Question

find the limit as x approaches 9 : WILL GIVE EQUATION IN RESPONSE

Answer 1

\[\lim_{x \rightarrow 9}(\frac{ 1 }{ \sqrt{x}-3 } - \frac{ 6 }{ x-9 })\]

Answer 2

do you know l hospital's rule?

Answer 3

please don't use lhopitals or anything im in basic calculus and haven't learned it yet. Suposed to be able to algebriaically change it to solve. I know the answer is suposed to be 1/6 i just don't know how to get there

Answer 4

ok by basics ,
rationalise first fraction

Answer 5

multiply by \(\sqrt{x}+3\) on numerator and denominator

Answer 6

so the left turns into right?

Answer 7

ugh didn't get the equation in

Answer 8

\[\frac{ \sqrt{x}+3 }{ x-3 }\]

Answer 9

try again ?

Answer 10

is that not right after multiplying by \[\sqrt{x}\]+3?

Answer 11

so i now have \[\frac{ \sqrt{x}+3 }{ x-9 } - \frac{ 6 }{ x-9}\]

Answer 12

ok now add them !! note denominator is same

Answer 13

well that gives me zero... and i know the answer is suposed to be 1/6

Answer 14

try again?

Answer 15

right, nevermind. thanks!

Answer 16

no i still get zero

Answer 17

\[\frac{ \sqrt{x}+3 }{ x-9 }- \frac{ 6 }{ x-9 }\]