Suppose $f(0)=0$ and $\lvert f(x)\rvert> \sqrt{\lvert x\rvert}$ for all x. Show that $f^\prime(0)$ does not exist. (Adam's Calculus).
Instead of going fot the definition of differentiation, I tried to show that the limit does not exist at $x=0$. By taking $\epsilon=\sqrt{\lvert h\rvert}$, there exists no $\delta>0$ such that $\lvert f(x)\rvert<\epsilon$. Is this approach correct? I somehow think I need to use the definition of differentiation.
Observe that if $h>0$, then:
$$ \left| \frac{f(h) - f(0)}{h - 0} \right| = \frac{|f(h)|}{h} > \frac{\sqrt{h}}{h} = \frac{1}{\sqrt{h}}$$
This shows that the following limit does not exist
$$\lim_{h \to 0} \frac{f(h) - f(0)}{h - 0}$$
So $f$ is not differentiable at $0$.
You have to show that $ \lim_{x \to 0} \frac{f(x)-f(0)}{x-0}= \lim_{x \to 0} \frac{f(x)}{x}$ does not exist.
To this end assume that the above limit exists and $=a$. Then we have
$$\frac{|f(x)|}{|x|} \to |a|$$
as $x \to 0.$
But for $x \ne 0$ we have
$$\frac{|f(x)|}{|x|} \ge \frac{\sqrt{|x|}}{|x|}= \frac{1}{\sqrt{|x|}},$$
which gives
$$\frac{|f(x)|}{|x|} \to \infty$$
as $x \to 0.$ A contradiction.
