# Homework Help: It seems easy but I'm confused - limit

1. Jan 16, 2005

### twoflower

Hi,

I've been trying to compute

$$\lim_{x \rightarrow \infty} \sin \left( 2\pi \sqrt{x^2+1} \right)$$

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.

2. Jan 16, 2005

### vincentchan

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

3. Jan 16, 2005

### twoflower

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

4. Jan 16, 2005

### vincentchan

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

5. Jan 16, 2005

### twoflower

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. Jan 16, 2005

### 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.

7. Jan 16, 2005

### Curious3141

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 ?

8. Jan 16, 2005

### 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

9. Jan 16, 2005

### Curious3141

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. Jan 16, 2005

### Gokul43201

Staff Emeritus
I second Curious,

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

Twoflower,

How did you get these "results" ?

11. Jan 16, 2005

### twoflower

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. Jan 16, 2005

### 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.

13. Jan 16, 2005

### twoflower

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)...

14. Jan 16, 2005

### 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.

15. Jan 16, 2005

### 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
$$\lim_{n \rightarrow \infty} \sin \left( 2\pi \sqrt{n^2+1} \right) = 0$$
sure, I have made an assumtion that n is integer...for an abitrary real number, the limit doesn't exist....

Last edited: Jan 16, 2005
16. Jan 16, 2005

### dextercioby

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

Daniel.

17. Jan 17, 2005

### 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?

18. Jan 17, 2005

### twoflower

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

19. Jan 17, 2005

### Hurkyl

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

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)

20. Jan 17, 2005

### dextercioby

If $$+\infty$$ is an integer,i agree.

Can u show that $$+\infty$$ is an integer????

Daniel.