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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.

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    $\begingroup$ ${x^2\over x-2}={x^2 -4+4\over x-2}=x+2+{4\over x-2}$ $\endgroup$ – Fermat Apr 10 '15 at 18:01
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    $\begingroup$ You have it correct. $\endgroup$ – ncmathsadist Apr 10 '15 at 18:03
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    $\begingroup$ If it were $x$ in the numerator, that would have a horizontal asymptote. $\endgroup$ – Mark Bennet Apr 10 '15 at 18:33
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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 $$ \lim_{x\to2^-}\frac{x^2}{x-2}=-\infty\\ \lim_{x\to2^+}\frac{x^2}{x-2}=\infty $$

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