Limit using L'Hopital's Rule

  • Thread starter phil ess
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  • #1
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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?
 

Answers and Replies

  • #2
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Try (cos x)/[(-1) / (xln x)2]
 
  • #3
Dick
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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
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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
Dick
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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?
 

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