# A trignometric limit going to infinity

## Homework Statement:

Problem -: lim cos(pi(n^2+n)^(1/2)) where n tends to infinity.

## Relevant Equations:

Costheta=cottheta/cosectheta
L'Hospital rule
I wrote cos(pi(n^2+n)^(1/2)) as cot(pi(n^2+n)^(1/2))/cosec(pi(n^2+n)^(1/2)) and as we know cot(npi)=infinity and cosec(npi)=infinity , so i applied L'Hospital.After i differentiated i again got the same form but this time cosec/cot which is again infinity/infinity.But if i differentiate it i will get the same form again.So what to do now.

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i dont think you have to use L'hospital here have you tried other methods

i dont think you have to use L'hospital here have you tried other methods
Can you suggest any other method i tried to use expansion also but it didn't worked

did you try expanding
##
\sqrt{n^2 +n}
##
for large n

Orodruin
Staff Emeritus
Homework Helper
Gold Member
You cannot apply l’Hopital. Your function is not a function of a continuous parameter.

did you try expanding
##
\sqrt{n^2 +n}
##
for large n
Tried it but it didn't helped.

You cannot apply l’Hopital. Your function is not a function of a continuous parameter.
So sir what should i do i am struggling over it.

Tried it but it didn't helped.
it does help how did you expand it

it does help how did you expand it
What is the use of treating it as capital N.After it what will i do.

epenguin
Homework Helper
Gold Member
Seems to me obvious what the answer is considering ##lim\text{ } cos(f(n)) = cos (lim(f(n)) ## as ##n→∞ ##

timetraveller123
Orodruin
Staff Emeritus
Homework Helper
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Seems to me obvious what the answer is considering ##lim\text{ } cos(f(n)) = cos (lim(f(n)) ## as ##n→∞ ##
This only works if f(n) has a well defined limit. This is not the case here. Instead, my suggestion is to follow the suggestion of post #4.

Tried it but it didn't helped.
Show us what you did. If you do not we cannot help you.

timetraveller123
This only works if f(n) has a well defined limit. This is not the case here. Instead, my suggestion is to follow the suggestion of post #4.

Show us what you did. If you do not we cannot help you.
I took out n^2 outside underroot and then opened the expansion of (1+1/n)^(1/2).Further i was left with cos(npi+pi/2) which will be equal to cos(pi/2) or -cos(pi/2) which gives me the answer as 0.
But i want to know is there any other method of solving it without expansion as we are not taught about this expansion in our school.

timetraveller123
i am not sure of this would work some one please correct me
and since op already knows answer just want another method i posting another "solution"

##
\sqrt{n^2+n} -n = x\\
0<x<1\\
0 = x^2 + 2nx -n\\
0=\frac{x^2}{n} + 2x - 1
##
first term goes arbitrarily close to zero
so x tends to ##
\frac{1}{2}
##

Physics lover
i am not sure of this would work some one please correct me
and since op already knows answer just want another method i posting another "solution"

##
\sqrt{n^2+n} -n = x\\
0<x<1\\
0 = x^2 + 2nx -n\\
0=\frac{x^2}{n} + 2x - 1
##
first term goes arbitrarily close to zero
so x tends to ##
\frac{1}{2}
##
Thanks for this one.It is far better than my previous attempt.

epenguin
Homework Helper
Gold Member
This only works if f(n) has a well defined limit. This is not the case here. Instead, my suggestion is to follow the suggestion of post #4.
Ah, true. Ulseful maybe as conjecture when stuck even if wrong, to get somewhere like #13

wait i am not understanding why does the function not have well defined limit doesn't it do this
##
\sqrt{n^2 + n} = n+\frac{1}{2}
##

Orodruin
Staff Emeritus
Homework Helper
Gold Member
wait i am not understanding why does the function not have well defined limit doesn't it do this
##
\sqrt{n^2 + n} = n+\frac{1}{2}
##
The limit of that is infinity. You are basically suggesting to compute ##f(\infty)## where ##f## is an oscillating function.

timetraveller123
but n is integer here which makes f(infinity) = 0
i am not understanding something here

wait i said the limit is
##
n + 0.5
##
and not infinity
ok this is probably wrong

Orodruin
Staff Emeritus
Homework Helper
Gold Member
but n is integer here which makes f(infinity) = 0
i am not understanding something here

wait i said the limit is
##
n + 0.5
##
and not infinity
ok this is probably wrong
That is the asymptote, not the limit. The limit is infinity. The appropriate way of doing this is noting that
$$\sqrt{n^2+n} = n \sqrt{1+1/n} = n+1/2+ \mathcal O(1/n).$$
Therefore
\begin{align*}
\cos(\pi\sqrt{n^2+n}) &= \cos(\pi[n+1/2+ \mathcal O(1/n)]) \\
&= \cos(\pi n + \pi/2 + \mathcal O(1/n)) \\
&= \cos(\pi n + \pi/2) \cos(\mathcal O(1/n)) - \sin(\pi n + \pi/2) \sin(\mathcal O(1/n)).
\end{align*}
The first term is zero and the second term is ##\mathcal O(1/n)## since the sine function is bounded. Thus, the limit of the expression is 0.

sysprog, Mark44 and timetraveller123
ah i see thanks