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Limit of a fraction

  1. Jan 26, 2008 #1
    1. The problem statement, all variables and given/known data
    [tex]\lim_{x\to 6}\frac{\sqrt{x+3}-3}{x-6}[/tex]


    2. Relevant equations

    Answer is 1/6.

    3. The attempt at a solution
    [tex]\lim_{x\to 6}\frac{\sqrt{x+3}-3}{x-6}=\lim_{x\to 6}\frac{(x+3-9)(x+6)}{(x^2-36)(\sqrt{x+3}+3)}[/tex]
     
  2. jcsd
  3. Jan 26, 2008 #2
    Look more carefully at your attempt at a solution. You actually have it.
     
  4. Jan 26, 2008 #3

    arildno

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    Do not bother with expanding with (x+6) as well:
    [tex]\frac{\sqrt{x+3}-3}{(x-6)}=\frac{x+3-9}{(x-6)(\sqrt{x+3}+3)}=\frac{(x-6)}{(x-6)(\sqrt{x+3}+3)}[/tex]
     
  5. Jan 26, 2008 #4
    "Look more carefully at your attempt at a solution. You actually have it."
    I don't see.
    [tex]\lim_{x\to 6}\frac{\sqrt{x+3}-3}{x-6}=\lim_{x\to 6}\frac{(x+3-9)(x+6)}{(x^2-36)(\sqrt{x+3}+3)}=\lim_{x\to 6}\frac{x^2+6x-6x-36}{x^2\sqrt{x+3}+3x^2-36\sqrt{x+3}-108}[/tex]

    I get it.
    [tex]\lim_{x\to 6}\frac{\sqrt{x+3}-3}{x-6}=\lim_{x\to 6}\frac{x+3-9}{(x-6)(\sqrt{x+3}+3)}=\lim_{x\to 6}\frac{1}{\sqrt{x+3}+3}=\frac{1}{6}[/tex]
     
    Last edited: Jan 26, 2008
  6. Jan 26, 2008 #5
    How can you write [itex]x^2-36[/itex]?
     
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