Vrftjkry

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} -Lright| < varepsilon $. Now, pick $x' in mathbb{R}setminusmathbb{Q}$ with $0<|x'| <delta$. Then we have $left| frac{f(x')-f(0)}{x'-0} -Lright| = |L| > varepsilon$. A contradiction!

Is this proof correct?

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 $n$