if i am given a vertical asymptote at x=3, how to i test limx->3+ and limx->3- (or from the right and the left)
A vertical asymptote will mean it will either, on one or both sides, diverge to infinity or to negative infinity. You have to take the limit as you would any other problem and see what you get, there's nothing special about it.
Consider,
$\displaystyle \lim_{ x \to 3^+} \frac{1}{x-3} $
Since the denominator will become infinitly small, but remain a posative quantity because we're approaching from the side of $\displaystyle 3^+ $ this will become infinity.
But on the opposite side
$\displaystyle \lim_{ x \to 3^-} \frac{1}{x-3} $
It happesn that $\displaystyle 3^- < 3 $ so $\displaystyle 3^- - 3 = $ a very small negative quantity. Thus, this limit is negative infinity.