How to show that there is no condition that can meet two inequalities? Here's an excerpt from Spivak's Calculus, 4th Edition, page 96:

If we consider the function
$$
f(x)= \left\{  \matrix{0, x \text{ irrational} \\
1, x \text{ rational}} \right.
$$
  then, no matter what $a$ is, $f$ does not approach an number $l$ near $a$. In fact, we cannot make $|f(x)-l|<\frac{1}{4}$ no matter how close we bring $x$ to $a$, because in any interval around $a$ there are number $x$ with $f(x)=0$, and also numbers $x$ with $f(x)=1$, so that we would need $|0-l| < \frac{1}{4}$ and also $|1-l| < \frac{1}{4}$.

How do I show that there is no condition that can meet the inequality $|0-l| < \frac{1}{4}$ and also $|1-l| < \frac{1}{4}$ ?
Thank you in advance for any help provided.

When dealing with equations including absolute values, it's often convenient to rewrite them as equations without absolute values.
In your case,
$$|0-l| < \frac{1}{4}$$
gives: 
$$-\frac{1}{4} < l < \frac{1}{4}$$ 
And
$$|1-l| < \frac{1}{4}$$
gives:
$$\frac{3}{4} < l < \frac{5}{4}$$
And now it's easy to spot the contradiction. 
It's easy to sometimes overlook cases when rewriting absolute valued equations, but it's good practice.  
