Calculate the limit of the series

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SUMMARY

The limit of the series (cos(a/n))^(n^3) as n approaches infinity, where a ≠ 0, can be calculated using the expression L = (n^3) * ln(cos(a/n)). This limit initially presents a 0*infinity form, necessitating the application of l'Hôpital's theorem. However, since n is a discrete variable (n = 1, 2, 3...), the function must be transformed to f(x) = ln(cos(a/x)) * x^3 for differentiability. The limit of f(x) as x approaches infinity will provide the necessary insight into the limit of L as n approaches infinity.

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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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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