# Any other way?

1. Dec 19, 2005

### dextercioby

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.

2. Dec 19, 2005

### arildno

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

3. Dec 19, 2005

### dextercioby

Ouch, that hurt :((

Daniel.

4. Dec 19, 2005

### TD

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

Last edited: Dec 19, 2005
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