Evaluating the limit of $\lim_{x \to 9} \frac {x-9} {\sqrt{x} - 3}$ How do you evaluate the limit of 
$$\lim_{x \to 9} \frac {x-9} {\sqrt{x} - 3}$$
by first principle? I was thinking of this way to evaluate but it seems to evaluate to $\frac {0} {0}$.
$$\lim_{x \to 9} \frac {x-9} {\sqrt{x} - 3} = \frac{\lim_{x \to 9} (x-9) }{\lim_{x \to 9}  {\sqrt{x} - 3}}$$
I mentioned that the limits do not exist because the function is not continuous at $ x=9$. Any idea on how to solve this question?

$$ \lim_{x \rightarrow 9} \frac{x-9}{\sqrt(x)-3} = \lim_{x \rightarrow 9} \frac{1}{\frac{1}{2\sqrt{x}}} = 6 $$
Here L'Hopital's rule is used.
