# Determine where f has a limit

1. Nov 13, 2008

### kathrynag

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

Define f:R--->R as follows:
f(x)=x-[x] if [x] is even
f(x)=x-[x+1] if [x] is odd

Determine the points where f has a limit and justify

2. Relevant equations

3. The attempt at a solution

I'm not even sure where to get started. I assume that f has a limit evrywhere except the integers, but not so sure. Maybe I should try each case seperately.

2. Nov 13, 2008

### HallsofIvy

Staff Emeritus
Start by calculating some values. What would f(x) be for x between 0 and 2?

3. Nov 13, 2008

### tiny-tim

Welcome to PF!

Hi kathrynag! Welcome to PF!

Yes, definitely do each of the three cases separately!

4. Nov 13, 2008

### kathrynag

If x =1, x is odd

Then f=1-1=0

If x=1.9999
Then f=1.9999-1=.9999

5. Nov 13, 2008

### HallsofIvy

Staff Emeritus
No. $f(x)= x-\lfloor x+ 1\lrloor$ if x is odd. Since 1 is odd, $f(1)= 1- \lfloor 2\rfloor= -1$

In fact, for any 0< x< 1, since 0 is even, f(x)= x-0= x while f(1)= -1. What does that tell you?

It should be obvious that, for any non-integer x between two integers, f(x) is x- some constant- and that's a linear function. What happens at x= 2? What is f(2)? What is f(x) if x is between 1 and 2?

6. Nov 13, 2008

### kathrynag

oops, i guess i did the wrong one by accident. But is 0, actually an even number? i thought it wasn't defined as even or odd?
So, we have x - some constant. f(2)=2-2=0.
f(x) between 1 and 2. f(1.5)=1.5-[1.5+1]=1.5-[2.5]=1.5-2=-0.5

7. Nov 13, 2008

### kathrynag

So, between any 2 integers the y value goes from -1 to 1.
So, this leads to something, i think....

8. Nov 13, 2008

### kathrynag

Ok, is there no limit at odd integers? So everywhere else there is a limit?

9. Nov 14, 2008

### HallsofIvy

Staff Emeritus
Yes, that is correct.
f(x) between 1 and 2. f(1.5)=1.5-[1.5+1]=1.5-[2.5]=1.5-2=-0.5[/QUOTE]
but the value at one number tells you nothing about the limit. If x is any number between 1 and 2, then $\lfloor x\rfloor= 1$, an odd number, so $f(x)= x- \lfloor x+1\rfloor= x-2$. What is the limit of that as x goes to 2?

For the limit to exist, it must be the same from both sides. If 2< x< 3, then $\lfloor x\rfloor= 2$, an even number so $f(x)= x- \lfloor x\rfloor= x- 2$. What is the limit of that as x goes to 0?

Finally, don't just guess for even and odd integers. Suppose a-1< x< a where a is an odd number. What does the formula reduce to? Suppose a< x< a+1 where a is an odd number. What does the formula reduce to? Now do the same for a an even number.

That is, by the way, a very nice little problem.

10. Nov 14, 2008

### kathrynag

but the value at one number tells you nothing about the limit. If x is any number between 1 and 2, then $\lfloor x\rfloor= 1$, an odd number, so $f(x)= x- \lfloor x+1\rfloor= x-2$. What is the limit of that as x goes to 2?

For the limit to exist, it must be the same from both sides. If 2< x< 3, then $\lfloor x\rfloor= 2$, an even number so $f(x)= x- \lfloor x\rfloor= x- 2$. What is the limit of that as x goes to 0?

Finally, don't just guess for even and odd integers. Suppose a-1< x< a where a is an odd number. What does the formula reduce to? Suppose a< x< a+1 where a is an odd number. What does the formula reduce to? Now do the same for a an even number.

That is, by the way, a very nice little problem.[/QUOTE]

If x is between 1 and 2, then [x] =1. Then f(x)=x-2. So as x goes to 2, the limit =0. If 2<x<3, then [x]=2. Then f(x)=x-2. So, the limit=0.
a-1<x<a where a is an odd number. Then [a]=a-1. Is this part correct?

11. Nov 15, 2008

### kathrynag

but the value at one number tells you nothing about the limit. If x is any number between 1 and 2, then $\lfloor x\rfloor= 1$, an odd number, so $f(x)= x- \lfloor x+1\rfloor= x-2$. What is the limit of that as x goes to 2?

For the limit to exist, it must be the same from both sides. If 2< x< 3, then $\lfloor x\rfloor= 2$, an even number so $f(x)= x- \lfloor x\rfloor= x- 2$. What is the limit of that as x goes to 0?

Finally, don't just guess for even and odd integers. Suppose a-1< x< a where a is an odd number. What does the formula reduce to? Suppose a< x< a+1 where a is an odd number. What does the formula reduce to? Now do the same for a an even number.

That is, by the way, a very nice little problem.[/QUOTE]

if 1<x<2, [x]=1 Then f(x)=x-2. The limit=0
If 2<x<3 [x]=2 Then f(x)=x-2 The limit =0

a-1<x<a where a is an odd number. [x]=a-1. Then f(x)=x-(a-1). Limit as x goes to a=1
a<x<a+1 where a is an odd number. [x]=a. Then f(x)=x-(a-1+1). Limit as x goes to a= 0
a-1<x<a where a is an even number. [x]=a-1. Then f(x)=x-(a-1+1). Limit=0
a<x<a+1 where a is an even number. [x]=a. Then f(x)=x-a. Limit = 0

So the limit exists for even numbers, it seems.

12. Nov 15, 2008

### kathrynag

Ok, I think I have a better idea. I drew a graph and that helps. It appears there is no limit at the integers obviously because of the holes in the graph and no limit at odd numbers because as x approaches 3 from the left it goes to 1, but as x approaches 3 from the left it goes to -1.
So the limit exists at all real numbers except for the integers and odd numbers.

13. Nov 16, 2008

### HallsofIvy

Staff Emeritus
?? What exactly do you MEAN by "odd numbers"? I was under the impression that even and odd were only defined for integers!

But what "holes" in the graph are you talking about? This function, f, is defined for all real numbers. The limit exists, and is 0, for all even numbers and f(x)= 0 for x any even number.

14. Nov 16, 2008

### kathrynag

Ok, I guess, there aren't holes. I was just thinking odd numbers as going toward an odd number like 3 from each side. I wasn't quite sure how to define that.

15. Nov 16, 2008

### kathrynag

Or could I just say there is no limit at the odd integers, but a limit everywhere else?