Today I ran into the following problem:
Let $f$ be a function such that $$ f(x) = \begin{cases} \frac{1}{4^n}, &x=\frac{1}{2^n}, \;\; n\in \{1,2,3,\ldots\} \\ 0, &x\neq\frac{1}{2^n} \end{cases} $$ Is the function $f$ differentiable at $0$?
So far, I have set the limit to show that the function is differentiable as:
$$ \lim_{h\to0} \frac{1}{h} f(h) $$
I have a hunch that tells me that this limit is zero, as $f(h)$ seems to approach to zero faster than $h$. However, I have no idea how to prove it.
Note that $f(h)$ is either $0$ or $h^{2}$. Hence $\frac {f(h)} h$ is either $0$ or $h$. Of course this tends to $0$ as $ h \to 0$ so $f'(0)=0$.