Defining $\epsilon$ in terms of $|x|$ Is it valid to define $\epsilon$ in terms of $|x|$ ?
Context:
Let $f(x) =
\begin{cases}
a, &\text{x rational}\\
x, &\text{x irrational}
\end{cases}$
$\hspace{1cm}$
Show that $\lim \limits_{x \to y} f(x)$ doesn't exist, where $y \neq a$ 
Suppose otherwise,
$$\begin{array}
\f \forall \delta \hspace{0.5cm} |x - y| < \delta &\Rightarrow |a - l| < \epsilon &&\wedge \qquad |x - l| < \epsilon \\
&\Rightarrow |l| < |a| - \epsilon &&\wedge \qquad  |l| < \epsilon - |x| \\
&\Rightarrow |a| - \epsilon  < \epsilon - |x| \\
&\Rightarrow |a| + |x| < 2\epsilon \\
\end{array}$$
Let $\epsilon = \dfrac{|a| + |x|}{2}$
$$\Rightarrow |a| + |x| < |a| + |x|$$
Contradiction.
P.S. How do you add vertical space with MathJax? The \vspace command does not seem to work. 
 A: I think this is similar to your proposal, but set out a little differently.  Apologies if too pedantic.  I don't know about \vspace for MathJax.
Suppose otherwise so that for a given $ y \neq a $ we have $ \lim_{x \to y} = l $.  Then for every $ \varepsilon > 0 $ there exists $ \delta > 0 $ with the property that whenever  $ x $ satisfies $ | x - y | \leq \delta ~ (*) $, we have $ | f(x) - l | \leq \varepsilon $.  But there always exists both a rational, $ x_r $ and an irrational $ x_i $ that satisfies $ (*) $ with function values $ a $ and $ x_i $ respectively, and so $ |a - l| \leq \varepsilon $ and $ | x_i - l | \leq \varepsilon $.
Now $ x_i $ can be chosen as close to $ y $ as we like, so we may a well fix on a value that satisfies the stronger requirement $ | x_i - y | \leq \min( \varepsilon, \delta ) $.
We then have three inequalities.  The value $ \varepsilon $ can be chosen freely, so the inequality $ | a - l | \leq \varepsilon $ implies $ a = l $.  Use the triangle inequality with the other two to obtain $ | l - y | \leq 2\varepsilon $, and by the same reasoning we must also have $ y = l $.
Thus $ y = a $ which contradicts our requirement $ y \neq a $.
