Understanding the Continuity of Real Functions on R^1

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For a real function f on R^1, the condition lim [f(x+h)-f(x-h)] = 0 for every x does not guarantee continuity at x. The discussion highlights that while the left-hand limit and right-hand limit may be equal, this does not confirm that they equal f(x). Counterexamples, such as f(x)=0 for x≠0 and f(0)=1, illustrate this point effectively. Participants emphasize that continuity requires the limits to equal the function value at that point. The thread also includes a request for forums on Economics, indicating a shift in the discussion topic.
Ka Yan
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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!
 
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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.
 
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.
 
Ka Yan said:
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.

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)! :smile:
™​
 
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!
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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