Determine existence of a two-sided limit from one-sided pieces

For $x < 0$, $b(x) = -x - 1$, so as $x \to 0^-$, we substitute values close to $0$ from the left: $-x - 1 \to -0 - 1 = -1$. Thus $\lim_{x \to 0^-} b(x) = -1$.

For $x \ge 0$, $b(x) = x^2$, so as $x \to 0^+$, $x^2 \to 0$. Thus $\lim_{x \to 0^+} b(x) = 0$. Since the left-hand limit ($-1$) and right-hand limit ($0$) are not equal, the two-sided limit $\lim_{x \to 0} b(x)$ does not exist.

The correct answer is 'unequal' (or 'mismatch'), referring to the one-sided limits.

The left-hand limit was negative one while the right-hand limit was zero, so the two sides disagreed.

Since a two-sided limit exists only when both one-sided limits agree, the limit does not exist here.

This is the first of the three standard failure cases: unequal one-sided limits.