Let $f:[0,\infty)\to\mathbb{R}$ be defined by: $$\begin{cases}&x\sin(\frac{1}{x}) \, \, &\text{if}\, \, x > 0\\ & 0, &\text{if} \, \, x = 0\end{cases}$$

Show that $f$ is continuous on $[0,\infty)$ and differentiable on $(0,\infty)$. Also, show that $f$ has no local maximum or minimum in the endpoint $x = 0$ of the domain of $f$.

I can manage to prove continuity on a single point using the epsilon-delta technique, although the intervals here were a surprise, how do you go about proving such a thing? And any hints about the second part of the problem would also be appreciated.

  • $\begingroup$ What does the phrase "local maximum or minimum in the endpoint" mean? $\endgroup$ – BigbearZzz Dec 8 '18 at 10:10
  • $\begingroup$ According to the definition in the book am using, "A local maximum (minimum) can occur as an end point of the domain of $f$: a point $x$ in the domain of $f$ that does not belong to an open interval lying completely inside the domain". Yeah am equally as confused too :) $\endgroup$ – kareem bokai Dec 8 '18 at 10:19
  • $\begingroup$ I don't quite understand what you want to prove here. Perhaps you want to show that $x=0$ is neither a local maximum nor a local minimum? $\endgroup$ – BigbearZzz Dec 8 '18 at 10:21
  • $\begingroup$ Yeah, I think the way they formulated the question is horrible, we just have to show that $x = 0$ is neither a local max or min $\endgroup$ – kareem bokai Dec 8 '18 at 10:25

An alternative way to show that $f$ is continuous at $x=0$ is to use the sandwich theorem with $$ -|x| \le f(x) \le |x|, $$ where $\pm|x|\to 0 $ as $x\to 0$. Differentiability of $f$ on $(0,\infty)$ is obvious.

For the second part, we want to show that for any $\varepsilon>0$, the point $x=0$ is neither a maximum nor minimum on $[0,\varepsilon)$. Choose $n$ large enough so that $\frac 1{2\pi n+\pi/2}<\varepsilon$, then $$ f\left(\frac 1{2\pi n+\pi/2}\right) = \frac 1{2\pi n+\pi/2}>0 = f(0) $$ so $x=0$ is not a maximum. You should be able to modify this a bit to prove that $x=0$ is not a minimum either. (Hint: we want $\sin(1/x)=-1$)


It'v very easy to show that for any bounded function $g$, defined on a neighbourhood of $0$, the function $f(x)=x\cdot g(x)$ is continuous at $x=0$.

For the differentiability of $f$ notice that $(f(x)-f(0))/(x-0)=g(x)$. Now if $\lim_{x\to0}g(x)$ doesn't exist, $f$ isn't differentiable at $x=0$.


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