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A trignometric limit going to infinity

Problem 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.
 
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

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You cannot apply l’Hopital. Your function is not a function of a continuous parameter.
 
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.
 

epenguin

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

Orodruin

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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.
 
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.
 
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}
##
 
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

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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. :blushing: 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

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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.
 
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

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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.
 

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