Show that limit doesn't exists using $\epsilon, \delta$ 
Let $f : \mathbb{R} \to \mathbb{R}$ be defined by
$$f(x) = \begin{cases}1 & x \text{ rational} \\ 0 & x \text{ irrational} \end{cases}$$
Show that $$\lim\limits_{x \to 0} f(x)$$ does not exist
and if $f$ is defined
$$\lim\limits_{x \to 0} xf(x) = 0$$

my take:
Suppose limit exists and $= 0$, for any $\epsilon >0$ there exists $\delta >0$ so that $|f(x)-L| < \epsilon$ then if $f(x)=1$ we have $|1-L| < \epsilon$ and if if $f(x)=0$ we have $|L| < \epsilon$
Let $\epsilon = \frac{1}4$ (for no reason), then $|1-L| < \frac{1}4$ meaning $\frac{3}4 \le L \le \frac{5}4$ and $|L| < \frac{1}4$, then we have a contradiction, so limit doesn't exist. Did I approach this question correctly??
and finally, how do we show if $f$ is defined, limit of $xf(x) =0$??
 A: Choosing $\epsilon = \frac{1}{4}$ is OK, but you need to show that for all $\delta > 0$, there exists some $|x| < \delta \mid |f(x) - L| \geq \epsilon$. Only one $\delta$ is not sufficient.
To show that the limit does not exist, use the fact that there exists a rational and a irrational inbetween every two different real numbers, and take $\epsilon$ to be within $]0,1[$.
As for the limit $\lim_{x \to 0} xf(x) = 0$, use squeeze theorem with upper bound $x$ and lower bound $0$.
A: Suppose $\lim_{x\rightarrow0}f(x)$ exists. We can find a sequence
$(x_{n})_{n}$ of positive irrational numbers satisfying $x_{n}\rightarrow0$
so that $\lim_{n\rightarrow\infty}f(x_{n})=\lim_{n\rightarrow\infty}0=0$.
Similarly, picking a sequence of positive rational numbers, we get
$\lim_{n\rightarrow\infty}f(x_{n})=1$, a contradiction.
As for the second limit, note that $f(x)$ is bounded. Taking the
$\limsup$ and $\liminf$ of $xf(x)$ as $x\rightarrow0$ gives you
zero, and hence the limit must exist and be equal to zero.
