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A classic!
Problem 2554 from Demidovich's "Problems in Mathematical Analysis"
Here's the problem:
Prove that the function
$$f(x) = \begincases x \sin \frac1x, & x \neq 0 \ 0, & x = 0 \endcases$$
is differentiable at $x=0$ and find $f'(0)$.
Solution
To prove differentiability at $x=0$, we must show that demidovich calculus
$$\lim_h \to 0 \fracf(0+h) - f(0)h = \lim_h \to 0 \fracf(h)h$$
exists.
For $h \neq 0$,
$$f(h) = h \sin \frac1h$$
So,
$$\fracf(h)h = \sin \frac1h$$
Now,
$$\lim_h \to 0 \sin \frac1h$$
does not exist, but
$$\left| \sin \frac1h \right| \leq 1$$
for all $h \neq 0$. Hence,
$$\lim_h \to 0 \fracf(h)h = 0$$
exists and equals $0$. Therefore, $f'(0) = 0$.
The function $f(x)$ is differentiable at $x=0$, and $f'(0) = 0$.
However, $f(x)$ is not continuously differentiable at $x=0$ since $f'(x)$ does not exist for $x \neq 0$ or is not continuous at $x=0$ in a certain sense;
we could add more!
Week 1 — Foundations & limits
Week 2 — Continuity & monotonicity
Week 3 — Derivatives & applications
Week 4 — Integration & techniques
Week 5 — Sequences and series of functions
Week 6 — Advanced techniques & inequalities
Week 7 — Multivariable basics
Week 8 — Synthesis & proofs
You cannot compare Demidovich to standard textbooks like Stewart or Thomas. Those are texts with problems attached. Demidovich is a problem bank with no hand-holding.
Ask any survivor for their least favorite Demidovich problem, and they will name one. Common candidates: