Is $f$ differentiable at $0$, where $f(x) = x$ if $x$ is rational and $f(x) = 0$ otherwise? This is the function:
$$f(x)=\begin{cases}x& \text{if $x$ is rational}\\0 &\text{if $x$ is irrational}\end{cases}$$
My attempt:
It's easy to verify that $f$ is continuous at $x=0$ using the sequential definition of continuity. I claim that $f$ is not differentiable at $x=0$. Assume the contrary and let $f'(0)=L$. Now, we pick an $\varepsilon$ such that $0<\varepsilon < |L|$. For this choice of $\varepsilon$ there is a $\delta >0$ such that if $0<|x-0|<\delta$ then we have $\left| \frac{f(x)-f(0)}{x-0} -L\right| < \varepsilon $. Now, pick $x' \in \mathbb{R}\setminus\mathbb{Q}$ with $0<|x'| <\delta$. Then we have $\left| \frac{f(x')-f(0)}{x'-0} -L\right| = |L| > \varepsilon$. A contradiction! 
Is this proof correct?
 A: EDIT: As @DanielWainfleet pointed out in the comments, your proof does not apply in the case $L=0$. The proof basically goes exactly like the one you have written: first you choose an $0<\varepsilon<1$ small enough, then you take $\delta$ as you please and look at the difference quotient for any $|x|<\delta$. If $f$ was differentiable, then it should hold that 
$$\left|\frac{f(x)-f(0)}{x-0}\right|<\varepsilon$$
but by density of the irrationals, you can find an irrational $x$ such that $|x|<\delta$ and for that one
$$\left|\frac{f(x)-f(0)}{x-0}\right|=\left|\frac{x-0}{x-0}\right|=1>\varepsilon$$
contradicting the assumption.

Another working approach is showing that approaching $0$ via irrationals and via rationals leads to different difference quotients:
Indeed :


*

*choose a sequence of irrational numbers ($a_n=\sqrt{2}/n$) tending to zero. Then $\frac{f(a_n)-0}{a_n-0}=0$ for all $n$

*choose a sequence of rational numbers ($a_n=1/n$) tending to zero. Then $\frac{f(a_n)-0}{a_n-0}=1$ for all $n$
