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

[knowlewj01]

Homework Statement

find the limit of:

\frac{lim}{x\rightarrow0} ln(x)sin(x)

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

Homework Equations

L'hopitals rule?

The Attempt at a Solution

i rearranged to get this

\frac{lim}{x\rightarrow0} \frac{\frac{ln(x)sin(x)}{x}}{\frac{1}{x}}

i know that \frac{sin(x)}{x} \rightarrow 1

and \frac{1}{x} \rightarrow \infty

does ln(x) \rightarrow -\infty ?

I might need to use L'hopitals rule after this but i don't 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.

 

[Mark44]

ln(x) \rightarrow -\infty
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
x~ln(x) \frac{sin(x)}{x}

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.