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Homework Help: Find the limit

  1. Oct 8, 2011 #1
    1. The problem statement, all variables and given/known data

    Find the lim of [itex]\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}[/itex]when x approaches 2, without using L'hopital rule or the definition.



    2. Relevant equations



    3. The attempt at a solution
    I try to multiply both the numerator and the denominator to get [itex]\frac{(\sqrt{6-x}-2)(\sqrt{3-x}-1)}{2-x}[/itex], but when I substitute x = 2, the term becomes undefined.

    Any suggestion? Thanks.

    I am thinking of de'rationalize' the rational surd fraction, but nothing works.
     
    Last edited: Oct 8, 2011
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  3. Oct 8, 2011 #2

    arildno

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    1. First off:
    There should be a "+1" rather than "-1" in the remaining factor after you myliplied both denominator and numerator with the 2conjugate".

    2. secondly:
    do the same "conjugate trick", with the surd expression that is contained in the numerator.
     
  4. Oct 8, 2011 #3
    Oops, sorry. Now I have

    [itex]\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}[/itex]

    =[itex]\frac{(\sqrt{6-x}-2)(\sqrt{3-x}+1)}{(\sqrt{3-x}-1))(\sqrt{3-x}+1)}[/itex]

    =[itex]\frac{(\sqrt{6-x}-2)(\sqrt{3-x}+1)}{2-x}[/itex]
     
  5. Oct 8, 2011 #4

    arildno

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    Correct.
    Now do the same trick with the other surd expression.
     
  6. Oct 8, 2011 #5
    Yay, I get it. I multiply by another conjugate and eventually the (2-x) term will cancel out. Then I will get 2/4 which is equal to 1/2. Thanks so much for the help. :)

    [itex]\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}[/itex]

    =[itex]\frac{(\sqrt{6-x}-2)(\sqrt{3-x}+1)}{(\sqrt{3-x}-1))(\sqrt{3-x}+1)}[/itex]

    =[itex]\frac{(\sqrt{6-x}-2)(\sqrt{3-x}+1)}{2-x}[/itex]

    =[itex]\frac{(\sqrt{6-x}-2)(\sqrt{3-x}+1)(\sqrt{6-x}+2)}{(2-x)(\sqrt{6-x}+2)}[/itex]

    =[itex]\frac{(\sqrt{3-x}+1)(2-x)}{(2-x)(\sqrt{6-x}+2)}[/itex]

    =[itex]\frac{\sqrt{3-x}+1}{\sqrt{6-x}+2}[/itex]

    So if I replace x = 2, I will get 2/4 = 1/2. :)
     
  7. Oct 8, 2011 #6

    arildno

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  8. Oct 8, 2011 #7
    Another question:

    Find the limit of [itex]\frac{sin^{-1}x}{x}[/itex] when x approaches 0.

    The answer given is 1.

    Seems like I can't express arcsin explicitly. Any clue on how to start? :)
     
  9. Oct 8, 2011 #8
    Can I use the series expansion?

    [itex]\frac{sin^{-1}x}{x}[/itex]

    =[itex]\frac{x+\frac{X^3}{6}+\frac{3x^5}{40}+...}{x}[/itex]

    =1+[itex]\frac{x^2}{6}+\frac{3x^4}{40}+...[/itex]

    Then I can get limit = 1 when I substitute x = 0.

    Can we get the same answer without using the series expansion thingy?
     
  10. Oct 8, 2011 #9

    arildno

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    Or, you could make the substitution x=sin(u), which is valid in the neighbourhood of x=0.
     
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