Why Does the Limit of Sine at Infinity Confuse Mathematicians?

[twoflower]

Hi,

I've been trying to compute

<br /> \lim_{x \rightarrow \infty} \sin \left( 2\pi \sqrt{x^2+1} \right)<br />

But don't have any idea. According to the results it should be 0, but I can't see the way I should get it...

Thank you.

 

[vincentchan]

for x>>1.. the square root approches to an integer... (even if it is going to infinite)

 

[twoflower]

for x>>1.. the square root approches to an integer... (even if it is going to infinite)

Yes, the square root approaches to infinity I think, but I still don't get it.

 

[vincentchan]

it is also an integer... sin(2pi n) =0

 

[twoflower]

it is also an integer... sin(2pi n) =0

Well, that's the thought I got instantly after the first look, but I'm trying to produce some serious proof of that...If it were in test, do you think that your answer would be accepted?

 

[Curious3141]

The limit should be undefined. Sine is periodic and there is no way to define what's going to happen when you take such a limit.

If x is defined to be a "very large" integer, I can see an argument for the answer zero, but by default x should be taken as real, and there is no defined limit.

 

[Curious3141]

for x>>1.. the square root approches to an integer... (even if it is going to infinite)

That's only if you assume x to be an integer, and there's no basis for such an assumption.

What if x^2 = e^2(10^{16}) - 1, for example ?

 

[vincentchan]

because he said the result is zero... it is reasonable for me to make an assumtion that x is integer... otherwise... obviously, the function is undefine

 

[Curious3141]

because he said the result is zero... it is reasonable for me to make an assumtion that x is integer... otherwise... obviously, the function is undefine

But what 'results' is he talking about ? The natural tendency is to keep plugging in large integer values (like 10^16) into the calculator in which case of course, you'd get zero. It would also be wrong.

 

[Gokul43201]

I second Curious,

The function is well-defined, but the particular limit does not exist.

Twoflower,

How did you get these "results" ?

 

[twoflower]

I second Curious,

The function is well-defined, but the particular limit does not exist.

Twoflower,

How did you get these "results" ?

He gave us these limits as recommended homework along with the results...The only that comes to my mind is that I miswrote the original limit and replaced "n" with "x" :-(

 

[dextercioby]

Okay,if it was "n" instead of "x",then the limit still does not exist.Moreover,it can be even shown it cannot be zero.For it to be zero,it is necessary that
\lim_{n\rightarrow +\infty} \sqrt{n^{2}+1} =\ natural \ multiple \ of \frac{1}{2}

Now u have an equation to solve:are there natural "n"-s for which
\sqrt{n^{2}+1}=\frac{k}{2}
,for arbitrary "k",very large?
My answer is "no",therefore the limit cannot be "0".Moreover,it doesn't exist.

Daniel.

 

[twoflower]

Okay,if it was "n" instead of "x",then the limit still does not exist.Moreover,it can be even shown it cannot be zero.For it to be zero,it is necessary that
\lim_{n\rightarrow +\infty} \sqrt{n^{2}+1} =\ natural \ multiple \ of \frac{1}{2}

Now u have an equation to solve:are there natural "n"-s for which
\sqrt{n^{2}+1}=\frac{k}{2}
,for arbitrary "k",very large?
My answer is "no",therefore the limit cannot be "0".Moreover,it doesn't exist.

Daniel.

Thank you Daniel but I can't see what couldn't the whole expression in sin be natural multiple of \pi (and ie. \sqrt{n^2+1} any integer)...

 

[dextercioby]

Well,solve this equation
\sqrt{n^{2}+1}=k
for natural "k" and "n".And pick out the (possible,if any) solutions,the ones for very large"n".

Daniel.

 

[vincentchan]

you confused the concept of limit again... \sqrt{n^{2}+1} approach to an integer while n is large... but \sqrt{n^{2}+1} itself is not an integer... proof:
\lim_{n \rightarrow \infty} \sqrt{n^{2}+1} = n where n is an integer... done
<br /> \lim_{n \rightarrow \infty} \sin \left( 2\pi \sqrt{n^2+1} \right) = 0<br />
sure, I have made an assumtion that n is integer...for an abitrary real number, the limit doesn't exist...

