Lipschitz continuity of $x\cdot\sin(1/x)$

Question

I have to proof, that $$f(x) = \begin{cases} x\cdot\sin(\frac{1}{x}) & x \neq 0 \\ 0 & x = 0 \\ \end{cases} $$ isn't Lipschitz continuous.

I have started with

$|f(x)-f(y)|$ and tried to get to $ \leq L \cdot | x-y| $ but without any success.

$|(1+\frac{y}{x-y})\cdot\sin(\frac{1}{x}) - \frac{y}{x-y}\cdot\sin(\frac{1}{y})| \cdot |x-y|$ was best rearranging, but I don't know at this point how to simplify the $\sin$

I'm a mathematics student in first semester, so I am not allowed to use differential calculus.

Hints or solutions are welcome :)

Answer 1

You should try to show that the ratio $$ \frac{|f(x) - f(y)|}{|x - y|} $$ has no upper bound for $x,y \in \Bbb R$. In particular, consider taking $$ x = \frac{1}{2 \pi n + \pi/2}\\ y = \frac{1}{2 \pi n + 3\pi/2} $$ for an arbitrary integer $n$.

Answer 2

you know that the graph of $y=f(x)$ is contained between the lines $y=±x$ and that at every inverse odd multiples of $\pi/2$ it hits either of the boundaries. at these points $dx=O(1/n^2)$ and $dy=O(1/n)$ so the slope of the sedan line $dy/dx$ goes to infinity. you may want to use the MVT to see that there are points with these slopes on the graph at intermediate points.