gabie1121
find the limit as x approaches 9 : WILL GIVE EQUATION IN RESPONSE
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gabie1121
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\[\lim_{x \rightarrow 9}(\frac{ 1 }{ \sqrt{x}-3 } - \frac{ 6 }{ x-9 })\]
AravindG
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do you know l hospital's rule?
gabie1121
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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
AravindG
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ok by basics ,
rationalise first fraction
AravindG
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multiply by \(\sqrt{x}+3\) on numerator and denominator
gabie1121
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so the left turns into right?
gabie1121
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ugh didn't get the equation in
gabie1121
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\[\frac{ \sqrt{x}+3 }{ x-3 }\]
AravindG
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try again ?
gabie1121
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is that not right after multiplying by \[\sqrt{x}\]+3?
gabie1121
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so i now have \[\frac{ \sqrt{x}+3 }{ x-9 } - \frac{ 6 }{ x-9}\]
AravindG
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ok now add them !! note denominator is same
gabie1121
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well that gives me zero... and i know the answer is suposed to be 1/6
AravindG
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try again?
gabie1121
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right, nevermind. thanks!
gabie1121
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no i still get zero
gabie1121
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\[\frac{ \sqrt{x}+3 }{ x-9 }- \frac{ 6 }{ x-9 }\]
AravindG
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gabie1121
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\[\frac{ \sqrt{x}-3 }{ x-9 }\] and if i plug in 9 it will be 0 over 0
AravindG
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please dont jump into putting vale ! we are not over yet!
AravindG
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*value
gabie1121
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ok, what do i do if i dont put the value in then
gabie1121
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because your calculations give me \[\lim_{x \rightarrow 9}\frac{ \sqrt{x}-3 }{ x-9 }\]
gabie1121
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do i split the bottom up? into its factors?
gabie1121
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found it out! thanks!!
AravindG
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yw : )
RadEn
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