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Let $f$ be defined on an open interval (a,b) and assume $x\in(a,b)$. Consider the two statements

(a) $$\lim_{h\to0} f(x+h)-f(x)=0$$

(b)$$ \lim_{h\to0} f(x+h)-f(x-h)=0$$.

Prove that (a) always implies (b) and give an example in which (b) holds but (a) does not .

My solution : (a) Shows that the function $f$ is continuous at $x$ . So both $$\lim_{h\to0} f(x+h)-f(x)=0$$ and $$\lim_{h\to0} f(x)-f(x-h)=0$$ exists . You add both of the limits and that proves (b) .This proves the first part of the problem .

For second part:

Now lets consider the function $f(x)=1$ when $x=0$ and $f(x)=0$ otherwise .

For this function $\lim_{h\to 0} f(h)-f(-h)=\lim_{h\to0}0-0=0$ but $f$ is not continuous .

I was also thinking about the function $f(x)=0 ,x\in\mathbb{Q}$ and $f(x)=1$ . I think this function too two satisfy the limit in (b) but not (a) as it is not a continuous .

I would just like to clear out my if my thinking process about continuous function is correct or not . Thank you

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The valid implication is perhaps best expressed by noting that if (a) exists, $$ 0 = \left( \lim_{h \to 0} f(x+h)-f(x) \right) - \left( \lim_{h \to 0} f(x-h)-f(x) \right) = \lim_{h \to 0} (f(x+h)-f(x)+f(x)-f(x-h)), $$ which is (b), by using the rule that $ \lim a + \lim b = \lim (a+b)$ if both limits on the left exist.

Any even function that is not continuous at zero will do as a counterexample for the other direction (indeed, you can see the problem in the calculation above: we didn't need to assume that the limit in (a) was zero).

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  • $\begingroup$ All too easy. (+1) $\endgroup$ – Mark Viola Mar 28 '17 at 3:22

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