1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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]

    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


    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook