## Calculus: Early Transcendentals 8th Edition

$\lim\limits_{x \to 0} f(x)$ does not exist.
$f(x) = 0$, if $x$ is rational $f(x) = 1$, if $x$ is irrational Assume $\lim\limits_{x \to 0} f(x) = c$ Let $\epsilon = \frac{1}{3}$ Choose any value for $\delta \gt 0$ Every interval $[0,\delta)$ contains a rational number $a$ and an irrational number $b$ $\vert a - 0 \vert \lt \delta$ and $\vert b - 0 \vert \lt \delta$ Then $\vert f(a) - c \vert \lt \epsilon = \frac{1}{3}$ and $\vert f(b) - c \vert \lt \epsilon = \frac{1}{3}$ Since $f(a) = 0$, then $-\frac{1}{3} \lt c \lt \frac{1}{3}$ Since $f(b) = 1$, then $\frac{2}{3} \lt c \lt \frac{4}{3}$ Clearly this is a contradiction since $c$ can not be in both of these intervals. Therefore, our assumption that $\lim\limits_{x \to 0} f(x) = c$ is false. Thus $\lim\limits_{x \to 0} f(x)$ does not exist.