# A function satisfying lim h->0[f(x+h-f(x-h)] = 0, and limit at 0 dne

1. Mar 15, 2013

### DotKite

1. The problem statement, all variables and given/known data
Give an example or show why it is impossible
A function f: R→R satisfying lim h->0[f(x+h-f(x-h)] = 0, and limit at 0 dne
2. Relevant equations

n/a

3. The attempt at a solution

Been trying different piecewise defined functions, with none of them working. I do not really know where to go. I can sit here for hours trying different functions, but i decided to see if anyone on here could help me out.

2. Mar 15, 2013

### jbunniii

 oops, misread the question. will update in a minute.

3. Mar 15, 2013

### Ray Vickson

So, in readable English are you saying that $\lim_{h \to 0} f(x+h)$ does not exist?

4. Mar 15, 2013

### jbunniii

Well, you know you need a function whose limit at zero does not exist. So start with a function which is only defined on $(0, \infty)$ and for which the limit does not exist as $h \rightarrow 0$. Then consider how you can extend this function to $(-\infty, 0)$ in such a way that the bad behavior will cancel out when you form the difference $f(x+h) - f(x-h)$ at $x = 0$.

5. Mar 15, 2013

### DotKite

Would this work,

f(x) = cos(1/x) x ≠ 0
f(x) = 0 , x = 0

6. Mar 15, 2013

### jbunniii

Yes, that works. $f(0+h) - f(0-h) = 0$ for all positive $h$ but $\lim_{x \rightarrow 0} f(x)$ does not exist. You should also explain why $\lim_{h \rightarrow 0} [f(x+h) - f(x-h)] = 0$ for $x \neq 0$.

7. Mar 15, 2013

### DotKite

lim h→0 [f(x+h) - f(x-h)] = 0 because as h→0 we get lim f(x+0) - f(x-0) = lim f(x) - f(x) = 0

right?

8. Mar 15, 2013

### jbunniii

That's true as long as $f$ is continuous at $x$. So the key is that your function is continuous at all $x \neq 0$.