Question

sqrt(6-x)-2 / sqrt(3-x)-1
find the limit as x->2

Answer 1

satellite go ahead and tell him that he might have to multiply bottom and top by the conjugate of both the top and the bottom

Answer 2

i did that already, but it was deleted!

Answer 3

as i was writing it, actually as i was trying to post it, it vanished!

Answer 4

so you go ahead because i got annoyed

Answer 5

\[\lim_{x \rightarrow ?}\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1} \cdot \frac{\sqrt{6-x}+2}{\sqrt{3-x}+1} \cdot \frac{\sqrt{3-x}+1}{\sqrt{6-x}+2}\]

Answer 6

\[=\lim_{x \rightarrow 2}\frac{(6-x)-2}{(3-x)-1} \cdot \frac{\sqrt{3-x}+1}{\sqrt{6-x}+2}\]

Answer 7

\[=\lim_{x \rightarrow 2}\frac{-x+4}{-x+2} \cdot \frac{\sqrt{3-x}+1}{\sqrt{6-x}+2}\]
maybe i should had tried just multiplying top and bottom by the conjugate of the bottom

Answer 8

no you made an arithmetic error

Answer 9

oh i see -2(2)=-4

Answer 10

i think

Answer 11

damn site is so slow it is not really worth the wait tonight. i post, and then noting happens
but the zeros will cancel because your numerator should be
\[2-x\]

Answer 12

\[\lim_{x \rightarrow ?}\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1} \cdot \frac{\sqrt{6-x}+2}{\sqrt{3-x}+1} \cdot \frac{\sqrt{3-x}+1}{\sqrt{6-x}+2} \]
\[=\lim_{x \rightarrow 2}\frac{(6-x)-4}{(3-x)-1} \cdot \frac{\sqrt{3-x}+1}{\sqrt{6-x}+2}\]

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

like this in the morning too. hope it is not a trend

Answer 17

tyvm

Answer 18

this site is going freakin' slow tonight :(

Answer 19

i knw