1. Mar 7, 2008

### Ka Yan

If f is a real function on R$$^1$$, and holds:lim [f(x+h)-f(x-h)] = 0 for every x belongs R$$^1$$. Does f continuous?

And I thought it no. Since I considered it mentioned only the left-hand and right-hand limit are equal, but whether or not equal to f(x) was not exactly known.

Will anybody provide me a suggestion? To whether my judgement is true or vice versa.

Thx!

Last edited: Mar 7, 2008
2. Mar 7, 2008

### Dick

Limit as h->0, right? Then your judgement is correct, though maybe not for the reason you think it is. The easy way to prove it's correct is to present a counterexample. Suppose f(x)=0 for x not equal to zero and f(0)=1.

3. Mar 7, 2008

### Shooting Star

To sharpen your thoughts, let me give another example. Consider the function f(x) = x/x. The RHL = LHL everywhere, but the function itself is not defined at x=0 and so not continuous at that point.

4. Mar 8, 2008

### tiny-tim

Hi Ka Yan!

No. Not equal to f(x) (unless f(x) = 0).

Just draw any graph, and look at f(3.1) - f(2.9), f(3.01) - f(2.99), f(3.001) - f(2.999), …

Even the first term, f(3.1) - f(2.9), is nowhere near f(3)!
™​

5. Mar 8, 2008

### Ka Yan

Thanks a lot, gentlemen!

Erm, may I BTW ask one more away-from-the-point question here? That, could anybody, if possible, provide me of any forums where Economics be discussed? Since I have just started the journey of self-learning of the Micro and Macroeconomics, but found questions nowhere to ask, and no one to discuss with.

I'll thank those who could help me for that favor!