Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**fusilli_jerry89****Member**- Registered: 2006-06-23
- Posts: 86

Fine the values of x for which f(x) = (x-2)/(x²-4) is discontinuous and label these discontinuities as removable or unremovable.

I just simple do not get what these words means, removable, unremovable and discontinuous. I can guess but I wanna be sure.

*Last edited by fusilli_jerry89 (2006-10-25 12:08:02)*

Offline

**polylog****Member**- Registered: 2006-09-28
- Posts: 162

Discontinuous at a point means: not continuous at that point, ie, this condition is not true:

In your example, see what happens at x = 2.

I'm not sure at this moment how to label these as removable or not, someone else will certainly post soon.

*Last edited by polylog (2006-10-25 13:19:01)*

Offline

**fgarb****Member**- Registered: 2006-03-03
- Posts: 89

A function is discontinuous at some point if there is a break in it at that point - this is mathematically represented by what polylog wrote. If there is a break at a point then you won't be able to find a limit for the function at that point, or in some cases you can but it won't be the same as the function evaluated at the point. Any point at which a function goes to infinity is guaranteed to be discontinuous.

The function listed here actually has two singularities, but they aren't both removable. This definition may not be formally correct, but the way I think about it a singularity is removable if you can "remove" it and make your function continuous by redefining the value of your function at that one point. For example,

[align=center]

[/align]technically has a singularity at x = 0, but you can "remove it" if you redefine f = 1 only at x = 0. In this case f is continuous at x = 0. You can logically see that it's smooth there, and if you formally evaluate the limit of sin(x)/x at x = 0 you will find it has 1 as its limit (this can be shown either through L'Hopital's rule or a Taylor expansion). A similar substitution can be made at one but not both of the singularities in the function you're trying to solve, but you should try a little more to think this through yourself before we give the answer away. Let us know if you're still stuck!

Offline

**fusilli_jerry89****Member**- Registered: 2006-06-23
- Posts: 86

so can you get the points that are discontinuous by simply finding what x cannot equal? For f(x) = (x-2)/(x² -4), and got 2(removable because it's smooth) and -2(non-removable because it goes off into infinite). Is this right or am I misunderstanding you guys?

Offline

**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Yup, that's right.

You get the undefined points when the denominator is 0, so that's +2 and -2.

At x=-2, the numerator is non-zero and so the function will be undefined.

But at x=2, the numerator is also 0 and so f(2) will have a finite limit, so the discontinuity can be removed.

Why did the vector cross the road?

It wanted to be normal.

Offline

Pages: **1**