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I have a periodic function $f$ with period $1$. I want to know if this limits exists $$ \lim_{h\to 0} f(x+h)+f(x-h) $$ implies that both left and right limit exist?

In other words, if limit of $f(x+h) + f(x-h) $ exists as $h \to 0 $ then $f(x+) $ and $f(x-) $ (which is limit of $f(x+h) $ and $f(x-h) $ as $h \rightarrow 0$) exists.

Powder
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KYJ
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  • No, you can not guarantee the existence of one-sided limits. Consider a function where the one sided limit doesn’t exist (at $x=0$, say). Try to modify the function to make the written limit exist, but by construction, the one-sided limits don’t exist. – Clayton Nov 20 '19 at 16:02
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    Look at $f(y)=\begin{cases}\sin(1/y),&x\neq0\0,&x=0\end{cases}$ on $[-1/2,1/2]$ and repeat it periodically. Take $x=0$. then $\lim_{h\to0}(f(h)+f(-h))=\lim_{h\to0}(\sin(1/h)-\sin(1/h))=0$. However, $\lim_{h\to0^+}f(h)=\lim_{h\to0^+}\sin(1/h)$ doesn't exist. – conditionalMethod Nov 20 '19 at 16:03
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    In fact, we can have $f(x+h)+f(x-h) = 2f(x)$ for each $x$ and each $h,$ which clearly implies "$\lim_{h \rightarrow 0} [f(x+h)+f(x-h)]$ exists for each $x$", and still have neither left nor right limit exist anywhere. In fact, the graph of such a function can even be dense in the plane, which is way, way worse than nonexistence at each point of a unilateral limit. See Graph of discontinuous linear function is dense. – Dave L. Renfro Nov 20 '19 at 17:10
  • If you guys are going to completely answer the question, why not do it as an answer so that the question doesn't appear as though it's unanswered hours later? – Daniel McLaury Nov 20 '19 at 22:53

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Define $$f(x)=\frac{1}{x}\forall x\in\Big(-\frac{1}{2},\frac{1}{2}\Big)\setminus\{0\}$$ $$f(0)=0$$ Extend this function over $\mathbb R$ periodically as desired. Observe that given limit is $$\lim_{h\to 0}\Big(\frac{1}{h}+\frac{1}{-h}\Big)=0$$ but neither left nor right limit exists finitely.

Martund
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