# Limit using L'Hopital's Rule

1. Nov 8, 2008

### phil ess

1. The problem statement, all variables and given/known data

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

2. Relevant equations

None

3. 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?

2. Nov 8, 2008

### vcarcu

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

3. Nov 8, 2008

### Dick

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.

4. Nov 9, 2008

### phil ess

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.

5. Nov 9, 2008

### Dick

Use more parentheses! I can't tell what sin(x)/-xcos(x) is supposed to mean. The closest it could be to being right is -(sin(x)/x)*cos(x). But that's not even right, the power of the sin is wrong. Can you fix it?