Any other way? (1 Viewer)

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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.
 

arildno

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Hmm.. shouldn't it work to evaluate the limit of the expression's logarithm first?
Just an idea..
 

dextercioby

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Ouch, that hurt :((

Daniel.
 

TD

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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:
 
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