# Basic definition of discontinuity

1. Mar 17, 2010

### FlO-rida

i am looking for a simple definition of discontinuity using the example 3/(x^2+x-6)

2. Mar 17, 2010

### rock.freak667

f(x)=3/(x2+x-6)

A simple definition would be that a point of discontinuity on the graph of f(x) occurs at x=a such that f(a) is undefined.

So where would your function be undefined for?

3. Mar 17, 2010

### FlO-rida

at x=2,-3. i understand that but when you say x=a what do u mean by that. can you show me a variable formula (using a, b and c) that would better explain this

Last edited: Mar 17, 2010
4. Mar 17, 2010

### rock.freak667

f(x)=1/(x-a)

Wouldn't you agree that $\lim_{x \rightarrow a} = \infty$ and so f(a) is undefined?

5. Mar 17, 2010

### FlO-rida

so you are saying that a would be the constant. like in my example if we factor x^2+x-6 we get (x-2)(x+3), so how would that be undefined

6. Mar 17, 2010

### rock.freak667

your f(x) is 3/(x-2)(x+3), as x→2, f(x)→∞, thus f(2) is undefined, similarly for f(-3)

7. Mar 17, 2010

### FlO-rida

sorry but i still dont get it, its not like we have a denominator of zero

8. Mar 17, 2010

### rock.freak667

but if x=2, don't you have one?

$$f(x)=\frac{3}{(x-2)(x+3)}$$

9. Mar 17, 2010

### FlO-rida

wat i ment was it dosent cancel out or anything

10. Mar 17, 2010

### FlO-rida

ok then how would you compare that with a function that is not discontinuous like 1/(x+3)

11. Mar 17, 2010

### rock.freak667

cancel out with what?

f(x)=1/(x+3) is discontinuous at x=-3.

Last edited: Mar 17, 2010
12. Mar 17, 2010

### jav

A function is discontinuous at a point c if the limit as x->c of f(x) does not equal f(c).

In this case, consider your c=2. f(c) is not defined. The limit of f(x) as x->2 from the positive side is +infinity. The limit of f(x) as x->2 from the negative side is -infinity.

The first way you can know this function is discontinuous at 2 is that it is not defined at 2. Then no matter what the limit of f(x) is as x->2, it is not equal to f(2) (even if the limit does not exist).

The second way you can know this function is discontinuous at 2 is that the limit of f(x) as x->2 from the positive side does not equal the limit of f(x) as x->2 from the negative side. Therefore the limit does not exist. If the limit does not exist, then it does not equal f(2) (even if f(2) is not defined).

Remember to keep in mind that a function is discontinuous at a point c if the limit as x->c of f(x) does not equal f(c). f(c) could exist, it just happens to not exist in this example.

13. Mar 17, 2010

### jav

Actually the limit of f(x) as x->a from the left is negative infinity, while the limit of f(x) as x->a from the right is positive infinity.

Therefore the limit of f(x) as x->a doesn't exist.

I think you mean x = -3.

14. Mar 17, 2010

### FlO-rida

so does that mean all rational functions are discontiuous.

15. Mar 17, 2010

### jav

Not necessarily, a rational function just means that it can be written as the ratio of two polynomials. But if, for example, the "denominator polynomial" is a constant, then the function is continuous over its domain.

16. Mar 17, 2010

### FlO-rida

do you mean it would be contiuous if the rational function was a polynomial divied by a constant like for ex. (x+1)/3. but wouldnt that give you a linear function

17. Mar 17, 2010

### jav

If you choose your numerator to be a first order polynomial, then yes. But what about f(x) = [(x+1)^2]/3

18. Mar 18, 2010

### FlO-rida

would (x)/(x^2-4) be a discontinuity

19. Mar 18, 2010

### jav

That function would be discontinuous at x=2 and x=-2