Limit of a quotient with a radical in the numerator I have a limit but I'm so confused in how to rationalize the numerator because it has two numbers separated. How should I change the signs, please help me out.
$$\lim \limits_{x \to 6}{\sqrt{x+10}-x+2\over 3x-18}$$

The numerator is $\sqrt{x+10} - (x-2)$, so let's multiply the numerator and denominator by $\sqrt{x+10} + (x-2)$. This gives us: 
$\dfrac{\sqrt{x+10} - (x-2)}{3x-18}$ 
$= \dfrac{\sqrt{x+10} - (x-2)}{3x-18} \cdot \dfrac{\sqrt{x+10} + (x-2)}{\sqrt{x+10} + (x-2)}$ 
$= \dfrac{(x+10)-(x-2)^2}{(3x-18)(\sqrt{x+10} + (x-2))}$ 
$= \dfrac{-x^2+5x+6}{(3x-18)(\sqrt{x+10} + (x-2))}$ 
$= \dfrac{-(x+1)(x-6)}{3(x-6)(\sqrt{x+10} + (x-2))}$
Can you finish from here?
