Question

what is the limit of sqrt(6-x) -2 / sqrt(3-x) -1 as x approaches 2?

Answer 1

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

Answer 2

as x -> 2

Answer 3

does it mean x=2

Answer 4

well, it means as x approaches 2, so no then

Answer 5

Multiply by the conjugates:
\[\frac{ \sqrt{6-x} -2 }{ \sqrt{3-x}-1 }\cdot\frac{\sqrt{6-x} +2}{\sqrt{6-x} +2}\cdot\frac{\sqrt{3-x}+1}{\sqrt{3-x}+1}\]

Answer 6

okay, I did that

Answer 7

Right, so you have
\[\frac{ 6-x -4 }{ 3-x-1 }\cdot\frac{{\sqrt{3-x}+1}}{\sqrt{6-x} +2}\]
\[\frac{ 2-x }{ 2-x }\cdot\frac{{\sqrt{3-x}+1}}{\sqrt{6-x} +2}\]
\[\frac{{\sqrt{3-x}+1}}{\sqrt{6-x} +2}\]
Now directly substituting should work.

Answer 8

ya, so i crossed out the first part of the denominator with the first part of the numerator

Answer 9

and then I became stuck

Answer 10

do I just plug 2 into x now is my question

Answer 11

Yeah, that's what I meant by directly substituting.

Answer 12

okay, ya. So i was on the right track, but I doubted myself.
Thank you so much :)

Answer 13

You're welcome