Here's a problem on a function that has non-zero derivative at the point $0,$ but is not monotone in any interval around the point $0$. I'm stating the problem and presenting my solution to it.
I'd greatly appreciate if someone checks the solution and tell me if there's any gap in my arguments. Also, is there any method/trick that may enable me produce a shorter solution? Thank you.
The Problem : Let $f:\mathbb{R} \to \mathbb{R}$ be given by $f(x)= \begin{cases} x+2x^2sin\Big(\frac{1}{x}\Big) & x\neq 0 \\ 0 & x=0 \end{cases} $
To show that $f'(0)=1,$ but $f$ is not monotonic in any interval around $0$.
My Solution : We see that
$\lim_{x \to 0} \frac{f(x)-f(0)}{x-0}=\lim_{x \to 0;~x \neq 0} \frac{f(x)}{x}=\lim_{x \to 0;~x \neq 0} \frac{x+2x^2sin\big(\frac{1}{x}\big)}{x}=1+2\lim_{x \to 0}xsin\Big(\frac{1}{x}\Big)$
We first find out $\lim_{x \to 0}xsin\Big(\frac{1}{x}\Big)$. Let $\epsilon>0.$ Choose $\delta=\epsilon$. Then $$x \in (-\delta,\delta)\setminus\{0\} \implies \left| xsin\Big(\frac{1}{x}\Big)-0 \right| \leq |x| < \delta=\epsilon$$ Since $\epsilon>0$ is arbitrary, $\lim_{x \to 0}xsin\Big(\frac{1}{x}\Big)=0$. And hence,
$\lim_{x \to 0} \frac{f(x)-f(0)}{x-0}$ exists and is equal to $1+(2 \times 0)=1$. Thus $f'(0)=1$.
Now, let $\alpha>0$ be arbitrary. By Archimedean Property $2,$
$\exists N \in \mathbb{N}$ such that $\frac{1}{N}<2\pi\alpha,$ i.e. $\frac{1}{2\pi N}<\alpha$.
Choose $x_1=\frac{1}{2\pi N},~x_2=\frac{1}{2\pi N+\frac{\pi}{2}},~x_3=\frac{1}{2\pi N+\frac{3\pi}{2}}$. Then, $x_1>x_2>x_3$. Now,
$f(x_1)=\frac{1}{2\pi N}+2\Big(\frac{1}{2\pi N}\Big)^2sin\Big(2\pi N\Big)=\frac{1}{2\pi N}$
$f(x_2)=\frac{1}{2\pi N+\frac{\pi}{2}}+2\Big(\frac{1}{2\pi N+\frac{\pi}{2}}\Big)^2sin\Big(2\pi N+\frac{\pi}{2}\Big)=\frac{1}{2\pi N+\frac{\pi}{2}}+2\Big(\frac{1}{2\pi N+\frac{\pi}{2}}\Big)^2$
$f(x_3)=\frac{1}{2\pi N+\frac{3\pi}{2}}+2\Big(\frac{1}{2\pi N+\frac{3\pi}{2}}\Big)^2sin\Big(2\pi N+\frac{3\pi}{2}\Big)=\frac{1}{2\pi N+\frac{3\pi}{2}}-2\Big(\frac{1}{2\pi N+\frac{3\pi}{2}}\Big)^2$
Claim : $f(x_3)<f(x_1)<f(x_2)$
We see that $f(x_3)=\frac{1}{2\pi N+\frac{3\pi}{2}}-2\Big(\frac{1}{2\pi N+\frac{3\pi}{2}}\Big)^2 < \frac{1}{2\pi N+\frac{3\pi}{2}} < \frac{1}{2\pi N} = f(x_1)$
Suppose $f(x_1) \geq f(x_2)$. Then, \begin{align} f(x_1) \geq f(x_2) &\iff \frac{1}{2\pi N} \geq \frac{1}{2\pi N+\frac{\pi}{2}}+2\Big(\frac{1}{2\pi N+\frac{\pi}{2}}\Big)^2\\ &\iff \frac{1}{2\pi N} - \frac{1}{2\pi N+\frac{\pi}{2}} \geq 2\Big(\frac{1}{2\pi N+\frac{\pi}{2}}\Big)^2\\ &\iff \frac{\frac{\pi}{2}}{(2\pi N)(2\pi N+\frac{\pi}{2})} \geq 2\Big(\frac{1}{2\pi N+\frac{\pi}{2}}\Big)^2\\ &\iff \frac{\frac{\pi}{2}}{2\pi N} \geq \frac{2}{2\pi N+\frac{\pi}{2}}\\ &\iff \frac{2\pi N+\frac{\pi}{2}}{2\pi N} \geq \frac{2}{\frac{\pi}{2}}\\ &\iff 1+\frac{1}{4N} \geq \frac{4}{\pi}\\ &\iff \frac{1}{4N} \geq \frac{4}{\pi}-1=\frac{4-\pi}{\pi}\\ &\iff 4N \leq \frac{\pi}{4-\pi} < 4\\ &[\text{If not, let} \frac{\pi}{4-\pi} \geq 4 \iff \pi \geq 16-4\pi \iff 5\pi \geq 16 \iff \pi \geq 3.2, \text{ contradiction.}]\\ & \implies N<1 \end{align}
And we reach a contradiction. Hence $f(x_1)<f(x_2)$.
Thus the claim is true. We have $x_1,x_2,x_3 \in (0,\alpha)$ such that $x_1>x_2>x_3$ and $f(x_3)<f(x_1)<f(x_2)$. We conclude that $f$ is not monotone in $(0,\alpha),$ and hence in $(-\alpha,\alpha)$.
Since $\alpha>0$ is arbitrary, $f$ is not monotone in any interval around $0$. $\blacksquare$
One tiny little detail, when you write $$\lim_{x\to 0, x\neq 0}$$
the second part under the limit, i.e. $x\neq 0$, is unnecesary. When calculating $\lim_{x\to a} f(x)$, we always look only at values $f(x)$ for $x\neq a$. That's the whole point of a limit:
(note the $0<|x-a|$ demands that $x\neq a$.
Also, I would prove that $f(x_1)<f(x_2)$ simply by showing that $$f(x_2)-f(x_1) = \frac{N\left(4\pi - \pi^2\right) - \frac{\pi^2}4}{\text{something}^2}$$
and that this number, because $4\pi - \pi^2 > 0$, is obviously positive for a large enough value of $N$.
Other than that, the proof is very well written and as far as I can see it, perfectly correct. Well done!