What is the limit as x approaches 0 from the positive side of (sin x)(ln x)?

[phil ess]

Homework Statement

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

Homework Equations

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?

 

[vcarcu]

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

 

[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.

 

[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.

 

[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?