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So I am trying to find the behaviour of this function around an asymptote at $x=0$.

$a>0$

$y=\frac{(x-a)(x^2+a)}{x^2}$

I know that as $x \to 0^+$, $y \to -\infty $ and $x \to 0^-$, $y \to -\infty $.

I've tried dividing the top and bottom of $y$ by $x^2$ and ended with:

$x-a+\frac{a}{x}-\frac{a^2}{x^2}$

Which does not help me as I end with $\infty -\infty$. Any help would be greatly appreciated.

  • 0
    What description of the behavior are you looking for? From your equation $y=x-a+\frac{a}{x}-\frac{a^2}{x^2}$ you can see that as $x$ approaches $0$ the last term will dominate. The fact that the numerator of that term (including the leading sign) is negative supports your claim that it goes to $-\infty$ whichever side of $0\ \ x$ is on.2011-10-05

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Expanding my comment above.

I know that as $x \to 0^+$, $y \to -\infty $ and $x \to 0^-$, $y \to -\infty $.

Only the first assertion is correct.

Since $\left( x-a\right) (x^{2}+a)=x^{3}-ax^{2}+ax-a^{2}$, we have $\begin{eqnarray*} \lim_{x\rightarrow 0}\frac{\left( x-a\right) (x^{2}+a)}{x^{2}} &=&\lim_{x\rightarrow 0}\left( x^{3}-ax^{2}+ax-a^{2}\right) \times \lim_{x\rightarrow 0}\frac{1}{x^{2}} \\ &=&-a^{2}\times \lim_{x\rightarrow 0}\frac{1}{x^{2}}=-a^{2}\left( +\infty \right) =-\infty, \qquad\text{ for }a\ne 0. \end{eqnarray*}$

So as $x$ approaches $0$, either from the left or from the right, $y\rightarrow -\infty $. The limit does not depend on the sign of $a$.

Plot of $y$ for $a=1$ (black) and $a=-1$ (green).

enter image description here

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The fact that $x-a$ is negative when $x$ is near $0$ and $x^2 + a$ positive is positive when $x$ is near zero, and the denominator is positive, will mean that the function must approach $-\infty$ at the vertical asymptote, unless the numerator also approaches $0$, in which case you've got more work to do. In this case the numerator approaches $-a^3$, which is not $0$, so that's it.

You're quite right about the fact that breaking it into $\infty-\infty$ won't help. If you had that, I'd tell you to write it as a single fraction by using a common denominator. That would take you to what you started with.