# Permissibility of Limit Technique

1. Feb 3, 2010

### Yitzach

The following came from a step in my EM homework. I came up with the answer that all of my resources (calculator, WolframAlpha, and a friend) were indicating as the correct answer, so it is not about the homework. This is about the permissibility of what was done.
1. $$x=\lim_{n\rightarrow\infty}n\sin\frac{\pi}{n}$$
2. $$x=\lim_{n\rightarrow\infty}\frac{n}{\csc\frac{\pi}{n}}$$
3. $$x=\lim_{n\rightarrow\infty}\frac{1}{\frac{\pi\cos\frac{\pi}{n}}{n^2\sin^2\frac{\pi}{n}}}$$
4. $$x=\lim_{n\rightarrow\infty}\frac{n^2\sin^2\frac{\pi}{n}}{\pi\cos\frac{\pi}{n}}$$
5. $$x=\lim_{n\rightarrow\infty}\frac{x^2}{\pi\cos\frac{\pi}{n}}$$
6. $$\frac{1}{x}=\lim_{n\rightarrow\infty}\frac{1}{\pi\cos\frac{\pi}{n}}$$
7. $$x=\lim_{n\rightarrow\infty}\pi\cos\frac{\pi}{n}$$
8. $$x=\pi\cos0=\pi$$
Given that the exponent on x in step 5 is not 1, x is not zero or infinity, and that x is not the entirety of the argument of the limit, can we/I conclude that this will work in general?
Other possible solutions between 1 and 8 include a change in variable, the limit if the series expansion at infinity, the squeeze theorem, or using the inverse of step two instead.

2. Feb 3, 2010

### Hurkyl

Staff Emeritus
If x is known to exist and be a positive real number, then this method works, although it is presented mildly sloppily -- remember that lim A/B is (lim A)/(lim B) if both limits on the right hand side are known to exist, and the denominator is nonzero.

Note that $x = +\infty$ and x = 0 are also solutions to the extended real number equation
$$x = \frac{x^2}{\pi}$$

3. Feb 3, 2010

### Yitzach

By what you said here, I take it to mean that once an answer has been arrived at by this method, it would be advisable/required to check 0 and infinity to make sure that they are not the answer. In this case I knew that neither of those could be the answer based on the nature of the question and the fact that the answer had to exist or the question would not have been asked.
Why did you limit the answer to positive real numbers?

4. Feb 3, 2010

### Hurkyl

Staff Emeritus
Really, I just wanted to avoid zero and +infinity. I could have said "nonzero real". (which avoids -infinity too, but ah well)

5. Feb 3, 2010

### LCKurtz

So you are aware or do you care that much simpler more direct arguments are available?

6. Feb 3, 2010

### Yitzach

Fair enough. So is there a reason for real numbers?

Yes I was aware of other methods. I was wondering about the permissibility of what I did for if I wanted to do it again later. Notice that I mentioned four other methods in the initial post.

7. Feb 4, 2010

### Hurkyl

Staff Emeritus
We are looking at a limit of real numbers, so the limit, if it exists, must be an extended real number.