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Limit of (sqrt(x) - 2)(x-4) as x approaches 4 from the left

  1. Jul 10, 2008 #1
    1. The problem statement, all variables and given/known data
    [tex]\lim_{x\rightarrow 4^{-}}\frac{\sqrt{x}-2}{x-4}[/tex]

    3. The attempt at a solution
    Well, I used a calculator to substitute arbitrary values to see what it appears to approach.

    input | output
    2 .293
    3 .268
    3.5 .258
    3.9 .252

    So I concluded that the the limiting value is in fact .25.. and I was wrong. How do I evaluate this?

    Last edited: Jul 10, 2008
  2. jcsd
  3. Jul 10, 2008 #2


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    Homework Helper

    Re: Limit

    Use L'Hospital's rule.
  4. Jul 10, 2008 #3
    Re: Limit

    Thank you, I got it.
  5. Jul 10, 2008 #4


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    Staff Emeritus
    Science Advisor

    Re: Limit

    A fairly standard way of handling a problem like that, and simpler than apply L'Hopital's rule is to rationalize the numerator. Multiplying both numerator and denominator by [itex]\sqrt{x}+ 2[/itex] you get
    [tex]\frac{(\sqrt{x}-2)(\sqrt{x}+2)}{(x-4)(\sqrt{x}+2)}= \frac{x- 4}{(x-4)(\sqrt{x}+ 2)}[/tex]
    which, for x> 4, is
    [tex]\frac{1}{\sqrt{x}+ 2}[/tex]
    Of course, the limit of that, as x goes to 2, is the same as the limit of the original problem.
  6. Jul 10, 2008 #5
    Re: Limit

    Yeah, that is what I did. I'm not up to the rule yet. Thanks again halls of ivy.
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