# Homework Help: Discontinous functions question.

1. Oct 21, 2012

### peripatein

Hi,
For a function to be continuous, three conditions must be met - the function must be defined at a point x0, its limit must exist at that point, and the limit of the function as x approaches x0 must be equal to the value of the function at x0.
Now, assuming my function is f(x)=[x], which assigns an integer smaller than x to f(x). Thus, if 1≤x<2, f(x)=1.
The instructor claimed that this function is discontinuous for every integer x, which is perfectly clear just from looking at the graph of f(x), which is a step graph. He also mentioned that two of the three requirements for continuity are unmet in this case.
Which two requirements of the three above are unmet in this case?

2. Oct 21, 2012

### Staff: Mentor

For the limit to exist, the left- and right-side limits have to exist and be equal. You're dealing with a step function that has jumps at integer values. Take a look at the three conditions at these integer values.

3. Oct 21, 2012

### peripatein

The instructor claimed that for a limit to exist at a point x0, the limit has to be equal both from above and below, unless the limit from one direction does not exist and then the limit at x0 DOES exist. Hence, that requirement is met for the step function as described above, I believe. Is it not?
Another requirement that is met is that the function is defined for any x0. Is it not?
So two requirements are already met and that alone disagrees with the instructor's claim that two conditions are unmet.
Could someone please clarify? What am I missing?

4. Oct 21, 2012

### Staff: Mentor

$$\lim_{x \to 1^-} f(x)?$$
and
$$\lim_{x \to 1^+} f(x)?$$

Do both exist? If so, are they equal?

5. Oct 21, 2012

### Ray Vickson

No, the function f(x) = [x] does not have a limit as x → n, for every integer n. The limits exist in BOTH directions, however. (Why do these two statements not contradict each other?)

RGV

6. Oct 21, 2012

### peripatein

Mark, is the limit from below equal to 0, whereas the limit from above is equal to 1?

7. Oct 21, 2012

### peripatein

Do both limits indeed exist and whereas the first is equal to 0, the second is equal to 1?

8. Oct 21, 2012

### Ray Vickson

What is f(0.9)? What is f(0.99)? What is f(1.1)? What is f(1.01)? Can you see now what is happening?

RGV

9. Oct 21, 2012

### peripatein

Well, as x approaches 1 from below f(x) is still 0 (or does that limit not exist?). When x approaches 1 from above, f(x) is 1. Is that correct?

10. Oct 21, 2012

### Staff: Mentor

Yes. So
$$\lim_{x \to 1^-} f(x) = 0$$
and
$$\lim_{x \to 1^+} f(x) = 1$$

So does $\lim_{x \to 1} f(x)$ exist?