Classifying the discontinuity of a piecewise function at x = 0

Since $x \le 0$ uses the rule $x^3$, we have $b(0) = 0^3 = 0$, and $\lim_{x\to 0^-} b(x) = \lim_{x\to 0^-} x^3 = 0$. For the right-hand limit, since $x>0$ uses $x^2+1$, we get $\lim_{x\to 0^+} b(x) = 0^2+1 = 1$.

The left-hand limit is $0$ and the right-hand limit is $1$; since these are unequal finite numbers, $\lim_{x\to 0} b(x)$ does not exist, even though both one-sided limits exist. This means condition two of the continuity test fails in the specific way where the two sides disagree, so $b(x)$ has a jump discontinuity at $x=0$.

No. In part a, the discontinuity at zero was classified as a jump discontinuity, since the left-hand and right-hand limits are unequal finite numbers, meaning the two-sided limit does not exist at all. Redefining b(0) only changes the function's value at that point, but it cannot make the left- and right-hand limits agree with each other. This fix only works for removable discontinuities, where the limit exists but simply does not match the function value.