solving a tricky infinite limit analytically

May 2010
8
0
Hi,
I've been trying to solve this limit analytically, but I can't figure out how to manipulate the expression in such a way so that I can plug in infinity for n without getting an indeterminate form:

lim n * sqrt(2-2cos(2pi/n))
n ->infinity

From entering this limit into the calculator, I get 2pi, which I know is the correct answer. So how might I solve this limit analytically?
Also, bear in mind that this is in the precalc section, so I'm familiar only with with limit properties and such.

Thanks a lot!
 
Jun 2008
292
87
1) 2-2cos(2pi/n)=4(sin(pi/n)^2)
2) sqrt[2-2cos(2pi/n)]=2mod(sin(pi/n))
now
3) Lt n*2mod(sin(pi/n))=2pi[(sin(pi/n)]/(pi/n)] (limit n approaches to inf)
n->inf
but since
4) [sinx]/x=1 if x->inf therfor
5) Lt(2pi)[(sin(pi/n))/(pi/n)]=2pi
n->inf
 
May 2010
8
0
Hi nikhil,
Thanks a lot for your help, but you're going to fast for me! I'm unsure as to how you got that 2 - 2cos (2pi/n) = 4(sin(pi/n))^2. In addition, I'm unfamiliar with the mod notation. Can you please explain each step, and is there a way to solve the limit without using mod?
Thanks very much,
-Synergy777
 
Jun 2008
292
87
1) cos2x=1-2(sinx)^2 its an identity
2) |x|(mod x)=x if x>0 and -x if x<0
now sqrt(x^2)=|x|
but since sin(pi/n) >0 therfor |sin(pi/n)|=sin(pi/n)
 
May 2010
8
0
Okay, I now understand the first part, and I get 2 - 2 cos(2pi/n) = 4(sin(pi/n))^2, and after n*sqrt(4(sin(pi/n))^2), I get 2n*sin(pi/n), assuming that everything is positive since sin(pi/n) > 0. But how did you then get 2pi from the expression we now have:
lim 2n*sin(pi/n)
n->inf

Also, later on you say that sin x/x = 1 if x -> inf, but isn't that = 0, not 1? And also, I know that lim of sin (n)/ n as n->0 is 0, but I'm unfamiliar with the identity that sin(pi/n)/(pi/n) as n -> inf is 1.
 
Jun 2008
292
87
sorry i did a mistake in hurry
sinx/x=1 if x->0
now
consider [sin(1/n)]/(1/n) where n->inf
let 1/n=t
if n->inf then t->0
so [sin(1/n)]/(1/n) where n->inf becomes
sin(t)/t where t->0
which is =1
in my steps
[sin(pi/n)/(pi/n)] n->inf
now let pi/n=t
if n->inf then t->0 therfor
[sin(pi/n)/(pi/n)] n->inf =sin(t)/t where t->0
 
May 2010
8
0
Ah, ok, that makes sense. But, how did you figure out to divide sin (pi/n) by (pi/n) in the step:

2n sin(pi/n) = 2pi (sin pi/n)/(pi/n)

In other words, had we not known beforehand that the answer to this limit was 2pi, how would we know to divide by pi/n ?
 
Jun 2008
292
87
no need to know solution beforehand

we know that sinx/x=1 as x->0 or
[sin(1/n)]/(1/n)=1 as n->inf
in the problem we had
n*sin(pi/n)
=[sin(pi/n)]/(1/n)
but we need the form sinx/x (angle and denominator both are x)
so to make angle and denominator same we multiply and divide numerator and denominator by pi so know we have
[sin(pi/n)]/(1/n)=pi*[sin(pi/n)]/(pi/n) [it is in sinx/x form as angle and denominator are same]
 
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