$f(x)=\begin{cases} x & \text{ if }x\text{ is rational}\\ 0 & \text{ if }x\text{ is irrational} \end{cases}$
Prove that this function is NOT differentiable at $c=0$.
My goal would be to find two sequences where $x_n \rightarrow c$, $x_n \not = c$ and $y_n \rightarrow c$, $y_n \not = c$ where $\frac{f(x_n)-f(c)}{x_n-c}$ and $\frac{f(y_n)-f(c)}{y_n-c}$ approach different numbers.
Well the obvious choice for sequences converging to $0$ would be $x_n=\frac{1}{n}$ and $y_n=-\frac{1}{n}$ which would give me $\frac{\frac{1}{n}-0}{\frac{1}{n}-0}=1$ where for $y_n$ you get $-1$.
Let $\pi_n$ denote the number which agrees with $\pi$ for the first $n$ digits and then has zero decimal expansion, e.g. the sequence will look like $$3, 3.1, 3.14, 3.141, 3.1415, \ldots $$ and consider $y_n = \pi - \pi_n$.