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Limit problem solution

  1. Mar 24, 2009 #1
    1. The problem statement, all variables and given/known data

    lim as x goes to 0 of

    [x*csc(2x)] / cos(5x)


    3. The attempt at a solution

    The book solution says the answer is 1/2. I keep getting zero as an answer because of the x in the numerator and am unsure of how else to go about the problem. I'm pretty sure I'm supposed to use the limit as x goes to 0 of sinx / x, but even when I get to that point the limit of x as x goes to 0 always gives me an answer of zero. What am I doing wrong? Thanks!
     
  2. jcsd
  3. Mar 24, 2009 #2

    Mark44

    Staff: Mentor

    The part to be concerned with is x*csc(2x), since the limit of cos(5x) is 1, as x goes to 0. If you can establish a limit for x*csc(2x), then lim x*csc(2x)/cos(5x) will be = lim x*csc(2x).

    x*csc(2x) = x/sin(2x), so all you need now is 2x in the numerator instead of the x that is there. Can you turn x into 2x by multiplying by 1 in some form?
     
  4. Mar 24, 2009 #3

    lanedance

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    Homework Helper

    so you have

    [tex] \frac{x.csc(2x)}{cos(5x)} = \frac{x.}{sin(2x).cos(5x)}[/tex]

    so the denmoinator goes to zero as well, so you can't say it is zero...
     
  5. Mar 24, 2009 #4
    So the lim as x goes to 0 of x/sin(x) is also equal to 1? I assume that is the case because multiplying by 2/2 gives the solution of 1/2. Thanks a lot, I really appreciate it.

    Just out of curiosity now, is the lim as x goes to 0 of x/(1-cos(x)) = infinity?
     
  6. Mar 24, 2009 #5

    lanedance

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    lim as x goes to 0 sin(x)/x is one, as is sin(x)/x

    think about this, if
    [tex]\stackrel{lim}{x \rightarrow 0} f(x) = c [/tex]

    then what is
    [tex]\stackrel{lim}{x \rightarrow 0} \frac{1}{f(x)} = ? [/tex]

    the second limit is true
    [tex]\stackrel{lim}{x \rightarrow 0} \frac{x}{(1-cos(x))} = \infty [/tex]
    can you show why?

    (l'hopitals rule is good for all of these if you know it...)
     
  7. Mar 25, 2009 #6

    Mark44

    Staff: Mentor

    No, the limit doesn't exist.
    [tex]\lim_{x \rightarrow 0^+} x/(1 -cos(x)) = \infty[/tex]
    while

    [tex]\lim_{x \rightarrow 0^-} x/(1 -cos(x)) = -\infty[/tex]
     
  8. Mar 25, 2009 #7

    lanedance

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    Homework Helper

    good pickup - my mistake
     
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