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\[\lim_{x \rightarrow 9}\frac{ x-9 }{ \sqrt{x}-3 }\]
Multiply top and bottom by the conjugate of the denominator. You should then be able to cancel and find the limit.
ohhh thank you @dylanhouse :)
or try factoring the numerator as \[(\sqrt{x}+3)(\sqrt{x}-3)\]and then cancelling
Oh yeah both would work right ?
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yes
you will get \(\sqrt{x}+3\) not matter what then replace \(x\) by 9
\[\frac{ x-9 }{\sqrt{x}-3 } \times \frac{ \sqrt{x}+3 }{ \sqrt{x}+3 }\]
Now just expand.
=6
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Absolutely :)
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