# A trignometric limit going to infinity

#### Physics lover

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.

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#### timetraveller123

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

#### Physics lover

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

#### timetraveller123

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.

#### Physics lover

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

#### Physics lover

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.

#### Physics lover

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

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

• timetraveller123

#### 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.

• timetraveller123

#### Physics lover

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

#### 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

#### 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

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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. Ulseful maybe as conjecture when stuck even if wrong, to get somewhere like #13

#### timetraveller123

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.

• timetraveller123

#### 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

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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.

#### timetraveller123

ah i see thanks

"A trignometric limit going to infinity"

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