# Show that there is no neighborhood of 0 on which h is increasing

1. Feb 10, 2013

### richyw

1. The problem statement, all variables and given/known data

I have the function
$$h(x) = \begin{cases} x^2\sin{\left(\frac{1}{x}\right)}+\frac{x}{2} & \text{if } x \neq 0 \\ 0 & \text{if } x =0 \end{cases}$$

and need to show that $h'(0)>0$ but there is no neighborhood of 0 on which h is increasing

2. Relevant equations

N/A

3. The attempt at a solution

well I can show that $h'(0)=1/2>0$ and if $x\neq 0$ then $$h'(x)=2x\sin{\left(\frac{1}{x}\right)}-\cos{\left(\frac{1}{x}\right)}+\frac{1}{2}$$ I'm a little lost on how I can show that there is no neighborhood of 0 on which h is increasing. This means that $h'(x)>0$ right? I think i'm a bit confused as to what this is asking me I guess.

Last edited: Feb 10, 2013
2. Feb 10, 2013

### Dick

It looks like you meant to type x^2*sin(1/x)+x/2 for the definition of your function. And what you should be thinking is "can I find values of x as close to 0 as I want where h'(x)<0".

Last edited: Feb 10, 2013
3. Feb 10, 2013

### richyw

thanks that was a type. I'm very confused as to what you are saying though.

4. Feb 10, 2013

### Zondrina

Using this, think about where your function is defined if your coming in from -∞.

5. Feb 10, 2013

### richyw

isn't it defined everywhere?

6. Feb 10, 2013

### Dick

I'm saying that if h'(x)<0 then h is decreasing in a neighborhood of x so h isn't increasing there. Take x=1/(2pi). What's h'(1/(2pi))?

7. Feb 10, 2013

### richyw

-1/2. so yes h isn't increasing there. but I don't get how this shows that there is no neighbourhood on 0 where h is increasing. what does that even actually mean?

8. Feb 10, 2013

### Dick

h is increasing in a neighborhood of 0 means h'(x)>=0 in some interval (-a,a). If 1/(2pi) is in that interval then it's not increasing on that interval. What about h'(1/(2pi*n)) where n is some integer? You can make 1/(2pi*n) as small as you want by picking n to be large.

9. Feb 10, 2013

### richyw

so since I can make x=1/(2*pi*n) as small as I want, so x is getting really close to zero. So there has to be like an interval that has 0 as an interior point for there to be a neighborhood of 0. and this neighborhood always contains h'(x)<0 no matter how small I make this interval?

I think that makes sense to me now. I just have no idea how to "show" it

10. Feb 10, 2013

### Dick

Pick an interval (-a,a). How large does n have to be to make 1/(2pi*n)<a?

11. Feb 10, 2013

### richyw

uh n>1/(2*pi*a)?

12. Feb 10, 2013

### richyw

can I just say that the limit of 1/(2*pi*n) as n-> infinity is equal to zero? so any interval with zero as an interior point must also contain 1/(2*pi*n) as long as n is big enough?

anyways thanks for all of your help. I'm going to give up on this one now. I know that math is really important, but it gets to the point of not being fun when I get this stuck. I like getting stuck on physics questions but not this. This is math that has been done before and there is someone who can just show it to me haha.

13. Feb 10, 2013

### Zondrina

Here's a graph of what's going on :
http://www.wolframalpha.com/input/?i=theta^2sin%281%2Ftheta%29+%2B+theta%2F2+from+-1%2F3+to+1%2F3

Here's a graph of the derivative as it tends to zero :
http://www.wolframalpha.com/input/?i=derivative+of+theta^2sin%281%2Ftheta%29+%2B+theta%2F2+from+-1%2F10+to+0

You've shown that h'(0) > 0 by already computing it.

Recall a few posts ago :

The question posed to you :

All he was asking was how large does n have to be for 1/(2pi*n) to be inside the interval (-a,a).

You replied :
Which is absolutely correct. So choosing n > 1/(2pi*a) implies 1/(2pi*n) is inside (-a,a) no matter what interval (-a,a) you choose ( There are infinitely many of them actually ).

So that shows there is a neighborhood around zero such that h'(x) < 0 since choosing n > 1/(2pi*a) implies 1/(2pi*n) is inside (-a,a).

This shows h′(0)>0, but there is no neighborhood of 0 in which h is increasing since there are infinitely many neighborhoods around it we can choose where it isn't.

Also don't quit so easily. Getting frustrated with it only ensures failure in the end.

14. Feb 10, 2013

### Dick

As Zondrina said, you already have gotten it. Time to move onto another problem. Not give up.

15. Feb 10, 2013

### richyw

oh awesome I totally understand it now. thanks for all your help and the motivation to not give up haha. It's easier said than done when there are other deadlines though!