How many asymptotes does $y=\frac{x^2}{x-2}$ have? In a question I came upon, the answer insisted that there were three; one was apparently a horizontal asymptote, which I do not agree with. There are only 2 asymptotes, correct? One is $y=x+2$ and the other is obviously $x=2$. I used limits and long division to make this conclusion.

You're right:
$$
\lim_{x\to\infty}\frac{x^2}{x-2}=\infty\\
\lim_{x\to-\infty}\frac{x^2}{x-2}=-\infty
$$
Moreover
$$
\lim_{x\to\infty}\frac{x^2}{x-2}\frac{1}{x}=1
$$
and
$$
\lim_{x\to\infty}\left(\frac{x^2}{x-2}-x\right)=
\lim_{x\to\infty}\frac{2x}{x-2}=2
$$
(the same at $-\infty$), so $y=x+2$ is an oblique asymptote at $\infty$ and at $-\infty$.
There cannot be a horizontal asymptote.
The line $x=2$ is a vertical asymptote as
$$
\lim_{x\to2^-}\frac{x^2}{x-2}=-\infty\\
\lim_{x\to2^+}\frac{x^2}{x-2}=\infty
$$
