# limit problem (with roots)

Is it possible to evaluate this limit without graphing or guessing (ie to replace it by a simpler function)

$$\lim_{x\to 2} \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$$

I tried normalizing by multiplying by the conjugate (both denominator and numerator) didn't work.

## 2 Answers

$$\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}=\frac{(\sqrt{6-x}-2)\times (\sqrt{6-x}+2)\times(\sqrt{3-x}+1)}{(\sqrt{3-x}-1)\times(\sqrt{3-x}+1)\times (\sqrt{6-x}+2)}$$

$$=\frac{\left((\sqrt{6-x})^2-4\right)\times(\sqrt{3-x}+1)}{\left((\sqrt{3-x})^2-1\right)\times (\sqrt{6-x}+2)}=\frac{\overbrace{\left((6-x)-4\right)}^{(2-x)}\times(\sqrt{3-x}+1)}{\underbrace{\left((3-x)-1\right)}_{(2-x)}\times (\sqrt{6-x}+2)}$$

$$=\frac{\sqrt{3-x}+1}{\sqrt{6-x}+2},~~x\ne 2$$

You can use L'Hopital's Rule:

$\lim_{x\rightarrow c} \frac{f(x)}{g(x)}=\lim_{x\rightarrow c} \frac{f'(x)}{g'(x)}$

In your case, $\frac{f'(x)}{g'(x)}$ is equal to $\frac{\sqrt{3-x}}{\sqrt{6-x}}$, so the limit when $x$ tends to $2$ is equal to $\frac{1}{2}$.