# Interchange of limit operations

1. Dec 9, 2010

### ait.abd

1. The problem statement, all variables and given/known data
Find a sequence of continuous functions $$f_n: R \rightarrow R$$ such that $$lim_{x \rightarrow 0}lim_{n \rightarrow \infty}f_n(x)$$ and $$lim_{n \rightarrow \infty}lim_{x \rightarrow 0}f_n(x)$$ exist and are unequal.

2. Relevant equations
N/A

3. The attempt at a solution
I think I need a sequence of continuous functions that has a limit function which is continuous at zero but discontinuous at some other point. In that case, the sequence of functions will not be uniformly convergent and we will not have these limits equal. But I dont know what function can fulfill this criteria.

2. Dec 9, 2010

### micromass

Staff Emeritus
Well, can you give a sequence which is not uniformly convergent?

3. Dec 9, 2010

### ait.abd

Yeah I can but the limits turn out to be equal. For example,
$$f_n(x) = (cos(x))^n$$
$$lim_{n \rightarrow \infty}lim_{x \rightarrow 0} f_n = 1$$
$$lim_{x \rightarrow 0}lim_{n \rightarrow \infty} f_n = 1$$
This sequence is not uniformly convergent but the two limits are equal. The question requires them to be UNEQUAL.

4. Dec 9, 2010

### ait.abd

In above post, there is a mistake.
$$lim_{x \rightarrow 0} lim_{n \rightarrow \infty} f_n(x)$$ doesn't exist. The question requires them to EXIST.

5. Dec 9, 2010

### micromass

Staff Emeritus
Are you certain that it doesn't exist? I think that the limit does exist and equals zero.

You do have to restrict the function to [-pi,pi], otherwise there is no limit...

6. Dec 9, 2010

### ait.abd

Let's say we restrict the sequence to [-pi,pi] and say it is zero everywhere else (because we have to define it on R and not a subset of it).

Yes it doesn't exist. As you can see the limit function is following
$$lim_{n \rightarrow \infty}f_n(x) = 1 for x=0$$
$$lim_{n \rightarrow \infty}f_n(x) = 0 , elsewhere}$$

So the limit function is discontinuous at x=0 and $$lim_{x \rightarrow 0} f(x)$$ doesn't exist. By the way, it is the very first question of the exercise so it should be easy (I think so ) I dont know where I'm missing the point.

7. Dec 9, 2010

### micromass

Staff Emeritus
So, the limit function is

$$f(0)=1~\text{and}~f(x)=0~\text{if}~x\neq 0$$.

Then the limit $$\lim_{x\rightarrow 0}{f(x)}$$ certainly exists (and it equals 0)! You can easily prove this with the definition of limit...

8. Dec 9, 2010

### ait.abd

Don't mind my asking very basic questions because i think my definition of limit is flawed.
What I know is that limit is defined, in above case, when f is defined on C{0} i.e. R - {0}. In this case, we can say limit x approaches 0 is zero as you described. But, since function is defined on x=0, shouldn't we take f(0) as limit ? Also, if we take it as a limit as you say, shouldn't the limit be different if we approach it from either side i.e. (-pi,0) and [0, pi). So a limit is not really defined here?

9. Dec 9, 2010

### micromass

Staff Emeritus
No, the limit of f is independent of f(0). The limit of a function is what the function value should be to make the function continuous. In our situation, we have that the function is 0, except in the point zero. So if f(0)=0 (which is not the case), then the function f would be continuous. This means that the limit of f equals 0.

10. Dec 9, 2010

### ait.abd

I got it now ...and in that case limit will be zero irrespective of the direction we use to approach x=0. Thank you very much micromass it really helped. Thanks again.