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anonymous

  • one year ago

Please help

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  1. anonymous
    • one year ago
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    \[\lim_{x \rightarrow 2}(\sqrt{x+2}-\sqrt{2x})\div(x^2-2x)\]

  2. anonymous
    • one year ago
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    @phi

  3. welshfella
    • one year ago
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    direct substitution gives 0/0 i would apply l'hopitals rule

  4. anonymous
    • one year ago
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    I haven't learned about that

  5. anonymous
    • one year ago
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    this week we just learned about intro to limits

  6. phi
    • one year ago
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    one thought is to multiply top and bottom by \( \sqrt{x+2}+ \sqrt{2x} \)

  7. welshfella
    • one year ago
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    ok so we need to find different way

  8. anonymous
    • one year ago
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    let me multiply it

  9. phi
    • one year ago
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    the top is a difference of squares

  10. anonymous
    • one year ago
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    |dw:1438036441844:dw|

  11. phi
    • one year ago
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    and don't multiply out the bottom you should get \[ \frac{ (x+2) - 2x} { x(x-2)(\sqrt{x+2}+\sqrt{2x})} \]

  12. phi
    • one year ago
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    simplify the top

  13. anonymous
    • one year ago
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    @phi did you multiply the whole equation by the conjugate?

  14. anonymous
    • one year ago
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    or by what factor?

  15. phi
    • one year ago
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    the top was (a-b)(a+b) = a^2 - b^2 where a and b are the square roots the bottom , I just show the multiplication

  16. phi
    • one year ago
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    I can't tell what you did , but it looks too complicated to be right.

  17. phi
    • one year ago
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    should I go slower?

  18. anonymous
    • one year ago
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    yes, please. I get where you said that they're difference of squares. But I don't get which factor did you multiply the whole equation to get that result

  19. phi
    • one year ago
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    \[ \frac{(\sqrt{x+2}-\sqrt{2x})}{(x^2-2x)} \\\frac{(\sqrt{x+2}-\sqrt{2x})}{x(x-2)} \] (factored the bottom, hopefully something cancels...

  20. phi
    • one year ago
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    \[ \frac{(\sqrt{x+2}-\sqrt{2x})}{x(x-2)} \cdot \frac{(\sqrt{x+2}+\sqrt{2x})}{(\sqrt{x+2}+\sqrt{2x})}\]

  21. phi
    • one year ago
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    it is easier to see the pattern if we call the first term a and the second b then we have (a-b)(a+b) which we know (right?) is a^2 - b^2

  22. anonymous
    • one year ago
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    yes

  23. phi
    • one year ago
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    or you can FOIL it out. either way the top becomes \[ (\sqrt{x+2})^2 - ( \sqrt{2x})^2 \\ x+2 - 2x \\ 2-x \] or -(x-2)

  24. anonymous
    • one year ago
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    okay, I got this part.

  25. phi
    • one year ago
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    so we now have the limit of \[ \frac{-(x-2)}{x(x-2)(\sqrt{x+2}+\sqrt{2x})}\]

  26. phi
    • one year ago
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    as long as x is not *exactly* 2 , we can cancel (x-2) from top and bottom then we can take the limit

  27. anonymous
    • one year ago
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    -1/8 right?

  28. phi
    • one year ago
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    yes

  29. anonymous
    • one year ago
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    OMG, thank you so much. I really learned everything

  30. phi
    • one year ago
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    yw

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