From the comments, it appears that it would be useful to give a fairly elementary and pedantic discussion of using the factor theorem (from precalculus or college algebra, in the U.S.).
Factor Theorem: If $r$ is a zero of a polynomial $P(x)$ (i.e. a solution to $P(x) = 0$ is $x=r$), then $x-r$ is a factor of $P(x)$.
Example 1: Factor $x^3 - 8$.
First, solve $x^3 - 8 = 0$ to find a value for $r$. Solving gives $x^3 = 8$, or $x = 2$. Therefore, we can use $r = 2$, which tells us that $x - 2$ is a factor of $x^3 - 8$. Now use long division (or synthetic division) to determine the quotient when $x - 2$ is divided into $x^3 - 8$. The quotient will be $x^2 + 2x + 4$. Therefore, we can factor $x^3 - 8$ as $(x-2)(x^2+2x+4).$
Example 2: Factor $x^3 + 8$.
First, solve $x^3 + 8 = 0$ to find a value for $r$. Solving gives $x^3 = -8$, or $x = -2$. Therefore, we can use $r = -2$, which tells us that $x - (-2)$, or $x+2$ is a factor of $x^3 + 8$. Now use long division (or synthetic division) to determine the quotient when $x + 2$ is divided into $x^3 + 8$. The quotient will be $x^2 - 2x + 4.$ Therefore, we can factor $x^3 + 8$ as $(x+2)(x^2-2x+4).$
Example 3: Factor $x^5 - 100,000$.
First, solve $x^5 - 100,000 = 0$ to find a value for $r$. Solving gives $x^5 = 100,000$, or $x = 10$. Therefore, we can use $r = 10$, which tells us that $x - 10$ is a factor of $x^5 - 100,000$. Now use long division (or synthetic division) to determine the quotient when $x - 10$ is divided into $x^5 + 100,000$. The quotient will be $x^4 + 10x^3 + 100x^2 + 1000x + 10,000$. Therefore, we can factor $x^5 - 100,000$ as $(x-10)(x^4 + 10x^3 + 100x^2 + 1000x + 10,000).$
Example 4: Factor $x^5 + x^4 + x^3 + x^2 + x + 1$.
First, solve $x^5 + x^4 + x^3 + x^2 + x + 1 = 0$ to find a value for $r$. In this case, standard methods don't work (unless you know about cyclotomic equations). However, by using the rational root test (google it), you'll get $x = 1$ and $x = -1$ as possible solutions. You'll find by direct substitution that $x = -1$ is a solution to $x^5 + x^4 + x^3 + x^2 + x + 1 = 0$. Therefore, we can use $r = -1$, which tells us that $x - (-1)$, or $x + 1$ is a factor of $x^5 + x^4 + x^3 + x^2 + x + 1$. Now use long division (or synthetic division) to determine the quotient when $x + 1$ is divided into $x^5 + x^4 + x^3 + x^2 + x + 1.$ The quotient will be $x^4 + x^2 + 1$. Therefore, we can factor $x^5 + x^4 + x^3 + x^2 + x + 1$ as $(x+1)(x^4 + x^2 + 1).$
Here's the really nice thing about your situation. There's almost no work in trying to find a solution to $P(x)= 0$. For example, in your specific limit problem you know that $x^3 + 8 = 0$ when $x=-2$ (because you presumably already tried to find the limit by plugging in $x=-2$). So you already know a value that can be used for $r$ in the factor theorem. In other words, something "tricky" to solve like Example 4 won't come up.
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.
Best Answer
Helpful fact: $t^n-a^n = (t-a)(t^{n-1}+t^{n-2}a^1+t^{n-3}a^2+...+t^1a^{n-2}+a^{n-1})$