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?

Multiply the expression's numerator and denominator by the conjugate of $\;\sqrt x - 3 = \sqrt x + 3$
$$\frac{x-9}{\sqrt x-3}=\frac{(x-9)(\sqrt x + 3)}{(\sqrt x - 3)(\sqrt x + 3)} = \frac{(x-9)(\sqrt x+3)}{(x-9)}=\sqrt x+3\quad\text{ provided }\; x\ne 9$$ 
Note: we evaluate the limit of $f(x)$ as $x\to 9$; we are not evaluating $f(9)$. So there is no problem if it happens to be the case that the expression is valid for $x\ne 9$.  
Recall, evaluating a limit is determining the value of the expression as $x \to 9$, not finding its value AT $x = 9$. So the expression may be discontinuous AT 9, but its limit is the value we obtain when $x$ gets VERY CLOSE to 9. 
This gives us:
$$\lim_{x\to 9}\frac{x-9}{\sqrt x-3}=\lim_{x\to 9}(\sqrt x+3)=\sqrt 9+3=6$$
