Question

CALCULUS: how to do this without L'Hospital's rule
lim x->2 (sqrt(6-x)-2)/(sqrt(3-x)-1)

Answer 1

multiply by the conjugate. maybe twice

Answer 2

@satellite73 i did but it gave me this and im stuck this seems too complicated to resolve.. http://puu.sh/14T7n

Answer 3

plz help anyone? :S

Answer 4

@KingGeorge

Answer 5

anyone still on this question? o_O

Answer 6

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

\[=\lim_{x\to 2}\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}\frac{\sqrt{6-x}+2}{\sqrt{6-x}+2}\frac{\sqrt{3-x}+1}{\sqrt{3-x}+1}\]

Answer 7

simplify...take limit

Answer 8

ok i just did, and it returned to its original form...

Answer 9

\[=\lim_{x\to 2}\frac{6-x-4}{3-x-1}\frac{1}{\sqrt{6-x}+2}\frac{\sqrt{3-x}+1}{1}\]

Answer 10

ya you see, it returns to the original form..

Answer 11

http://puu.sh/14TOp

Answer 12

\[=\lim_{x\to 2}\frac{2-x}{2-x}\frac{1}{\sqrt{6-x}+2}\frac{\sqrt{3-x}+1}{1}\]

\[=\lim_{x\to 2}\frac{\cancel{2-x}}{\cancel{2-x}}\frac{1}{\sqrt{6-x}+2}\frac{\sqrt{3-x}+1}{1}\]

\[=\lim_{x\to 2}\frac{\sqrt{3-x}+1}{\sqrt{6-x}+2}\]

Answer 13

OMFG MY BAD

Answer 14

IT CAN ACTUALLY BE SOLVED NOW!! THANKS!!!!

Answer 15

please don't write 'OMFG'