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Limit problem

  1. Jan 8, 2006 #1
    How do you find the limit to:

    Lim
    x->0+ sqrtx/(1-cosx)

    I use the L'Hopital rule and i get

    lim
    x->0+ 1/(2sqrtx*sinx)

    and I can't get anywhere after this. I even tried multiplying by sqrtx/sqrtx to get rid of the sqrtx at the beginning, but it still doesn't work. Please help, thanks.
     
  2. jcsd
  3. Jan 8, 2006 #2

    shmoe

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    What's sqrt(x) going to as x->0+? sin(x)?
     
  4. Jan 9, 2006 #3

    dextercioby

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    Use the asymptotic expansion of cosine

    [tex] \cos x \simeq 1-\frac{x^{2}}{2} [/tex] when "x" is close to zero.

    Daniel.
     
  5. Jan 9, 2006 #4
    it's still in the form infinity * infinity isn't it? got to use l'hopital's rule one more time. i think the answer is 0.
     
  6. Jan 9, 2006 #5

    Hurkyl

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    Check your list of indeterminate forms.
     
  7. Jan 14, 2006 #6
    it is simple I guess...use the
    "when top is going to 0 and bottom is going to soem where else."......I forgot the rule...But the answer is +0
     
  8. Jan 14, 2006 #7

    arildno

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    Just rewrite your expression as
    [tex]\frac{\sqrt{x}}{1-\cos(x)}=\frac{1}{x^{\frac{3}{2}}}(\frac{x}{\sin(x)})^{2}(1+\cos(x))[/tex]
    Then it is easy to see what you should get.
    Alternatively, use dexter's approach..
     
  9. Jan 14, 2006 #8
    No you dont have to make it hard..beleive me it so simple....we did in calc 1 I forgot the rule name.
     
  10. Jan 14, 2006 #9

    arildno

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    So, why is your "answer" dead wrong, then? :wink:
     
  11. Jan 14, 2006 #10
    opps didnt noticed bottom was going to 0 too...haha
    then you will have to use l'hopital rule.
     
  12. Jan 14, 2006 #11

    mathwonk

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    well i just graphed them and looked at the graphs. sqrt(x) is really big compared to x, while 1-cos(x) is really small, so I think the limit is infinity. but now i will do the math, and no doubt come back chagrined.
     
  13. Jan 14, 2006 #12

    mathwonk

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    no i seem to be right, i.e. one use of l'hopital gives it, yes?
     
  14. Jan 14, 2006 #13

    mathwonk

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    or just write it as (sqrt(x)/x). (x/[1-cos(x))]) and both factors go to + infinity, the first trivially and the second by l'hopital.
     
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