Types of undefined for removable discontinuities and vertical asymptotes Consider this rational function:
$$
f(x) = \frac{x^2 - 2x - 24}{x^2 + 10x + 24}
$$
I have been taught that to solve for a removable discontinuity, I find the $x$ values such that both the numerator and denominator are equal to $0$; and to solve for vertical asymptotes, I find the $x$ values that make just the denominator equal to zero. So:
$$
f(x) = \frac{(x-6)(x+4)}{(x+6)(x+4)} = \frac{x-6}{x+6}, \quad x \neq -4
$$
What this means is that we have a vertical asymptote at $x = -6$ and a removable discontinuity at $x = -4$. Great. I can compute these.
But why?
We have two kinds of undefined here, $f(x_0) = \frac{0}{0}$ and $f(x_0) = \frac{g(x)}{0}$. Why do these result in different types of undefined behavior?

We have that


*

*at $x=-4$ the function is not defined but we don't have any vertical asymptote since for  $x\to -4$ we have $f(x) \to -5$

*at $x=-6$ also the function is not defined but we have a vertical asymptote since for  $x\to -6$ we have $f(x) \to \pm \infty$
Note that in the case of $x=-4$ we can remove the discontinuity defining $f(-4)=-5$ but for $x=-6$ we can't since $|f(x)| \to \infty$.
Note also that not always the discontinuity in the form $f(x_0) = \frac{0}{0}$ can be removed as for example for
$$f(x)=\frac{\log (1+x)}{x^2}$$
at $x=0$ since $|f(x)| \to \infty$.
