Finding a limit help.

  • Thread starter Eats Dirt
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  • #1
Eats Dirt
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Homework Statement



Finding the limit of (1/n)^(1/ln(n))

as n-> infinity

Homework Equations



Log rules

The Attempt at a Solution



so I take the ln of both sides and get

(1/ln(n))*ln(1/n) in an attempt to get it into the proper condition for l'hopitals rule.

1/ln(infinity) is zero but ln(1/n) is undefined and I have the same problem when trying to multiply by ((1/n)/(1/n)) because it is still an indeterminate form and I can not apply l'hopitals rule.
 

Answers and Replies

  • #2
36,711
8,711

Homework Statement



Finding the limit of (1/n)^(1/ln(n))

as n-> infinity

Homework Equations



Log rules

The Attempt at a Solution



so I take the ln of both sides and get

(1/ln(n))*ln(1/n) in an attempt to get it into the proper condition for l'hopitals rule.

1/ln(infinity) is zero but ln(1/n) is undefined and I have the same problem when trying to multiply by ((1/n)/(1/n)) because it is still an indeterminate form and I can not apply l'hopitals rule.

Let y = (1/n)(1/ln(n))
Then ln y = 1/(ln(n)) * ln (1/n) = 1/(ln(n)) * (-ln(n)) = (-ln(n))/ln(n)

Now take the limit, noting that for all finite n, the right side above equals -1. Note also that you can switch the order of the lim operation and the ln operation under certain conditions.

Does that get you started?
 
  • #3
Eats Dirt
92
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thanks!
 
  • #4
Eats Dirt
92
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Let y = (1/n)(1/ln(n))
Then ln y = 1/(ln(n)) * ln (1/n) = 1/(ln(n)) * (-ln(n)) = (-ln(n))/ln(n)

Now take the limit, noting that for all finite n, the right side above equals -1. Note also that you can switch the order of the lim operation and the ln operation under certain conditions.

Does that get you started?

Hey thanks you this does help a lot but I didn't even know about that rule for ln!
so ln(1/n) = -ln(n) ?!

I wonder how many other rules like this there are that I don't even know about haha.

Thanks so much.
 
  • #5
36,711
8,711
Hey thanks you this does help a lot but I didn't even know about that rule for ln!
so ln(1/n) = -ln(n) ?!
Yes. It's a special case of ln(A/B) = ln(A) - ln(B), with A = 1.
I wonder how many other rules like this there are that I don't even know about haha.

Thanks so much.
 

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