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Limit of [( x^1/2 ) - x^2] / (1- x^1/2)

  1. Dec 27, 2005 #1
    limit of [( x^1/2 ) - x^2] / (1- x^1/2)

    Can anyone help me to solve this question?

    I have tried all the method but I still cant get the answer.
     
  2. jcsd
  3. Dec 27, 2005 #2
    is x going to infinity or what?
     
  4. Dec 27, 2005 #3

    HallsofIvy

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    I think the only difficulty would be if x is going to 1. Standard trick: multiply both numerator and denominator by 1- x1/2, then factor x1/2- x out of the numerator.
     
  5. Dec 27, 2005 #4
    I get -1 as the answer if limit is x->1
    I think you can use L'hopitals rule.

    blumfeld0
     
  6. Dec 27, 2005 #5
    Yes..x is approaching to 1
     
  7. Dec 27, 2005 #6
    I get the answer of 3 by using hallsofivy. So, which one correct actually?
     
  8. Dec 27, 2005 #7

    VietDao29

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    Err, I don't really get this... You are actually multiplying both numerator and denominator by a factor (i.e 1 - x1 / 2), which also tends to 0 as x tends to 1.
    I thought we should cancel out as many factors that tend to 0 as possible?
    Am I missing something?
    [tex]\lim_{x \rightarrow 1} \frac{\sqrt{x} - x ^ 2}{1 - \sqrt{x}} = \lim_{x \rightarrow 1} \sqrt{x} \ \frac{1 - \sqrt{x ^ 3}}{1 - \sqrt{x}} = \lim_{x \rightarrow 1} \sqrt{x} \ \frac{1 - (\sqrt{x}) ^ 3}{1 - \sqrt{x}}[/tex]
    Yup. That looks good. You can always test your result by using your calculator. Pick up an x that is close to 1 from the left (say 0.99), and from the right (say 1.01). Plug that in the expression:
    [tex]\frac{\sqrt{x} - x ^ 2}{1 - \sqrt{x}} = \frac{\sqrt{0.99} - 0.99 ^ 2}{1 - \sqrt{0.99}} = 2.97002 \approx 3[/tex]
    [tex]\frac{\sqrt{x} - x ^ 2}{1 - \sqrt{x}} = \frac{\sqrt{1.01} - 1.01 ^ 2}{1 - \sqrt{1.01}} = 3.03002 \approx 3[/tex].
    Are 2.97002, and 3.03002 close to 3 or close to -1?
    So you are correct! Congratulations. :smile:
     
    Last edited: Dec 27, 2005
  9. Dec 27, 2005 #8

    Hurkyl

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    Sometimes, you have to make an expression more complicated before you can simplify. :smile: This is a common trick, though: you multiply something by 1 in a clever way that simplifies the terms involved.

    I think HOI meant to use [itex]1 + \sqrt{x}[/itex], though. (to rationalize one of the terms)


    You could also do this with a change of variable, though they don't seem to teach how to do those in elementary calc classes. :frown: But you can still do it the "hard" way by writing the whole thing as a function of [itex]\sqrt{x}[/itex].
     
  10. Dec 27, 2005 #9

    BobG

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    L'Hopital's rule is the easiest way to solve this, but you get 3 with L'hopital's rule, as well. (1/2-2)/(-1/2)
     
  11. Dec 27, 2005 #10
    Opps my bad. you're all right it is 3. i dont know how i got -1.
    sorry for any confusion

    blumfeld0
     
  12. Dec 27, 2005 #11
    more basic way is

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

    Isn't it clear? no law or rules...

    if you remember [tex]1-y^n = (1+y+y^2+ \cdots +y^{n-1}) (1-y)[/tex]
    or simply
    [tex]1-y^3 = (1+y+y^2)(1-y)[/tex]

    "back to basic" is often useful :smile:
     
    Last edited: Dec 28, 2005
  13. Dec 27, 2005 #12
    Thanks...calculus is exciting..!!
    By the way, is [tex]\frac{\sqrt{x} - x ^ 2}{1 - \sqrt{x}} = \frac{\sqrt{1.01} - 1.01 ^ 2}{1 - \sqrt{1.01}} = 3.03002 \approx 3[/tex]
    a general answer checking method?
     
  14. Dec 28, 2005 #13

    VietDao29

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    Ooops... This approach looks familiar. Have you had a look at post #7??
    Hmm... :tongue2:
    And so, my approach is classified as a hard way? :cry: :cry: :cry:
    In case you just have your calculator with you, then yes, it is. I often use it to check whether my answer is correct or not.
    It's based on the idea that when you choose an x near 1, the expression will evaluate to some value very close to 3 (since x approaches 1, the expression will approach 3).
    But, watch out for rounding error. For example:
    [tex]\lim_{x \rightarrow 0} \frac{1}{\sin ^ 2 x} - \frac{1}{x ^ 2} = \frac{1}{3}[/tex]
    If you choose x = 0.01, plug that in your calculator (in radian mode), you'll have:
    [tex]\frac{1}{\sin ^ 2 0.01} - \frac{1}{0.01 ^ 2} = 0.33334 \approx \frac{1}{3}[/tex]
    But if you choose x = 0.00001:
    [tex]\frac{1}{\sin ^ 2 0.00001} - \frac{1}{0.00001 ^ 2} \approx 0[/tex]
    So choose the x wisely so that it's close enough, but not too close to 1 (or 0, or whatever value x tends to.), so the calculator can evaluate the expression with the smallest rounding error.
     
  15. Dec 28, 2005 #14
    It is obvious the numerator overwhelm the denominator.
     
  16. Dec 28, 2005 #15
    Thanks a lot buddy...:)
     
  17. Dec 28, 2005 #16

    HallsofIvy

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    What exactly do you mean by that?
     
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