 

[dextercioby]

So what's your point??Is that limit zero,or it doesn't exist??

Daniel.

 

[vincentchan]

if n is integer, the limit exist, if not,then not...
but you do agree with me that \lim_{x \rightarrow \infty} \sqrt{x^2+1} is integer, right?

 

[twoflower]

if n is integer, the limit exist, if not,then not...
but you do agree with me that \lim_{x \rightarrow \infty} \sqrt{x^2+1} is integer, right?

Yes, this seems more logical to me than the Daniel's one...

 

[Hurkyl]

if n is integer, the limit exist, if not,then not...
but you do agree with me that \lim_{x \rightarrow \infty} \sqrt{x^2+1} is integer, right?

No. The limit doesn't exist, so it can't be an integer.

However, \lim_{x \rightarrow \infty} \mathrm{frac} \sqrt{x^2+1} = 0

If x is only allowed to range over integers. (frac is the fractional part, so frac 1.2 = .2)

 

[dextercioby]

if n is integer, the limit exist, if not,then not...
but you do agree with me that \lim_{x \rightarrow \infty} \sqrt{x^2+1} is integer, right?

If +\infty is an integer,i agree.

Can u show that +\infty is an integer?

Daniel.

 

[dextercioby]

Hurkyl agrees with me,so i must be right...

Daniel.

EDIT:Nice quote...

 

[vincentchan]

Can u show that LaTeX graphic is being generated. Reload this page in a moment. is an integer?

this is the assumtion i made earlier... no need to show

 

[Curious3141]

No. The limit doesn't exist, so it can't be an integer.

However, \lim_{x \rightarrow \infty} \mathrm{frac} \sqrt{x^2+1} = 0

If x is only allowed to range over integers. (frac is the fractional part, so frac 1.2 = .2)

Is there a proper mathematical basis for saying "x is only allowed to range over integers" for a limit ? I'm not referring to this question in particular, but can you say that for a general limit, e.g. does the statement

\lim_{\substack{x \rightarrow \infty}} f(x), where x \in {Z^+}

have valid mathematical meaning ? Thanks.

 

[Hurkyl]

That is a deficiency in notation -- the domain is left implicit. It's one of those things you're supposed to infer from context (and will be stated explicitly when such an inference can't be made)

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

 

[vincentchan]

first show there exist infinitely many positive integer (using induction),
by the reason above, it is reasonable to say a positive integer x approach to infinite..

 

[Curious3141]

That is a deficiency in notation -- the domain is left implicit. It's one of those things you're supposed to infer from context (and will be stated explicitly when such an inference can't be made)

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

If the original question is restated allowing x to vary only over the integers, might the limit of zero be correct ? The fractional part tends to zero after all.

 

[dextercioby]

Well,then yes.It would be
\lim_{n\rightarrow +\infty} \sin(2\pi[\sqrt{n^{2}+1}]) =0

Daniel.

 

[twoflower]

Well,then yes.It would be
\lim_{n\rightarrow +\infty} \sin(2\pi[\sqrt{n^{2}+1}]) =0

Daniel.

Ok I think the original limit was supposed to be a limit of sequence, but then I think it doesn't exist anyway...

 

[Curious3141]

Well,then yes.It would be
\lim_{n\rightarrow +\infty} \sin(2\pi[\sqrt{n^{2}+1}]) =0

Daniel.

So you've changed your mind from this previous post ?

Okay,if it was "n" instead of "x",then the limit still does not exist.Moreover,it can be even shown it cannot be zero.For it to be zero,it is necessary that
\lim_{n\rightarrow +\infty} \sqrt{n^{2}+1} =\ natural \ multiple \ of \frac{1}{2}

Now u have an equation to solve:are there natural "n"-s for which
\sqrt{n^{2}+1}=\frac{k}{2}
,for arbitrary "k",very large?
My answer is "no",therefore the limit cannot be "0".Moreover,it doesn't exist.

Daniel.

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

 

[twoflower]

I finally have it. I didn't become aware that if "n" goes over integers only, it really has the limit zero.

 

[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!