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

Mathematics
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\[\lim_{x \rightarrow 9}(\frac{ 1 }{ \sqrt{x}-3 } - \frac{ 6 }{ x-9 })\]
do you know l hospital's rule?
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

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Other answers:

ok by basics , rationalise first fraction
multiply by \(\sqrt{x}+3\) on numerator and denominator
so the left turns into right?
ugh didn't get the equation in
\[\frac{ \sqrt{x}+3 }{ x-3 }\]
try again ?
is that not right after multiplying by \[\sqrt{x}\]+3?
so i now have \[\frac{ \sqrt{x}+3 }{ x-9 } - \frac{ 6 }{ x-9}\]
ok now add them !! note denominator is same
well that gives me zero... and i know the answer is suposed to be 1/6
try again?
right, nevermind. thanks!
no i still get zero
\[\frac{ \sqrt{x}+3 }{ x-9 }- \frac{ 6 }{ x-9 }\]
|dw:1359220621580:dw|
\[\frac{ \sqrt{x}-3 }{ x-9 }\] and if i plug in 9 it will be 0 over 0
please dont jump into putting vale ! we are not over yet!
*value
ok, what do i do if i dont put the value in then
because your calculations give me \[\lim_{x \rightarrow 9}\frac{ \sqrt{x}-3 }{ x-9 }\]
do i split the bottom up? into its factors?
found it out! thanks!!
yw : )
|dw:1359220790623:dw|

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