# Intermediate Value Property

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1. Jun 27, 2016

### JasMath33

1. The problem statement, all variables and given/known data
'
Here is the given problem

2. Relevant equations

3. The attempt at a solution

a. For part a, I felt it was not continuous because of the sin(1/x) as it gets closer to 0, the graph switches between 1 and -1. Then I felt it might be continuous, therefore I am not sure.

b. For part b, I felt it has the Intermediate Value Property (IVP), because I can do something with the IVT. Those were my thoughts and ideas.

Last edited by a moderator: Jun 27, 2016
2. Jun 27, 2016

### BiGyElLoWhAt

Can you show that the limit as approached positively is different from that as approached negatively? Or that it's different than f(0)? So in other words,
$\lim_{x \to 0} f(x) \neq f(0)$?

For b, I'm not really sure. I thought one of the requirements was that f(x) was continuous...
**
Perhaps this will be of some use.
http://math.stackexchange.com/quest...te-value-property-and-discontinuous-functions
Looks like you need to look at the derivatives near zero.

3. Jun 28, 2016

### NihalRi

For part a you would have to do what BiGyElLoWhAt suggested. For b I believe you would have to prove that the function in either monotone increasing or decreasing. IVP says that for any x value between two other x values, the y value will be in between the y values for the other two x values.

Last edited: Jun 28, 2016
4. Jun 28, 2016

### NihalRi

IVP theorem

Last edited: Jun 28, 2016
5. Jun 28, 2016

6. Jul 2, 2016

### pasmith

If there exist sequences $x_n$ and $y_n$ such that $\lim_{n \to \infty} x_n = \lim_{n \to \infty} y_n = 0$ but $\lim_{n \to \infty} f(x_n) \neq \lim_{n \to \infty} f(y_n)$ then $f$ is not continuous at zero.