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Limit question.
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\[\lim_{x \rightarrow 9}\frac{ x-9 }{ \sqrt{x} -3}\]
factor the numerator
First you can see that it will equal 0, so you have to do a different rule.
conjugate
nvm
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\[\frac{x-9}{\sqrt{x}-3}=\frac{(\sqrt{x}+3)(\sqrt{x}-3)}{\sqrt{x}-3}\] cancel replace \(x\) by \(9\)
Multiply both the top and bottom by sqrt(x) +3 This will give you (x-9)(sqrtx - 3)/x-9
You can cancel out (x-9) from both the top and bottom Leaving you with (sqrt(x) - 3) Does that make sense?
yeah that works as well either way
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