Q.Examine the continuity at $x = 2$ of the function $f(x) = \begin{cases} x + 1, & x \le 2 \\ 3x - 1, & x > 2 \end{cases}$ and, if discontinuous, state the type.

This is a piecewise function, so compute the left-hand limit, the right-hand limit and $f(2)$ separately.

LHL $=f(2)=3$ (from $x+1$), but RHL $=3(2)-1=5$; the one-sided limits differ, so the limit does not exist — a jump discontinuity.

For a piecewise function the joining point $x = 2$ must be tested with one-sided limits.

Value at the point. Since the rule for $x \le 2$ applies at $x = 2$, $f(2) = 2 + 1 = 3$.

Left-hand limit. As $x \to 2^-$ (values just below $2$), the rule $x + 1$ applies: $\displaystyle\lim_{x\to 2^-} f(x) = 2 + 1 = 3$.

Right-hand limit. As $x \to 2^+$ (values just above $2$), the rule $3x - 1$ applies: $\displaystyle\lim_{x\to 2^+} f(x) = 3(2) - 1 = 5$.

Compare. The left-hand limit is $3$ and the right-hand limit is $5$. They are both finite but not equal, so the two-sided limit $\displaystyle\lim_{x\to 2} f(x)$ does not exist. