It seems easy but I'm confused - limit

The conversation is about the limit of the function \lim_{x \rightarrow \infty} \sin \left( 2\pi \sqrt{x^2+1} \right)and whether it exists or not. The original poster is trying to find a proof for the limit being 0, but others in the conversation argue that it does not exist. One person suggests that if x is considered to be a "very large" integer, the limit could be 0, but this is not a valid assumption. Another person suggests that the limit cannot be 0 because it would require the square root to approach a natural multiple of 1/2, which is not possible for very large values of x. The conversation ends
  • #1
twoflower
368
0
Hi,

I've been trying to compute

[tex]
\lim_{x \rightarrow \infty} \sin \left( 2\pi \sqrt{x^2+1} \right)
[/tex]

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.
 
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  • #2
for x>>1.. the square root approches to an integer... (even if it is going to infinite)
 
  • #3
vincentchan said:
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.
 
  • #4
it is also an integer... sin(2pi n) =0
 
  • #5
vincentchan said:
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?
 
  • #6
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.
 
  • #7
vincentchan said:
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 [itex]x^2 = e^2(10^{16}) - 1[/itex], for example ?
 
  • #8
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
 
  • #9
vincentchan said:
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.
 
  • #10
I second Curious,

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

Twoflower,

How did you get these "results" ?
 
  • #11
Gokul43201 said:
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" :-(
 
  • #12
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
[tex] \lim_{n\rightarrow +\infty} \sqrt{n^{2}+1} =\ natural \ multiple \ of \frac{1}{2} [/tex]

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

Daniel.
 
  • #13
dextercioby said:
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
[tex] \lim_{n\rightarrow +\infty} \sqrt{n^{2}+1} =\ natural \ multiple \ of \frac{1}{2} [/tex]

Now u have an equation to solve:are there natural "n"-s for which
[tex] \sqrt{n^{2}+1}=\frac{k}{2} [/tex]
,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 [itex]\pi[/itex] (and ie. [itex]\sqrt{n^2+1}[/itex] any integer)...
 
  • #14
Well,solve this equation
[tex] \sqrt{n^{2}+1}=k [/tex]
for natural "k" and "n".And pick out the (possible,if any) solutions,the ones for very large"n".

Daniel.
 
  • #15
you confused the concept of limit again... [tex] \sqrt{n^{2}+1}[/tex] approach to an integer while n is large... but [tex] \sqrt{n^{2}+1}[/tex] itself is not an integer... proof:
[tex] \lim_{n \rightarrow \infty} \sqrt{n^{2}+1} = n [/tex] where n is an integer... done
[tex]
\lim_{n \rightarrow \infty} \sin \left( 2\pi \sqrt{n^2+1} \right) = 0
[/tex]
sure, I have made an assumtion that n is integer...for an abitrary real number, the limit doesn't exist...
 
Last edited:
  • #16
So what's your point??Is that limit zero,or it doesn't exist?? :confused:

Daniel.
 
  • #17
if n is integer, the limit exist, if not,then not...
but you do agree with me that [itex] \lim_{x \rightarrow \infty} \sqrt{x^2+1} [/itex] is integer, right?
 
  • #18
vincentchan said:
if n is integer, the limit exist, if not,then not...
but you do agree with me that [itex] \lim_{x \rightarrow \infty} \sqrt{x^2+1} [/itex] is integer, right?

Yes, this seems more logical to me than the Daniel's one...
 
  • #19
vincentchan said:
if n is integer, the limit exist, if not,then not...
but you do agree with me that [itex] \lim_{x \rightarrow \infty} \sqrt{x^2+1} [/itex] is integer, right?

No. The limit doesn't exist, so it can't be an integer. :tongue2:

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

If x is only allowed to range over integers. (frac is the fractional part, so frac 1.2 = .2)
 
  • #20
vincentchan said:
if n is integer, the limit exist, if not,then not...
but you do agree with me that [itex] \lim_{x \rightarrow \infty} \sqrt{x^2+1} [/itex] is integer, right?

If [tex] +\infty [/tex] is an integer,i agree.

Can u show that [tex] +\infty [/tex] is an integer?

Daniel.
 
  • #21
Hurkyl agrees with me,so i must be right... :tongue2:

Daniel.

EDIT:Nice quote... :rofl: :rofl: :rofl:
 
Last edited:
  • #22
dextercioby said:
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
 
  • #23
Hurkyl said:
No. The limit doesn't exist, so it can't be an integer. :tongue2:

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

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

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

have valid mathematical meaning ? Thanks.
 
  • #24
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.
 
  • #25
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..
 
  • #26
Hurkyl said:
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.
 
