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Limit of a sine function problem.

  1. Oct 10, 2006 #1
    Part a)I need help understanding a step in solving [tex]\lim_{x\rightarrow \infty}xsin(\frac{1}{x})[/tex]

    The textbook is suggesting I replace [tex]\frac{1}{x}[/tex] with y, so that I can get a limit in the form

    [tex]\lim_{y\rightarrow 0}\frac{siny}{y}[/tex] which is understandably easy to solve. The part I don't understand is how the limit changes from x approaches inf to y approaches 0. I was running into the same confusion with this problem as well so perhaps I should post it.

    Part b)Solve [tex]\lim_{x\rightarrow 0}\frac{sin4x}{sin3x}[/tex]

    They then proceed in their next step to write it as [tex]\frac{\lim_{4x\rightarrow 0}4x(\frac{sin4x}{4x})}{\lim_{3x\rightarrow 0}3x(\frac{sin3x}{3x})}[/tex]

    How do they get the 4x approaches 0 and 3x approaches 0, I understand it as a "this is the way to do it" and can do it easily, but I'd like to know what they are doing. Anyways, thanks :smile:
     
    Last edited: Oct 10, 2006
  2. jcsd
  3. Oct 10, 2006 #2

    Office_Shredder

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    Forthe first question: If x is approaching infinity, what is 1/x approaching? Therefore, what is y approaching?

    For the second question, I can't see your latex image, but I think I can help you with your question. If x approaches zero, what would you think 4x approaches? Specifically, try taking the limit of 4x as x goes to zero, and remember that if there's no problem, as a general rule of thumb you can just plug in what x approaches and do the limit real easy
     
  4. Oct 10, 2006 #3

    StatusX

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    If f(x)->b as x->a, then the limit of g(f(x)) as x goes to a is equal to the limit of g(y) as y goes to b. Proving this is a nice exercise in the epsilon delta definition of a limit.
     
  5. Oct 10, 2006 #4
    Thankyou both, :D
     
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