the steps that I would take to prove it, under the assumption that $p$ and $q$ are continuous (without that assumption, or something very like it, there really isn't much you can do in general):
Edit: Thorough working out:
We ultimately want to disprove that $\lim_{x\to a}\frac{p(x)}{q(x)} = L$, so we just need to find a single $\epsilon>0$ that makes a contradiction. I pick $1$, because I like it (and because I actually know that they will all fail, so it doesn't matter which one I pick, so I go for one that is easy to work with). Since we assumed that the limit existed, that must mean that there is a $\delta>0$ that fulfills the definition $\lim_{x\to a}\frac{p(x)}{q(x)} = L$ for this specific value of $\epsilon$. In other words, for any $x\in (a-\delta, a+\delta)\setminus \{a\}$, we have $$ \left|\frac{p(x)}{q(x)} - L\right|<1\\ \left|\frac{p(x) - Lq(x)}{q(x)}\right|<1\\ \frac{|p(x)-Lq(x)|}{|q(x)|}<1\\ |p(x) - Lq(x)| < |q(x)| $$ Now let's use that $p$ and $q$ are continuous. $p$ being continuous means that there is a $\delta_p>0$ such that for any $x\in (a-\delta_p, a+\delta_p)$, we have $|p(x) - p(a)|<\frac{|p(a)|}2$ (here we use that $p(a)\neq 0$). This, in turn, means that $|p(x)|>\frac{|p(a)|}2$.
Similarily, we know that there is some $\delta_q>0$, such that for any $x\in (a-\delta_q, a+\delta_q)$ we have $|q(x)-q(a)| = |q(x)| <\frac{|p(a)|}{2(|L| + 1)}$ (here we also use that $p(a) \neq 0$, along with $q(a) = 0$). This, in turn, means that $(|L| + 1)|q(x)|<\frac{|p(a)|}{2}$.
If we now assume that $x\in (a-\delta_p, a+\delta_p)\cap (a-\delta_q, a+\delta_q)$, we can chain these two implications together to get $$ |Lq(x)| + |q(x)| < \frac{|p(a)|}{2} < p(x)\\ |q(x)|<|p(x)| - |Lq(x)|\\ |q(x)|<|p(x) - Lq(x)| $$ Now, let $\delta_f = \min(\delta, \delta_p, \delta_q)$, and pick an $x\in (a-\delta_f, a+\delta_f)\setminus\{a\}$, which is possible since $\delta_f>0$. For such an $x$, we have both $|q(x)|<|p(x) - Lq(x)|$ and $|p(x) - Lq(x)| < |q(x)|$, which is a contradiction.