Let $f(x)$ be
$f(x) = \begin{cases} \space\space\space 0 & \text{if } x \text{ is rational}\\ \space\space\space1 & \text{if } x \text{ is irrational} \end{cases}$
Prove that $\lim_{x\to 0}f(x)$ does not exist.
$\forall \varepsilon \gt 0$ $\exists \delta \gt 0$ s.t
$0 \lt |x - 0| \lt \delta$ $\implies$ $|f(x) - L| \lt \varepsilon$
I tried using contradiction.
Set $\varepsilon \lt \frac 12$. By definition, for any given $\varepsilon$ there exists a $\delta$ such that everything is satisfied, so if I show that there's is not such $\delta$ for a specific $\varepsilon$ the proof is done.
First case: let $x$ approach $0$ from rationals numbers. What I mean is that $x$ is approaching $0$ but "ignoring" any irrational number in the way.
$0 \lt |x| \lt \delta$ $\implies$ $|0 - L| \lt \frac 12$
$|0 - L| = |-L| = |L| \lt \frac 12\iff -\frac 12 \lt L \lt \frac 12$
Second case: let $x$ approach $0$ from irrational numbers, ignoring any rational number in the way.
$0 \lt |x| \lt \delta$ $\implies$ $|1 - L| \lt \frac 12$
$|1 - L| \lt \frac 12 \iff -\frac 12 \lt 1 - L \lt \frac 12 \iff \frac 12 \lt L \lt \frac 32$
There it is. From the first inequality $L \lt \frac 12$, but the second tells that $L \gt \frac 12$. Contradiction. There is no such $\delta$ because $L$ is not a fixed number. Is this valid?
Proof by contradiction:
Assume that the limit exists and equals $L.$ Take $\epsilon=\frac13$ and let $\delta>0$ be such that $|f(x)-L|<\epsilon$ whenever $0<|x-0|<\delta.$
Since the rationals are dense in $\mathbb R$ there is $x_{\text{rational}}\in(-\delta,\delta)\cap\mathbb Q.$ Then $0<|x_{\text{rational}}-0|<\delta$ so we shall have $|f(x_{\text{rational}})-L|<\epsilon,$ i.e. $|0-L|<\frac13.$ This implies that $-\frac13<L<\frac13.$
Likewise, since the irrationals are dense in $\mathbb R$ there is $x_{\text{irrational}}\in(-\delta,\delta)\cap(\mathbb R\setminus\mathbb Q),$ and $0<|x_{\text{irrational}}-0|<\delta$ so we shall have $|f(x_{\text{irrational}})-L|<\epsilon,$ i.e. $|1-L|<\frac13,$ which implies that $\frac23<L<\frac43.$
Thus, $-\frac13<L<\frac13$ and $\frac23<L<\frac43.$ But these two conditions are incompatible. Contradiction!
We can therefore conclude that the limit does not exist.