What is the limit of \(\cos^{x} \frac{\pi}{x}\) as \(x\) approaches infinity?

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Homework Help Overview

The discussion revolves around evaluating the limit of \(\cos^{x} \frac{\pi}{x}\) as \(x\) approaches infinity, focusing on the behavior of the cosine function as its argument approaches zero.

Discussion Character

  • Exploratory, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the idea of using the logarithm of the expression to evaluate the limit. There is mention of applying L'Hôpital's rule in the context of the limit evaluation.

Discussion Status

Some participants are considering different approaches to tackle the limit, including the logarithmic transformation and L'Hôpital's rule. There is an acknowledgment of previous responses, indicating an ongoing dialogue without a clear consensus.

Contextual Notes

Participants are discussing the behavior of \(\cos \frac{\pi}{x}\) as \(x\) approaches infinity, which involves assumptions about the approximation of cosine near zero.

dextercioby
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of doing this

[tex]\lim_{x\rightarrow +\infty }\cos^{x}\frac{\pi}{x}[/tex]

than using

[tex]\cos x \simeq 1- \frac{x^{2}}{2} \mbox {when "x" goes to zero}[/tex]

?

Daniel.
 
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Hmm.. shouldn't it work to evaluate the limit of the expression's logarithm first?
Just an idea..
 
That would work, together with applying L'Hopital once.

[tex]\mathop {\lim }\limits_{x \to \infty } \left( {\cos \frac{\pi }{x}} \right)^x = \mathop {\lim }\limits_{x \to \infty } \exp \left( {x\ln \left( {\cos \frac{\pi }{x}} \right)} \right) = \exp \mathop {\lim }\limits_{x \to \infty } \left( {\frac{1}{{1/x}}\ln \left( {\cos \frac{\pi }{x}} \right)} \right) = \exp \mathop {\lim }\limits_{x \to \infty } \left( { - \pi \tan \frac{\pi }{x}} \right) = e^0 = 1[/tex]

Edit: didn't see you already replied :smile:
 
Last edited:

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