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

## 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

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 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.

Mark44
Mentor
$ln(x) \rightarrow -\infty$
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.