  • #27
Well,then yes.It would be
[tex] \lim_{n\rightarrow +\infty} \sin(2\pi[\sqrt{n^{2}+1}]) =0 [/tex]

Daniel.
 
Last edited:
  • #28
dextercioby said:
Well,then yes.It would be
[tex] \lim_{n\rightarrow +\infty} \sin(2\pi[\sqrt{n^{2}+1}]) =0 [/tex]

Daniel.

Ok I think the original limit was supposed to be a limit of sequence, but then I think it doesn't exist anyway...
 
  • #29
dextercioby said:
Well,then yes.It would be
[tex] \lim_{n\rightarrow +\infty} \sin(2\pi[\sqrt{n^{2}+1}]) =0 [/tex]

Daniel.


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

dextercioby said:
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
[tex] \lim_{n\rightarrow +\infty} \sqrt{n^{2}+1} =\ natural \ multiple \ of \frac{1}{2} [/tex]

Now u have an equation to solve:are there natural "n"-s for which
[tex] \sqrt{n^{2}+1}=\frac{k}{2} [/tex]
,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.
 
  • #30
I finally have it. I didn't become aware that if "n" goes over integers only, it really has the limit zero.
 
  • #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.
[tex] \lim_{n\rightarrow +\infty} \sin(2\pi\sqrt{n^{2}+1}) [/tex]

does not exist...

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

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

Daniel.
 
  • #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... :tongue2: 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... :tongue2:

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
[tex] \lim_{n\rightarrow +\infty} \sin(2\pi [\sqrt{n^{2}+1}]) =0 [/tex]
That's ofcourse a pretty lame limit, since [itex][\sqrt{n^{2}+1}],[/itex] and [itex]\lfloor \sqrt{n^{2}+1} \rfloor[/itex] and [itex]\lceil \sqrt{n^{2}+1} \rceil[/itex] are all integers and [itex]\sin(2m\pi)=0 \, \forall \, m \in \mathbb{N}[/itex]. So the function is the constant zero function.
 
<h2>1. What is a limit in mathematics?</h2><p>A limit in mathematics is a fundamental concept that describes the behavior of a function as its input approaches a specific value. It is used to determine the value that a function approaches as its input gets closer and closer to a particular value.</p><h2>2. How do you calculate a limit?</h2><p>To calculate a limit, you need to evaluate the function at values that are very close to the specific value that the input is approaching. This can be done by plugging in values that are slightly larger and slightly smaller than the specific value and observing the output. If the outputs approach a specific value, then that value is the limit.</p><h2>3. Why is it important to understand limits?</h2><p>Understanding limits is crucial in mathematics because it allows us to analyze the behavior of functions and make predictions about their values. It is also used to define important concepts such as continuity and differentiability, which are essential in calculus and other branches of mathematics.</p><h2>4. What are some common types of limits?</h2><p>There are several types of limits, including one-sided limits, infinite limits, and limits at infinity. One-sided limits only consider the behavior of a function from one side of the specific value, while infinite limits involve functions that approach positive or negative infinity. Limits at infinity are used to describe the behavior of a function as its input approaches positive or negative infinity.</p><h2>5. How can limits be used in real-world applications?</h2><p>Limits have various real-world applications, such as in physics, engineering, and economics. They can be used to describe the behavior of physical systems, such as the speed of an object as it approaches a specific point. In economics, limits are used to analyze the behavior of markets and make predictions about future trends.</p>

1. What is a limit in mathematics?

A limit in mathematics is a fundamental concept that describes the behavior of a function as its input approaches a specific value. It is used to determine the value that a function approaches as its input gets closer and closer to a particular value.

2. How do you calculate a limit?

To calculate a limit, you need to evaluate the function at values that are very close to the specific value that the input is approaching. This can be done by plugging in values that are slightly larger and slightly smaller than the specific value and observing the output. If the outputs approach a specific value, then that value is the limit.

3. Why is it important to understand limits?

Understanding limits is crucial in mathematics because it allows us to analyze the behavior of functions and make predictions about their values. It is also used to define important concepts such as continuity and differentiability, which are essential in calculus and other branches of mathematics.

4. What are some common types of limits?

There are several types of limits, including one-sided limits, infinite limits, and limits at infinity. One-sided limits only consider the behavior of a function from one side of the specific value, while infinite limits involve functions that approach positive or negative infinity. Limits at infinity are used to describe the behavior of a function as its input approaches positive or negative infinity.

5. How can limits be used in real-world applications?

Limits have various real-world applications, such as in physics, engineering, and economics. They can be used to describe the behavior of physical systems, such as the speed of an object as it approaches a specific point. In economics, limits are used to analyze the behavior of markets and make predictions about future trends.

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