# Limit as 'constant' approaches infinity

1. Dec 4, 2013

### omlatte1

1. The problem statement, all variables and given/known data

Determine limA→∞(sinc(x/2A))

2. Relevant equations

I would use L'Hôpital's rule, but I'm not sure if it is valid in this case as the function is that of x, not A. I want to know if it's valid to treat x as constant and take derivatives with respect to A and evaluate as A->inf.

3. The attempt at a solution

limA→∞(sinc(x/2A))=limA→∞(sin(x/2A)/(x/2A))
=limA→0(sin((x/2)A)/((x/2)A))
=limA→0(cos((x/2)A))
=1

Is this correct usage of L'Hôpital's rule?

2. Dec 4, 2013

### omlatte1

Perhaps I should include that the sinc function is the Fourier transform of the ractangle function of height A and width 1/A, hence why A is 'constant'...but not...in this question.

3. Dec 4, 2013

### cp255

Wouldn't the just be zero. If you treat x as a constant then no matter its value, having A in the denominator will push it toward zero.

4. Dec 4, 2013

### cpscdave

When you have a sin(x/A) what changes with the sin function as you vary A?

5. Dec 4, 2013

### omlatte1

I believe you are correct cp255, I am just lacking in formal proof.

cpscdave - x is inversely proportional to A, so the sin function approaches zero, but, the denominator of the sinc function (it is the sinc function not the sin function) approaches zero also...need some sort of formal analysis to determine which dominates.

6. Dec 4, 2013

### omlatte1

hence l'hopitals rule, as sinc(x/2A)=sin(x/2A)/(x/2A)

7. Dec 4, 2013

### cpscdave

Well I'm trying to propose a way for you to verify that the application of L'Hôpital's

With the Sinc function changing A alters the period So when A hits ∞ I'll have an infinite period no? Which in turn verify's your answer

Or conversly in the time domain setting A to ∞ would result in a dirac pulse, the fourier transform of which is 1

8. Dec 4, 2013

### omlatte1

Yes that is a good point, i like the analysis. I guess what I would like to know now then, is whether I can indeed use l'hopitals rule in the way I have, using A as the variable in the analysis of the limit(taking derivatives with respect to A), even though sinc was originally a function of x.

9. Dec 4, 2013

### cpscdave

I don't see any reason why you can't as you are doing the derrivate of the varible thats changing in this case :)
The fact that you can verify the answer in a few other methods lends credance to this as well :)

10. Dec 4, 2013

### Staff: Mentor

omlatte1, you title is misleading. In your limit A is not a constant - it is a variable. In the limit process, x is held constant.

11. Dec 4, 2013

### omlatte1

It it? I think many other people like me who don't exactly understand how it works would phrase the question in a similar manner, hence the quotes for constant.

12. Dec 4, 2013

### Staff: Mentor

The operative phrase here is "don't exactly understand how it works."

In this limit, limA→∞(sinc(x/2A)), A is what is changing, not x.

This is exactly the same as writing
$$\lim_{x \to \infty} sinc(\frac C {2x})$$

I don't believe you would be fooled into thinking x is a constant (or 'constant' as you wrote it).

BTW, when you write x/2A, what you're really writing is (x/2) * A. If you mean x/(2A), you need parentheses.

13. Dec 4, 2013

### omlatte1

A is what I said was constant,and it is, with respect to the transform of the original rectangle function, others in this thread have quoted x as being constant. This also is valid when speaking of l'hopitals rule in part because the partial derivative of sinc with respect to A was taken.

Also you write: "A is what is changing, not x" then "I don't believe you would be fooled into thinking x is a constant"
a confusing post at best. By the way do not assume what you think I know and whether or not I am making true or purposely false statements with respect to semantics. Your post implies I am lying and is completely beyond the limits of your purpose in this forum. Keep it to yourself.

14. Dec 4, 2013

### Staff: Mentor

But none of that has anything to do with the limit you posted, in which A is changing. That's what $lim_{A \to \infty} <something>$ means. If something is changing, it is NOT constant.
If you take a derivative or partial derivative of some expression with respect to A, the clear implication is that A is changing (not constant), and any other quantities are held fixed.
It shouldn't be confusing if you read (and quoted) all of what I said.
You are completely misreading what I said. I did not say or imply or suggest in any way that you were making purposely false statements. Period.
My post does nothing of the kind, and I don't see how you can possibly infer that.

15. Dec 5, 2013

### omlatte1

This was solved a while back now.

16. Dec 5, 2013

### HallsofIvy

Staff Emeritus
He said that in post 10 in which x was held constant and A was "going to infinity".

He said that in post 12 where he had given a different example

It would have been if if it had been a single post- but it wasn't, it was two different posts saying different things.

We can only assume you know what you say. Mark44 was correcting your use of a word and trying to clarify for others. I cannot imagine how you think that implies you were "lying". It only implies that you were mistaken.

17. Dec 5, 2013

### omlatte1

Knowledge was in reference to the question, funnily enough I did not know the answer to my own question.