Limit of (1/n)^(1/ln(n)) as n-> infinity

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In summary, in order to find the limit of (1/n)^(1/ln(n)) as n->infinity, we can use logarithm rules to simplify the expression to -ln(n)/ln(n), which equals -1 for all finite values of n. This allows us to apply the limit operation, and using the rule that ln(1/n) = -ln(n), we can find the limit to be -1.
  • #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.
 
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  • #2
Eats Dirt said:

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
thanks!
 
  • #4
Mark44 said:
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
Eats Dirt said:
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.
Eats Dirt said:
I wonder how many other rules like this there are that I don't even know about haha.

Thanks so much.
 

FAQ: Limit of (1/n)^(1/ln(n)) as n-> infinity

What is a limit in mathematics?

A limit in mathematics is the value that a function or sequence approaches as the input or index approaches a certain point or infinity. It helps us understand the behavior of a function or sequence near a specific point.

How do I find a limit?

To find a limit, you can use various methods such as substitution, factoring, or using limit laws. You can also use a graph or a table to estimate the limit. It is important to remember that not all limits can be found algebraically and some may require more advanced techniques.

What does it mean when a limit does not exist?

If a limit does not exist, it means that the function or sequence does not approach a single value as the input or index approaches a certain point. This could happen if there are abrupt changes or oscillations in the function or sequence, or if the function or sequence approaches different values from different directions.

Can a limit be infinite?

Yes, a limit can be infinite. This can happen if the function or sequence approaches positive or negative infinity as the input or index approaches a certain point. An example of this is the limit of 1/x as x approaches 0, which is equal to positive infinity.

Why is finding a limit important?

Finding a limit is important because it helps us understand the behavior of a function or sequence near a specific point. It is also used in many mathematical applications, such as calculating derivatives and integrals, and in determining the convergence or divergence of series.

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