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I stumbled upon this question in my course, and I am out of ideas. Let $f$ be a periodic function $$f(x)=f(x+l), \qquad l>0$$

Prove that if it is not constant, then $\lim_{x\to 0}f\left(\frac1x\right)$ does not exist.

I don't understand why it's true, let alone how to prove it.

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  • $\begingroup$ Can you prove that if the limit exists, then the function must be constant (and equal to the limit)? $\endgroup$ – Roland Dec 19 '15 at 12:08
  • $\begingroup$ Intuition: Try to picture the map $x\mapsto x^-1$ Where happens to infinity under this map? what happens to small neighbourhoods of 0? Can you see what happens to the function when the argument maps this way? $\endgroup$ – Ranc Dec 19 '15 at 12:53
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Assume the limit $\lim_{x \to 0} f(1/x) = a$ exists. Let $\varepsilon > 0$. Now there exist $\delta > 0$ such that \begin{align} |x| < \delta & \implies |f(1/x) - a| < \varepsilon\,. \end{align} This is equivalent with $$ |y| > 1/\delta \implies |f(y) - a | < \varepsilon\,. $$ Thus for all $y \in (1/\delta, \infty)$ we have $f(y) \in (a - \varepsilon, a + \varepsilon)$.

Because $f(y-l) = f(y)$ this implies that for all $y \in (1/\delta -l, \infty)$ we have $f(y) \in (a- \varepsilon, a + \varepsilon)$. Continuing this shows that $f(y) \in (a - \varepsilon, a + \varepsilon)$ for all $y \in \mathbb{R}$.

Since $\varepsilon$ was arbitrary, we must have $f(y) = a$ for all $y$.

So if the limit exists, $f$ is constant. Thus if $f$ is not constant, the limit does not exist.

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Let $\exists a, b :$ $f(a) \ne f(b)$.

Let $\exists \lim_{x \to 0} f(\frac{1}{x}) = c$.

Let $\epsilon > 0: |f(\frac{1}{x}) - c| < \frac{|f(a) - f(b)|}{2}\ \forall x \in (0, \epsilon).$

But $\frac{1}{a + nl}, \frac{1}{b + nl} \to 0$ if $n \to \infty$.

Then $\exists n: \frac{1}{a + nl}, \frac{1}{b + nl} \in (0, \epsilon).$

Then $$|f(a) - f(b)| = |f(a + nl) - f(b + nl)| =|f(\frac{1}{\frac{1}{a + nl}}) - f(\frac{1}{\frac{1}{b + nl}})| \le $$$$|f(\frac{1}{\frac{1}{a + nl}}) - c| + |f(\frac{1}{\frac{1}{b + nl}}) - c| < \frac{|f(a) - f(b)|}{2} + \frac{|f(a) - f(b)|}{2} = |f(a) - f(b)|. $$

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