1. ## Finding a limit involving two different radicals

$\lim_{x\rightarrow4} \frac{\sqrt{8-x}-2} {\sqrt{29-x}-5}$

Every attempt at multiplying either the top or bottom by the negated counterpart ends up leading to a 0/0 answer. I know there's some trick I'm missing to solve this, could some one point me in the right direction without giving away the answer?

2. Originally Posted by penguinpwn
$\lim_{x\rightarrow4} \frac{\sqrt{8-x}-2} {\sqrt{29-x}-5}$

Every attempt at multiplying either the top or bottom by the negated counterpart ends up leading to a 0/0 answer. I know there's some trick I'm missing to solve this, could some one point me in the right direction without giving away the answer?
$\displaystyle \lim_{x\to4} \frac{\sqrt{8-x}-2} {\sqrt{29-x}-5} \cdot \frac{\sqrt{8-x}+2}{\sqrt{8-x}+2} \cdot \frac{\sqrt{29-x}+5}{\sqrt{29-x}+5}$

$\displaystyle \lim_{x\to4} \frac{[(8-x)-4] \cdot [\sqrt{29-x}+5]}{[\sqrt{8-x}+2] \cdot [(29-x)-25]}$

finish it

3. Originally Posted by skeeter
$\displaystyle \lim_{x\to4} \frac{\sqrt{8-x}-2} {\sqrt{29-x}-5} \cdot \frac{\sqrt{8-x}+2}{\sqrt{8-x}+2} \cdot \frac{\sqrt{29-x}+5}{\sqrt{29-x}+5}$

$\displaystyle \lim_{x\to4} \frac{[(8-x)-4] \cdot [\sqrt{29-x}+5]}{[\sqrt{8-x}+2] \cdot [(29-x)-25]}$

finish it
Gahhh my mind isn't abstract enough for calculus...