Finding vertical asymptotes 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.
 A: 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.
A: 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).

