How do I evaluate without using L'Hopital's rule
lim sqrt(x^2-9)/abs(3-x) as x->3-
I didn't manage to manipulate the algebric expression. I got to sqrt((x+3)/(x-3)) as x->3- and I don't know what to do from here
Since you are approaching $\displaystyle 3$ from the left, you need to take notice of what happens with the values of $\displaystyle x < 3$.
Notice that $\displaystyle |x - 3| = -(x - 3) = 3 - x$ if $\displaystyle x < 3$.
So $\displaystyle \lim_{x \to 3^{-}}\frac{\sqrt{x^2 - 9}}{|x - 3|} = \lim_{x \to 3^{-}} \frac{\sqrt{x^2 - 9}}{3 - x}$
$\displaystyle = \lim_{x \to 3^{-}}\frac{x^2 - 9}{(3 - x)\sqrt{x^2 - 9}}$
$\displaystyle = \lim_{x \to 3^{-}}\frac{-(3 - x)(x + 3)}{(3 - x)\sqrt{x^2 - 9}}$
$\displaystyle = \lim_{x \to 3^{-}}\frac{-(x + 3)}{\sqrt{x^2 - 9}}$.
This now tends to $\displaystyle -\frac{6}{0}$ which tends to $\displaystyle -\infty$.