# 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.

• ${x^2\over x-2}={x^2 -4+4\over x-2}=x+2+{4\over x-2}$ – Fermat Apr 10 '15 at 18:01
• You have it correct. – ncmathsadist Apr 10 '15 at 18:03
• If it were $x$ in the numerator, that would have a horizontal asymptote. – Mark Bennet Apr 10 '15 at 18:33

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 $-\infty$.
The line $x=2$ is a vertical asymptote as $$\lim_{x\to2^-}\frac{x^2}{x-2}=-\infty\\ \lim_{x\to2^+}\frac{x^2}{x-2}=\infty$$