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anonymous

  • one year ago

Limit of x-3/((root of x+6)-3) as x approaches 3?

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  1. freckles
    • one year ago
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    try multiply the conjugate of the denominator on top and bottom

  2. anonymous
    • one year ago
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    would the conjugate be \[\sqrt{x+6}+3\] ?

  3. freckles
    • one year ago
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    yep!

  4. anonymous
    • one year ago
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    thank you!

  5. freckles
    • one year ago
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    so you already got the limit?

  6. freckles
    • one year ago
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    \[\lim_{x \rightarrow 3}\frac{x-3}{\sqrt{x+6}-3} \\ \lim_{x \rightarrow 3}\frac{(x-3)(\sqrt{x+6}+3)}{(\sqrt{x+6}-3)( \sqrt{x+6}+3)}\] 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"

  7. anonymous
    • one year ago
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    And then the answer would be 6?

  8. freckles
    • one year ago
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    sqrt(3+6)+3 sqrt(9)+3 3+3 yes 6

  9. anonymous
    • one year ago
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    Thank you!

  10. freckles
    • one year ago
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    np

  11. idku
    • one year ago
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    \[\lim_{x \rightarrow 3}\dfrac{x-3}{\sqrt{x+6}-3}\]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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