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anonymous
 one year ago
Limit of x3/((root of x+6)3) as x approaches 3?
anonymous
 one year ago
Limit of x3/((root of x+6)3) as x approaches 3?

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freckles
 one year ago
Best ResponseYou've already chosen the best response.0try multiply the conjugate of the denominator on top and bottom

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0would the conjugate be \[\sqrt{x+6}+3\] ?

freckles
 one year ago
Best ResponseYou've already chosen the best response.0so you already got the limit?

freckles
 one year ago
Best ResponseYou've already chosen the best response.0\[\lim_{x \rightarrow 3}\frac{x3}{\sqrt{x+6}3} \\ \lim_{x \rightarrow 3}\frac{(x3)(\sqrt{x+6}+3)}{(\sqrt{x+6}3)( \sqrt{x+6}+3)}\] and you should see: \[(\sqrt{x+6}3) (\sqrt{x+6}+3)=(x+6)9=x3\] so a common factor from top and bottom will "cancel"

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And then the answer would be 6?

freckles
 one year ago
Best ResponseYou've already chosen the best response.0sqrt(3+6)+3 sqrt(9)+3 3+3 yes 6

idku
 one year ago
Best ResponseYou've already chosen the best response.0\[\lim_{x \rightarrow 3}\dfrac{x3}{\sqrt{x+6}3}\]when you plug in x=3, you get 0/0, thus you can differentiate on top and bottom. \[\lim_{x \rightarrow 3}{\Large \frac{10}{\frac{1}{2\sqrt{x+6}}0}}\] \[\lim_{x \rightarrow 3}{\Large \frac{1}{\dfrac{1}{2\sqrt{x+6}}}}\] \[\lim_{x \rightarrow 3}2{\sqrt{x+6}}\]\[2{\sqrt{3+6}}=6\]
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