Why Does the Limit of Sine at Infinity Confuse Mathematicians?

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Homework Help Overview

The discussion revolves around the limit of the sine function as its argument approaches infinity, specifically the expression \(\lim_{x \rightarrow \infty} \sin \left( 2\pi \sqrt{x^2+1} \right)\). Participants express confusion regarding whether this limit exists or equals zero, raising questions about the behavior of the sine function and the nature of the argument as \(x\) increases.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Some participants attempt to reason that as \(x\) becomes very large, \(\sqrt{x^2+1}\) approaches an integer, leading to the conclusion that \(\sin(2\pi n) = 0\) for integers \(n\). Others question the validity of assuming \(x\) is an integer and explore the implications of treating \(x\) as a real number. There are discussions about the periodic nature of the sine function and whether the limit can be defined under different assumptions about \(x\).

Discussion Status

The discussion is ongoing, with various interpretations being explored. Some participants suggest that the limit might not exist due to the periodicity of the sine function, while others propose that if \(x\) is restricted to integers, the limit could be zero. There is no explicit consensus, but productive dialogue continues around the assumptions and definitions involved.

Contextual Notes

Participants note the ambiguity in the original problem statement regarding whether \(x\) is to be considered as an integer or a real number. This distinction significantly impacts the conclusions drawn about the limit's existence and value.

  • #31
Curious3141 said:
So you've changed your mind from this previous post ?



Then we're in agreement now. For x ranging only over the integers, the limit is zero. For real x, there is no limit.


NO,nonononono,the limit does not exist.I didn't change my mind.
\lim_{n\rightarrow +\infty} \sin(2\pi\sqrt{n^{2}+1})

does not exist...

However
\lim_{n\rightarrow +\infty} \sin(2\pi [\sqrt{n^{2}+1}]) =0

Do you see the difference?
What do those square paranthesis represent? :wink:

Daniel.
 
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  • #32
twoflower said:
I finally have it. I didn't become aware that if "n" goes over integers only, it really has the limit zero.

It doesn't... :-p I inserted the "[,]" function to make it work...Do you know this function??

Daniel.

PS.HINT:it's related with the function Hurkyl used.
 
  • #33
Is it a nearest integer function ? I'm familiar with the ceiling and floor notation, but not this one.

But still, isn't this equivalent to saying the limit exists (and is zero) as long as n is allowed to range over the integers only ?
 
  • #34
Yes,that's the function... :-p

It can't b zero,because the sqrt is never an integer,even for natural "n".I'm sure u know that the squares of natural numbers don't form a dense subset in N.

Daniel.
 
  • #35
\lim_{n\rightarrow +\infty} \sin(2\pi [\sqrt{n^{2}+1}]) =0
That's ofcourse a pretty lame limit, since [\sqrt{n^{2}+1}], and \lfloor \sqrt{n^{2}+1} \rfloor and \lceil \sqrt{n^{2}+1} \rceil are all integers and \sin(2m\pi)=0 \, \forall \, m \in \mathbb{N}. So the function is the constant zero function.
 
  • #36
dextercioby said:
NO,nonononono,the limit does not exist.I didn't change my mind.
\lim_{n\rightarrow +\infty} \sin(2\pi\sqrt{n^{2}+1})

does not exist...

Sorry, but the limit who doesn't exist is
\lim_{n\rightarrow +\infty} \sqrt{n^2+1}

Take a look:

\lim_{n\rightarrow +\infty} \sin(2\pi \sqrt{n^{2}+1}) =

\lim_{n\rightarrow +\infty} \sin(2\pi (int(\sqrt{n^{2}+1}) + frac(\sqrt{n^{2}+1})) ) =

\lim_{n\rightarrow +\infty} \sin(2\pi frac(\sqrt{n^{2}+1}) ) = 0

:smile:
 
  • #37
dextercioby said:
Yes,that's the function... :-p

It can't b zero,because the sqrt is never an integer,even for natural "n".I'm sure u know that the squares of natural numbers don't form a dense subset in N.

Daniel.

Now, I'm confused. :confused:

Hurkyl asserted this :

\lim_{x \rightarrow \infty} \mathrm{frac} \sqrt{x^2+1} = 0 provided x is allowed to range only over the integers.

I think you're in agreement with Hurkyl's statement above ?

But \lim_{x \rightarrow \infty} (\sin {(2\pi(\sqrt{x^2 + 1})} = \lim_{x \rightarrow \infty}(\sin{2\pi(\lfloor \sqrt{x^{2}+1} \rfloor + \mathrm{frac} ( \sqrt{x^{2}+1))}}

Since \lim_{x \rightarrow \infty}\mathrm{frac} ( \sqrt{x^{2}+1}) = 0, the original limit becomes :

\lim_{x \rightarrow \infty}(\sin{2\pi(\lfloor \sqrt{x^{2}+1} \rfloor)}

which is zero. (for all integral x)
 
Last edited:
  • #38
Yes,i was wrong,for natural numbers,the limit is zero.See the proof above.2 posts above.

Daniel.

PS.Yes,from time to time i am wrong.I try to keep these moments as far apart as possible... :-p
 
Last edited:
  • #39
Curious3141 said:
\lim_{x \rightarrow \infty}(\sin{2\pi(\lfloor \sqrt{x^{2}+1} \rfloor)}

which is zero. (for all integral x)

That is zero for every real,because the function selects the closest natural number in the vecinity of \sqrt{x^{2}+1}.

I would say that the initial limit being zero it's true for all integers,but not for all reals,because the fraction part would not approach zero from the positive part...

Daniel.

PS.Analyze this limit
\lim_{x\rightarrow +\infty} \{\sqrt{x^{2}+1}\}
for x real.It goes to zero but with alternating values.So limit of sine would not be defined.
 
  • #40
dextercioby said:
NO,nonononono,the limit does not exist.I didn't change my mind.
\lim_{n\rightarrow +\infty} \sin(2\pi\sqrt{n^{2}+1})

does not exist...

However
\lim_{n\rightarrow +\infty} \sin(2\pi [\sqrt{n^{2}+1}]) =0

Do you see the difference?
What do those square paranthesis represent? :wink:

Daniel.

WRONG!Don't mislead people. :mad: The limit exists. It is ZERO.
I advise u read some Calculus.

Rogerio.
 
  • #41
I repeat, the limit only exists when n is constrained to range over integers.
 
  • #42
Yes, of course.

However this assumption is obvius from the question. As you said some posts ago,

"-- the domain is left implicit. It's one of those things you're supposed to infer from context"

and,

"you'll usually see n or m as the dummy variable when it's supposed to range over integers."

:smile:
 
Last edited:
  • #43
dextercioby said:
Yes,that's the function... :-p

It can't b zero,because the sqrt is never an integer,even for natural "n".I'm sure u know that the squares of natural numbers don't form a dense subset in N.

Daniel.

What?!

It is WRONG AGAIN! Don't mislead people. :mad:

Take the advice and read some Calculus!

 

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