# Any other way? (1 Viewer)

### Users Who Are Viewing This Thread (Users: 0, Guests: 1)

#### dextercioby

Homework Helper
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.

#### arildno

Homework Helper
Gold Member
Dearly Missed
Hmm.. shouldn't it work to evaluate the limit of the expression's logarithm first?
Just an idea..

#### dextercioby

Homework Helper
Ouch, that hurt :((

Daniel.

#### TD

Homework Helper
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:

### The Physics Forums Way

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving