Show that n^(logc)/c^(logn) =1 as n->inf

  • Thread starter bfpri
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In summary, the conversation discusses the use of logarithms to prove that n^(logc)/c^(logn) approaches 1 as n approaches infinity, where c is a constant greater than 1. The initial attempt at using L'Hopital's rule is hindered by the complexity of the logs. However, by considering the logarithm of the function, it can be simplified to equal 0. Ultimately, the conclusion is reached and the conversation ends with a resolution.
  • #1
bfpri
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Homework Statement


Show that n^(logc)/c^(logn) =1 as n->inf where c is a constant greater than 1

Homework Equations


The Attempt at a Solution



Tried L'hospitals. But the logs mess it up. Even if you assume that logc>1 then the top does eventually become a constant (second derivative). However the bottom gets too messy. Is there another method to start it?
 
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  • #2
If f(n) = n^log(c) / c^log(n), look carefully at log (f(n)).

RGV
 
  • #3
Thats log(n^logc/c^logn)= log(n^logc)-log(c^logn)=log(c)log(n)-log(n)log(c)=0.

Ok i got it

Thanks
 

1. What does the equation n^(logc)/c^(logn) represent?

The equation represents the limit of a function as n approaches infinity.

2. How can you prove that n^(logc)/c^(logn) approaches 1 as n increases?

This can be proven using the definition of a limit and properties of logarithms.

3. Can this equation be applied to any values of n and c?

Yes, as long as n and c are positive real numbers and c is not equal to 1.

4. What is the significance of this limit as n approaches infinity?

This limit shows the relationship between exponential and logarithmic functions as n increases.

5. How is this equation helpful in the field of mathematics or science?

This equation is helpful in solving problems involving exponential and logarithmic functions, and can be applied in various fields such as engineering, physics, and computer science.

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