# Limit using L'Hopital's Rule

## Homework Statement

Find the limit as x -> 0+ of (sin x)(ln x)

None

## The Attempt at a Solution

I rewrote this as (sin x) / (1/ln x), then using L'Hopital it becomes:

(cos x) / [(-ln x) / (ln x)2] = [(ln x)2 cos x] / (-ln x)

So I get limit as x -> 0+ of [(ln x)2 cos x] / (-ln x)

Which isn't any better. I'm not sure if I can even use L'Hopital here because it requires that substitution gives the indeterminate state, but I don't know if 0+/0+ really counts. Any thoughts?

Try (cos x)/[(-1) / (xln x)2]

Dick
Homework Helper
It's probably easier to try l'Hopital on ln(x)/(1/sin(x)). What does that give you? Remember lim sin(x)/x=1.

Doing it on ln(x)/(1/sin(x)) gives me sin(x)/-xcos(x). Now what do I do though? I can't solve using direct substitution because 0+ isn't really a number. Do I just substitute in something close like 0.0001?

I guess I could rewrite it -tan(x)/x but that doesn't help me much.

Dick