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Homework Help: Limit Evaluation

  1. May 7, 2012 #1
    1. The problem statement, all variables and given/known data

    I had an exam last night and I was stuck on this problem.
    I went home where I had more time (and resources) and still can't figure it out, and I won't get my exam back (it was a final).


    2. Relevant equations

    3. The attempt at a solution
    I have tried combining into one fraction, splitting up in to two separate limits, I can't use L'Hopital's as it is 1/inf.
    What am I missing? The rest of the exam was somewhat easy, but I didn't even know where to start!
    Can someone help me work though it? I think I just need help with a first step.
  2. jcsd
  3. May 7, 2012 #2
    If you combine them into one fraction, then it's 0/0 right? Only problem is that you have to use L'Hopital 3 times to get the result.
  4. May 7, 2012 #3
    Hello crybllrd write your expression in the following form:


    I think it should be clear from here on.
  5. May 7, 2012 #4


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    Homework Helper

    You can combine ito one fraction and then apply L'Hopitals repeatedly.

    Or you can do a series expansion.
  6. May 7, 2012 #5
    How did you get the sin(x)2 in the denominator? If I multiply x2by xsin(x) I get x3sin(x).
  7. May 7, 2012 #6
    I used l'hopital's rule 6 times, not 3 times.
  8. May 7, 2012 #7


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    Why would you do that? The "least common denominator" is [itex]x^2sin(x)[/itex]. Of course, you also do not get "[itex]x^2sin^2(x)[/tex]"!

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

    Now, you can use L'Hopital's rule on that, just once. Differentiating both numerator and denomimator separately you get
    [tex]\frac{cos(x)- 2x}{2x sin(x)+ x^2cos(x)}[/tex].

    You do NOT get "1/infinity".
  9. May 7, 2012 #8
    Ah, ok, I got -1/6.
    I don't remember why exactly I didn't get a good answer on the exam, but I think I was on like my 6th L'Hopital iteration and wasn't willing to spend any more time on a 4 point problem.
    Thanks guys, now I can rest easy this summer, as my brain is completely fried now!
  10. May 7, 2012 #9
    Hi. As already other colleagues have stated,


    [itex]=\lim_{x\to0}\frac{1}{x^{2}}[ 1 - \frac{x}{sinx}][/itex]

    [itex]=\lim_{x\to0}\frac{1}{x^{2}}[ 1 - \frac{1}{1 - \frac{x^{2}}{3!} + ...} ][/itex]

    [itex]=\lim_{x\to0}\frac{1}{x^{2}}[ 1 - 1 - \frac{x^{2}}{3!} + ...][/itex]
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