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Homework Help: Finding the limit of x tending to 0 of ln(x)sin(x)

  1. Feb 23, 2010 #1
    1. The problem statement, all variables and given/known data

    find the limit of:

    [itex]\frac{lim}{x\rightarrow0} ln(x)sin(x)[/itex]

    tip:
    you know that
    [itex] \frac{sin(x)}{x} \rightarrow 1[/itex]
    so rearrange to use


    2. Relevant equations

    L'hopitals rule?

    3. The attempt at a solution

    i rearranged to get this

    [itex]\frac{lim}{x\rightarrow0} \frac{\frac{ln(x)sin(x)}{x}}{\frac{1}{x}}[/itex]

    i know that [itex]\frac{sin(x)}{x} \rightarrow 1[/itex]

    and [itex]\frac{1}{x} \rightarrow \infty[/itex]

    does [itex]ln(x) \rightarrow -\infty[/itex] ?

    I might need to use L'hopitals rule after this but i dont think it will make anything any easier.
    Am i missing something major here, does the fact that sin(0)=0 mean that the limit is 0? because if it is, the question is fairly misleading.
     
  2. jcsd
  3. Feb 23, 2010 #2

    Mark44

    Staff: Mentor

    [itex]ln(x) \rightarrow -\infty[/itex]
    as x --> 0+

    You'll probably need to look at this as a right-side limit, since the ln function is not defined for x <=0.

    Using the hint, write the limit expression as
    [tex]x~ln(x) \frac{sin(x)}{x}[/tex]

    The limit of a product is the product of the limits, provided that both limits in the product exist. If you rewrite x*ln(x) appropriately, you can use L'Hopital's Rule on it.
     
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