A trignometric limit going to infinity

• Physics lover
In summary, the conversation discusses the use of L'Hospital's rule to solve a mathematical problem involving the function cos(pi(n^2+n)^(1/2)). The speaker suggests trying other methods, such as expanding the function or treating it as a capital N. Another suggestion is to use the fact that cos(f(n)) = cos(lim(f(n))) as n approaches infinity. However, it is pointed out that this only works if f(n) has a well-defined limit, which is not the case in this problem. Instead, the suggestion is made to follow the approach in post #4, which involves rewriting the function as a limit and using trigonometric identities to simplify it. In conclusion, the conversation highlights the importance of
Physics lover
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.

i don't think you have to use L'hospital here have you tried other methods

timetraveller123 said:
i don't 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

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

timetraveller123 said:
did you try expanding
##
\sqrt{n^2 +n}
##
for large n
Tried it but it didn't helped.

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

timetraveller123 said:
it does help how did you expand it
What is the use of treating it as capital N.After it what will i do.

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

timetraveller123
epenguin said:
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.

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

timetraveller123
Orodruin said:
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
timetraveller123 said:
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.

Orodruin said:
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}
##

timetraveller123 said:
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

timetraveller123 said:
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

1. What is a trigonometric limit going to infinity?

A trigonometric limit going to infinity is a mathematical concept where the value of a trigonometric function (such as sine, cosine, or tangent) approaches infinity as the input variable approaches a certain value, usually represented by the symbol "x". This means that the function continues to increase without bound as x gets closer and closer to the specified value.

2. How do you solve a trigonometric limit going to infinity?

To solve a trigonometric limit going to infinity, you can use various techniques such as factoring, rationalization, or substitution. It is important to first determine the type of limit and then apply the appropriate method to evaluate the limit. You may also need to use trigonometric identities or properties to simplify the expression before taking the limit.

3. What are the common types of trigonometric limits going to infinity?

The most common types of trigonometric limits going to infinity are sin(x)/x, cos(x)/x, and tan(x)/x as x approaches 0. Other common types include sin(x)/1-cos(x) and 1-cos(x)/x as x approaches 0. These types of limits often involve trigonometric identities and properties, and may require special techniques to solve.

4. What is the significance of trigonometric limits going to infinity?

Trigonometric limits going to infinity are important in understanding the behavior of trigonometric functions as the input variable approaches a certain value. They also have applications in calculus, physics, and engineering, where they can be used to model real-world situations and make predictions about the behavior of systems.

5. Are there any limitations to using trigonometric limits going to infinity?

Yes, there are limitations to using trigonometric limits going to infinity. These limits only apply to specific types of functions, and may not always accurately represent the behavior of a function as x approaches infinity. Additionally, the evaluation of these limits may become more complex as the expressions involved become more complicated, making it difficult to find an exact solution.

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