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

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The limit of \(\cos^{x} \frac{\pi}{x}\) as \(x\) approaches infinity can be evaluated by first taking the logarithm of the expression. This approach leads to the limit of the logarithm being expressed as \(\exp \left( -\pi \tan \frac{\pi}{x} \right)\). As \(x\) approaches infinity, \(\tan \frac{\pi}{x}\) approaches zero, resulting in the overall limit equating to \(e^0\), which is 1. The discussion highlights the effectiveness of using L'Hopital's rule in this evaluation. Ultimately, the limit is confirmed to be 1.
dextercioby
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of doing this

\lim_{x\rightarrow +\infty }\cos^{x}\frac{\pi}{x}

than using

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

?

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.

\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

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