There's a function $f(x)=\frac{(x+2)(x+1)}{(x-4)^2(x+1)}$
I'm a little confused: is this function considered to be defined at $x=-1$? On the one hand, if you plug in $-1$, you'll get $0/0$. But on the other hand, $(x+1)/(x+1)=1$... Also does $x=-1$ count as a zero of this function? I have the same confusion here.
And if somebody's asking "How many removable singularities does $f$" have" (the correct answer is 1), what does "removable singularity" mean in real analysis? If it means removable discontinuity, then why does it have one such discontinuity? If we ignore the $x+1$'s (whose quotient is 1), the only discontinuity is an infinite discontinuity.
The function is not defined at $x=-1$ due to division by zero. However, it does have a limit as $x\rightarrow -1,$ namely
$$\lim_{x\rightarrow -1}f(x)=\frac{(x+2)}{(x-4)^2}\Bigg|_{x=-1}=-1/25.$$
The end result is that the graph of $f$ has a hole (removable discontinuity) at the point $(x,y)=(-1,-1/25)$.
The only other point where $f$ is not defined due to division by zero is at $x=4$. The limit of $f$ as $x\rightarrow 4$ is $+\infty$. Thus, $f$ has a vertical asymptote at $x=4$.