$x\sin (1\mathrm{/}x)$ extended by zero is continuous but not differentiable at zero

Given: The function $f$ in the Statement.

For every real $u$, $\mathrm{\mid }\sin u\mathrm{\mid }\le 1$ (Parity and the Pythagorean identity for sine and cosine).

If a function is squeezed near a point between two functions having the same limit there, then it has that limit (If $f\le g\le h$ near $c$ and $f$ and $h$ have the same limit at $c$, then so does $g$).

The quarter-turn values and period give $\sin (\pi \mathrm{/}2+2m\pi )=1$ and $\sin (3\pi \mathrm{/}2+2m\pi )=-1$ for every integer $m$ (Quarter-turn values and shifts by pi/2 and pi, The zero sets of sine and cosine and the least positive common period 2 pi).

For every real $\epsilon >0$, there is a positive integer $N$ with $1\mathrm{/}N<\epsilon$ (For every $\epsilon >0$ in a complete ordered field there is a natural $n\ge 1$ with $1\mathrm{/}n<\epsilon$).

If two punctured-domain sequences approach a limit point while their images approach distinct real limits, then the function has no limit there (A function has no limit at $c$ as soon as two sequences in $A\setminus \{c\}$ tending to $c$ give different limits of the values).

The derivative at zero, if it exists, is the limit of $(f(x)-f(0))\mathrm{/}x$ as $x\to 0$ through nonzero reals (The derivative $f^{\mathrm{\prime }}(c)={\lim }_{x\to c}\frac{f(x)-f(c)}{x-c}$ of $f:A\to \mathbb{R}$ at a point $c\in A$ that is a limit point of $A$, and differentiability on a set).

Sine is continuous; sums and products of continuous functions are continuous; quotients are continuous where their denominators do not vanish; and composites of continuous real functions are continuous (The derivatives of sine and cosine are cosine and minus sine, A function differentiable at $c$ is continuous at $c$, Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function, A composite of continuous functions is continuous, with no side hypothesis of the kind the composition of limits needs).

The number $\pi =2\gamma$ is positive because the first positive cosine zero satisfies $\gamma \in (0,2)$ (Pi as twice the smallest positive zero of cosine, Cosine has a smallest positive zero, lying strictly between zero and two).