Let $f(x)=\{\begin{array}{ll}3x+1,& x\le 2\\ x^2-4,& x>2\end{array}$. Which statement about the continuity of $f$ at $x=2$ is correct?

Step-by-step solution

1. Compute the left–hand limit: ${\lim }_{x\to 2^{-}}f(x)=3(2)+1=7$.
2. Compute the right–hand limit: ${\lim }_{x\to 2^{+}}f(x)=2^2-4=0$.
3. Since the two one‑sided limits are not equal, $f$ would be discontinuous, but notice that the definition of $f$ gives $f(2)=3(2)+1=7$, which matches the left limit. The right limit equals $0$, so the function is not continuous at $x=2$.
4. Hence the correct statement is that $f$ has a jump discontinuity at $x=2$.
5. Therefore the correct option is B. $f$ has a jump discontinuity at $x=2$.

Concept tested: Continuity of piecewise functions

Published 4 September 2026 · How this solution was prepared and checked