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Limit question

  1. Aug 28, 2009 #1
    1. The problem statement, all variables and given/known data

    lim as x tends to zero of (x^(-2) - (cosecx)^2)

    2. Relevant equations

    L'Hospital's rule


    3. The attempt at a solution

    I keep getting things in the denominator that won't go away no matter how much I differentiate. I'm missing something here...
     
  2. jcsd
  3. Aug 29, 2009 #2

    Mark44

    Staff: Mentor

    Show us what you did.
     
  4. Aug 29, 2009 #3

    VietDao29

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    Homework Helper

    You mean:

    [tex]\lim_{x \rightarrow 0} \left( \frac{1}{x ^ 2} - \frac{1}{\sin ^ 2 x} \right)[/tex], right?

    If you really want to do it, by applying L'Hospital rule, then the first step is to group them into 1 fraction, like this:

    [tex]\lim_{x \rightarrow 0} \left( \frac{1}{x ^ 2} - \frac{1}{\sin ^ 2 x} \right) = \lim_{x \rightarrow 0} \frac{\sin ^ 2 (x) - x ^ 2}{x ^ 2 \sin ^ 2 x}[/tex]

    Now, just apply L'Hospital Rule, if you want us to check your work, you can always post it here, and we'll be more than willing to help you out. :)

    Btw, this is actually the hard way, you have to apply the rule four-folds to get to the final result.

    Have you covered Talor's expansion yet? This could make the problem way simpler.
     
  5. Aug 29, 2009 #4
    yeah we've learnt taylor polynomials. i guess i could make a taylor polynomial of appropriate degree for each one, though the question says to use l'hospitals.

    i went

    [tex]
    \lim_{x \rightarrow 0} \left( \frac{1}{x ^ 2} - \frac{1}{\sin ^ 2 x} \right)
    = \lim_{x \rightarrow 0} \left( \frac{\sin ^ 2 (x) - x ^ 2}{x ^ 2 \sin ^ 2 x} \right)
    = \lim{x \rightarrow 0} \left( \frac{sin x cos x - x}{x ^ 2 sin x cos x + x sin ^2 x} \right)
    = \lim{x \rightarrow 0} \left( \frac{-sin^2 x + cos ^2 x -1}{x ^ 2 cos (2x) + 3 x sin (2x) + sin ^2 x} \right)
    [/tex]

    then i gave up
     
    Last edited: Aug 29, 2009
  6. Aug 29, 2009 #5
    The denominator of the last expression on the right doesn't look correct. You should have a lot more terms.
     
  7. Aug 29, 2009 #6

    VietDao29

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    Homework Helper

    In the first post I told you to use it 4 times. You only used it twice. Since it's still in the form 0/0, just continue until you cannot go any further.
     
  8. Aug 29, 2009 #7
    yeah cos i gave up before i finished the denominator

    fix'd
     
  9. Aug 29, 2009 #8
    Why? If you don't want to use the product rule extended to products of more than two factors, then use the fact that sin(2x) = 2(sin x)(cos x) or sin(x)cos(x) = sin(2x) / 2 and just keep adding like terms together after differentiating.
     
  10. Aug 29, 2009 #9
    so i do it four times and get
    [tex]
    \lim{x \rithgtarrow 0} \left( \frac{-4 cos (2x)}{-20 sin (2x) + 16 cos (2x) - 4 x ^ 2 cos (2x)} \right)
    [/tex]
     
  11. Aug 29, 2009 #10

    VietDao29

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    Homework Helper

    You can continue using it for 2 more times, or by looking at it a little bit closely, your final expression is:

    [tex]\lim_{x \rightarrow 0} \left( \frac{-\sin^2 x + \cos ^2 x -1}{x ^ 2 \cos (2x) + {\color{red}2} x \sin (2x) + \sin ^2 x} \right) = \lim_{x \rightarrow 0} \left( \frac{-2\sin^2 x}{x ^ 2 \cos (2x) + {\color{red}2} x \sin (2x) + \sin ^2 x} \right)[/tex]

    Now, what will you get if you divide both numerator, and denominator by x2?

    --------------

    Oh, and btw, you have miscalculated it, the factor 3 in your step should read 2 instead. I have corrected it for you. :)
     
    Last edited: Aug 29, 2009
  12. Aug 29, 2009 #11

    VietDao29

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    Homework Helper

    Ohhhh.. :(

    Since you have differentiated it incorrectly in the denominator, in your second step. I'm sorry to inform you that this is wrong. :( You have to re-calculate everything from your second use of L'Hospital's rule. :(

    Or, you can try my hint in the post above. :)
     
  13. Aug 29, 2009 #12
    I don't think this is correct but as others have noted, it's hard to help when you don't show the intermediate steps.
     
  14. Aug 29, 2009 #13
    cheers =] i got the result

    i'm looking at trying to do the same problem with taylor polynomials... i get the taylor polynomial for [tex] sin ^2 (x) [/tex] to be[tex] \sum{\frac{-(-1)^k 2^{2k-1}x^{2k}}{(2k)!}} ... [/tex]
    inverting this to get the one for [tex] cosec ^2 (x) [/tex] looks a bit mean.
     
    Last edited: Aug 29, 2009
  15. Aug 29, 2009 #14

    VietDao29

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    Ummm.. How did you get that? It doesn't look quite correct to me. Could you should us the step, so that i can help you point out where it went wrong?

    No, you don't have to find the expansion for csc2(x). Just truncate it (the expansion for sin2(x)) wisely, then use that truncated expression to plug in the limit. Much simpler than to find its reciprocal, right? :)

    [tex]\lim_{x \rightarrow 0} \left( \frac{1}{x ^ 2} - \frac{1}{\sin ^ 2 x} \right) = \lim_{x \rightarrow 0} \frac{\sin ^ 2 (x) - x ^ 2}{x ^ 2 \sin ^ 2 x}[/tex]
     
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