Re: question

On 5/16/2024 7:41 PM, Peter Fairbrother wrote:

Is lim (cos pi/2n)^n = 1 as n -> infinity?

Yes, 1.
Take the logarithm of both sides,
Swap log and lim, because continuity.
Evaluate lim by L'Hôpital's rule .
Take the exponential of both sides.
lim(cos(1/n)^n
[n→inf])
exp(log(lim(cos(1/n)^n)))
[n→inf])))
exp(lim(log(cos(1/n)^n)))
[n→inf]))
continuity
exp(lim(
n*log(cos(1/n))
[n→inf]))
exp(lim(
log(cos(1/n))
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
1/n
[n→inf]))
exp(lim(
log(cos(x))
⎯⎯⎯⎯⎯⎯⎯⎯⎯
x
[x→0]))
exp(lim(
-sin(x)/cos(x)
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
1
[x→0]))
L'Hôpital
exp(0)
1

Any formula for calculating it from a given n
(other than the obvious)?

What is it that is obvious?