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Finding the limit of an indeterminate using series, not lhopitals

  1. Sep 20, 2009 #1
    1. The problem statement, all variables and given/known data
    My calc teacher says that we should do the problem "by series", even though it is in the section of the book where the teach L'Hopital's rule.

    For example, one of the questions is:

    [tex]

    \lim_{x\rightarrow 0} \frac{sin(x)}{\sqrt{x}}

    [/tex]


    3. The attempt at a solution

    Using L'Hopital's rule, you eventually get 0.

    I managed to make a series for it...
    [tex] \sum_{n=0}^\infty \frac{\frac{(-1)^n\*x^{2n+1}}{(2n+1)!}}{x^\frac{1}{2}} [/tex]

    but after that, i dont know what to do
     
  2. jcsd
  3. Sep 20, 2009 #2

    Office_Shredder

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    You can turn that into

    [tex] \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+.5}}{(2n+1)!}[/tex]

    by canceling the x in the denominator
     
  4. Sep 20, 2009 #3
    yeah, i just simplified it on my homework, but after that, would you simply just plug 0 into x? it may give 0 for this problem, but for the other problems (after find the power series for them) it wouldnt give the right answer... so what would i do?
     
  5. Sep 20, 2009 #4

    HallsofIvy

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    Another way to do this that does not use "L'Hopital" nor "series" is to write
    [tex]\frac{sin(x)}{\sqrt{x}}= \sqrt{x}\frac{sin(x)}{x}[/tex]
    and use the fact that
    [tex]\lim_{x\to 0}\frac{sin(x)}{x}= 1[/itex]
     
  6. Sep 20, 2009 #5
    it would work that way too, but i just dont know what to do, as the professor specifically asked for it to be done "by series". and i am at a loss as to knowing how it can be done that way.
     
  7. Sep 20, 2009 #6
    still need some help lol
     
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