Answer
See the proof below.
Work Step by Step
We shall prove that the limit $\lim_{x \to 0} f(x)$ does not exist via contradiction. Assume, for the sake of contradiction, that $\lim_{x \to 0} f(x) = L$. Since the limit exists, there exists $\delta$ such that $|f(x) - L| < 1/2$ for all $x$ such that $|x| < \delta$. But for such $x$, we must either have $|f(x)-L| = |0-L| = |L|$ or $|f(x) - L| = |1-L|$. Then
$$|L| + |1-L| < 1/2 + 1/2 = 1,$$
which contradicts the triangle inequality, which states that $|L|+|1-L| \geq L + (1-L) = 1$. Thus the limit can not exist.
