Continuity limit graph help

Please graph this piecewise function and answer the questions that follow.

F(x)= { 1/x if x<1

0 if x=1

1 if x>1

1.)Label any discontinuity and what type of discontinuity.

2.)What values of c does - lim as x approaches c of F(x)- NOT exist?

This is a question from my final test in my calc class. We don't get the test back .

I don't get to hear from the instructor maybe a week from today. I'm SO eager to know the answers and if I did it right. Thanks in advance!

Re: Continuity limit graph help

Hey Mathinik.

You are definitely going to get a discontinuity at x = 1: can you guess why (Hint: continuity implies lim x->a f(x) = f(a): where does this condition fail?)

Re: Continuity limit graph help

I answered, infinite discontinuity at x=0, jump discontinuity at x=1, and a removable discontinuity at x=1. Are those the right answers?

Quote:

Originally Posted by

**chiro** Hey Mathinik.

You are definitely going to get a discontinuity at x = 1: can you guess why (Hint: continuity implies lim x->a f(x) = f(a): where does this condition fail?)

Re: Continuity limit graph help

I'm not sure what removable means but you are right about the other properties.

Re: Continuity limit graph help

Removable or hole discontinuity Quote:

Originally Posted by

**chiro** I'm not sure what removable means but you are right about the other properties.

Re: Continuity limit graph help

Re: Continuity limit graph help

Need help still, from others?

Re: Continuity limit graph help

Quote:

Originally Posted by

**Mathinik** I answered, infinite discontinuity at x=0, jump discontinuity at x=1, and a removable discontinuity at x=1. Are those the right answers?

How can there be both jump discontinuity and removable discontinuity at x = 1? It should be removable. Concerning x = 0, Wikipedia stresses that a point of discontinuity must belong to the function's domain by definition, so x = 0 is not a point of discontinuity. This may not be a universal convention, so you need to check your sources.

Quote:

Originally Posted by

**Mathinik** 2.)What values of c does - lim as x approaches c of F(x)- NOT exist?

c = 0.

Re: Continuity limit graph help

Re: Continuity limit graph help

When I graph the function I also have the point (0,1) and together with 1/x from x<1 it has a jump discontinuity at x=1. X=0 is a point of discontinuity coz it has an asymptote/ break.

C=O DNE, because the function will have an asymptote over there, so it is no continuous over there too. I answered c=0 and c=1(bec of the jump)

Quote:

Originally Posted by

**emakarov** How can there be both jump discontinuity and removable discontinuity at x = 1? It should be removable. Concerning x = 0,

Wikipedia stresses that a point of discontinuity must belong to the function's domain by definition, so x = 0 is not a point of discontinuity. This may not be a universal convention, so you need to check your sources.

c = 0.

Re: Continuity limit graph help

Are there two discontinuity at x=1? Based on the function's domain?

Quote:

Originally Posted by

**Plato** Any other answer will depend upon which textbook you are following.

The discontinuity at

is most often called a

*removable discontinuity*.

The discontinuity at

is called several different names.

One being

*an essential discontinuity*. I have seen authors who make a different name because there is no limit at

as opposed to either

being the limit. So, I were you, I would read the text.

Re: Continuity limit graph help

Quote:

Originally Posted by

**Mathinik** Are there two discontinuity at x=1?

No, there is only one. As I said, what it is called depends upon your textbook.

Jump or removable are two impossibles.

Re: Continuity limit graph help

Quote:

Originally Posted by

**emakarov** How can there be both jump discontinuity and removable discontinuity at x = 1? It should be removable. Concerning x = 0,

Wikipedia stresses that a point of discontinuity must belong to the function's domain by definition, so x = 0 is not a point of discontinuity. This may not be a universal convention, so you need to check your sources.

c = 0.

Quote:

Originally Posted by

**Plato** No, there is only one. As I said, what it is called depends upon your textbook.

Jump or removable are two impossibles.

I answered wrong then, on my test.

Re: Continuity limit graph help

Quote:

Originally Posted by

**Mathinik** When I graph the function I also have the point (0,1) and together with 1/x from x<1 it has a jump discontinuity at x=1.

Do you mean you have (1,0) in the graph? The jump discontinuity is called this way not because there is a jump between the limit of the function (1 in this case) and the value of the function (0 in this case) at that point. The fact that there is a discontinuity already means that either the limit does not exist or it is not equal to the function value. Rather, it is called jump discontinuity because there are left and right limits and there is a jump between those. The value of the function does not matter. In this case, both left and right limits are 1, so this is a removable discontinuity (if f(1) were 1 instead of 0, the function would be continuous at x = 1).

Quote:

Originally Posted by

**Mathinik** C=O DNE, because the function will have an asymptote over there, so it is no continuous over there too. I answered c=0 and c=1(bec of the jump)

c = 0 is correct, but the limit when x -> 1 exists.

Quote:

Originally Posted by **Mathinik**

Are there two discontinuity at x=1?

Quote:

Originally Posted by **Plato**

No, there is only one. As I said, what it is called depends upon your textbook.

Jump or removable are two impossibles.

Do you mean, "two possibilities"?

Re: Continuity limit graph help

I was taught that you don't talk about a discontinuity in points were the function is not defined. As in this example the function is not defined in x = 0. Hence we don't have a discontinuity here.

But at x = 1 we have a discontinuity. The book told us it was a removable discontinuity, and in this point we don't have lim{x->1} = f(1).

For the second question we need to know that

lim{x->a} f(x) = L <=> lim{x->a+} f(x) = lim{x->a-} f(x) = L

to consider the function having a limit L at x = a.

Thus we have no limit att x = 0 and neither at x = 1.