Is the following function differentiable at $x=0$?

Question

I'm having trouble solving the following true/false statement (prove or disprove):

Let $f$ be the following function :

If $\exists n \in N$ s.t. $x = \frac{1}{2^n}$: $$ f(\frac1{2^n}) =\frac{1}{(2^2)^n} $$ Else: $$ f(x) = 0 $$

Then $f$ is differentiable at $x = 0$.

Answer 1

$$\lim_{x\to0}\frac{f(x)-f(0)}x=\begin{cases}\lim\limits_{x\to0,\,x\neq\frac1{2^n}}\cfrac{0-0}x=0\\{}\\ \lim\limits_{x=\frac1{2^n}, x\to0\iff n\to\infty}\cfrac{\frac1{2^{2n}}-0}{\frac1{2^n}}=\lim\limits_{x=\frac1{2^n}, x\to0\iff n\to\infty}\cfrac1{2^n}=0\end{cases}$$

thus the function'sdifferentiable at zero.

Answer 2

It is true. You can calculate $f'(0)$ using the differential quotient.