Why Does the Limit of Sine at Infinity Confuse Mathematicians?

[dextercioby]

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?

Daniel.

 

[dextercioby]

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... I inserted the "[,]" function to make it work...Do you know this function??

Daniel.

PS.HINT:it's related with the function Hurkyl used.

 

[Curious3141]

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 ?

 

[dextercioby]

Yes,that's the function...

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.

 

[Galileo]

\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.

 

[Rogerio]

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

 

[Curious3141]

Yes,that's the function...

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.

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)

 

[dextercioby]

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...

 

[dextercioby]

\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.

 

[Rogerio]

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?

Daniel.

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

Rogerio.

 

[Hurkyl]

I repeat, the limit only exists when n is constrained to range over integers.

 

[Rogerio]

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."

 

[Kittel Knight]

Yes,that's the function...

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.

Take the advice and read some Calculus!