anonymous
  • anonymous
Limit of x-3/((root of x+6)-3) as x approaches 3?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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freckles
  • freckles
try multiply the conjugate of the denominator on top and bottom
anonymous
  • anonymous
would the conjugate be \[\sqrt{x+6}+3\] ?
freckles
  • freckles
yep!

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anonymous
  • anonymous
thank you!
freckles
  • freckles
so you already got the limit?
freckles
  • freckles
\[\lim_{x \rightarrow 3}\frac{x-3}{\sqrt{x+6}-3} \\ \lim_{x \rightarrow 3}\frac{(x-3)(\sqrt{x+6}+3)}{(\sqrt{x+6}-3)( \sqrt{x+6}+3)}\] and you should see: \[(\sqrt{x+6}-3) (\sqrt{x+6}+3)=(x+6)-9=x-3\] so a common factor from top and bottom will "cancel"
anonymous
  • anonymous
And then the answer would be 6?
freckles
  • freckles
sqrt(3+6)+3 sqrt(9)+3 3+3 yes 6
anonymous
  • anonymous
Thank you!
freckles
  • freckles
np
idku
  • idku
\[\lim_{x \rightarrow 3}\dfrac{x-3}{\sqrt{x+6}-3}\]when you plug in x=3, you get 0/0, thus you can differentiate on top and bottom. \[\lim_{x \rightarrow 3}{\Large \frac{1-0}{\frac{1}{2\sqrt{x+6}}-0}}\] \[\lim_{x \rightarrow 3}{\Large \frac{1}{\dfrac{1}{2\sqrt{x+6}}}}\] \[\lim_{x \rightarrow 3}2{\sqrt{x+6}}\]\[2{\sqrt{3+6}}=6\]

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