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If $\lim_{x \rightarrow 0} f(x)+f(2x)=0$, prove or disprove with example, that $\lim_{x \rightarrow 0} f(x)=0$ for any function $f(x)$.

f(x) can be a piecewise functions as well.

I tried too disprove it considering several functions but I wasn't able to do so. So I guess that there statement is true but how do we prove it?

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    $\begingroup$ Hint: find a non-zero function that satisfies $f(2x)=-f(x)$. A good place to start might be with $\sin x$ since at least $\sin (x+\pi)=-\sin x$. $\endgroup$
    – lulu
    Sep 29, 2019 at 12:32
  • $\begingroup$ We would also require $f(x) \neq 0$ as $x \to 0$ for disproving the statement isn't? I tried all several functions such as (-1)^x *gif(greatest integer function, Signum function, and creating some piecewise function, but for all either the limit doesn't exist or the statement comes out to be true $\endgroup$
    – user600016
    Sep 29, 2019 at 13:02
  • $\begingroup$ Consider $f(x)=\sin\left(\pi\log_2(x)\right)$ $\endgroup$
    – robjohn
    Sep 29, 2019 at 13:04
  • $\begingroup$ @thewitness: define $f(0)=0$. $\endgroup$
    – robjohn
    Sep 29, 2019 at 13:07
  • $\begingroup$ @robjohn I mean $x \to 0^{-} $ for f(x) isn't defined. $\endgroup$
    – user600016
    Sep 29, 2019 at 13:08

1 Answer 1

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As a counterexample, consider the function

$$f(x) = (-1)^n \hspace{14 pt} 2^n \leq |x| < 2^{n+1}$$

for $n\in\mathbb{Z}$. Clearly $f(x) + f(2x)=0$ on $\mathbb{R}-\{0\}$ but $\lim_{x\to 0} f(x)$ does not exist since the function oscillates infinitely fast as you get closer to $0$.

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  • $\begingroup$ The sum of the limits has been assumed to exist in the premise. Does, $\lim_{x \to 0} (f(x) +f(2x))$ exist here? $\endgroup$
    – user0
    Sep 29, 2019 at 13:17
  • $\begingroup$ @DevashishKaushik It does, as I said $f(x)+f(2x)=0$ everywhere except $0$. $\endgroup$ Sep 29, 2019 at 13:19
  • $\begingroup$ Wow this solution is really wonderful. $\endgroup$
    – user600016
    Sep 29, 2019 at 13:25

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