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

  • Thread starter knowlewj01
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Homework Statement



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


Homework Equations



L'hopitals rule?

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.
 

Answers and Replies

  • #2
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[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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