Calculate the limit of the series

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Homework Help Overview

The discussion revolves around calculating the limit of the series (cos(a/n))^(n^3) as n approaches infinity, where a is a non-zero constant. Participants are exploring the behavior of the limit and the implications of using logarithmic transformations.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to find the limit by expressing it in terms of L = (n^3)*ln(cos(a/n)). Some participants suggest applying l'Hôpital's theorem to resolve the indeterminate form, while others question the applicability of this method due to the nature of n as a discrete variable.

Discussion Status

The discussion is ongoing, with participants providing different perspectives on the use of l'Hôpital's theorem and exploring alternative approaches to evaluate the limit. There is no explicit consensus on the best method to proceed, but some guidance has been offered regarding the differentiability of related functions.

Contextual Notes

Participants note the challenge of dealing with the discrete nature of n and the implications this has on the application of calculus techniques. The original poster also emphasizes the condition that a is not equal to zero.

sedaw
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need to calculate the limit of the series

(cos(a/n))^(n^3) n goes to infinity and a=!0


i try that :

e^L


L= the limit of (n^3)*ln(cos(a/n))


what now ?


TNX !
 
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L has the form 0*infinity. Use l'Hopital's theorem on it to try and figure out the limit. You may have to use l'Hopital more than once.
 


Dick said:
L has the form 0*infinity. Use l'Hopital's theorem on it to try and figure out the limit. You may have to use l'Hopital more than once.

hello Dick !

first of all tnx.

i can't use l'Hopital's theorem cause n =1,2,3...

so F(n) not differentiable .
 


f(x)=ln(cos(a/x))*x^3 is differentiable. If you know the limit of f(x) as x->infinity, then you know the limit of L as n->infinity.
 

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