Limit Question- lim[x->2] (sqrt(6-x)-2)/(sqrt(3-x)-1)

[Wandoes]

2) (sqrt(6 - x) - 2)\/(sqrt(3 - x) - 1) = 1\/2" title="lim_(x->2) (sqrt(6 - x) - 2)/(sqrt(3 - x) - 1) = 1/2" data-attribution="">


I know

 

[tkhunny]

Does it help to ponder whether the expression is continuous AT x = 2?

 

[lookagain]

2) (sqrt(6 - x) - 2)\/(sqrt(3 - x) - 1) = 1\/2" title="lim_(x->2) (sqrt(6 - x) - 2)/(sqrt(3 - x) - 1) = 1/2" data-attribution="">


I know

If you wanted to use conjugates of the numerator and the denominator, you could begin with these steps:


\(\displaystyle \displaystyle\lim_{x\to \ 2} \ \dfrac{\sqrt{6 - x} \ - \ 2}{\sqrt{3 - x} \ - \ 1} \ = \)

\(\displaystyle \displaystyle\lim_{x\to \ 2} \ \dfrac{(\sqrt{6 - x} \ - \ 2)(\sqrt{6 - x} \ + \ 2)(\sqrt{3 - x} \ + \ 1)}{(\sqrt{3 - x} \ - \ 1 )(\sqrt{6 - x} \ + \ 2)(\sqrt{3 - x} \ + \ 1)} \ = \)

\(\displaystyle \displaystyle\lim_{x\to \ 2} \ \dfrac{(\sqrt{6 - x} \ - \ 2)(\sqrt{6 - x} \ + \ 2)(\sqrt{3 - x} \ + \ 1)}{(\sqrt{3 - x} \ - \ 1 )(\sqrt{3 - x} \ + \ 1)(\sqrt{6 - x} \ + \ 2)} \ = \)