Limit of a quotient with a radical in the numerator

Question

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.

$$\lim \limits_{x \to 6}{\sqrt{x+10}-x+2\over 3x-18}$$

Answer 1

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?

Answer 2

Hint: you can write $-x+2$ as $-(x-2)$

Another hint: Multiply by the conjugate: $$\frac{\sqrt{x+10}+(x-2)}{\sqrt{x+10}+(x-2)}$$ And recall that $(\alpha - \beta)(\alpha + \beta) = \alpha^2 - \beta^2$.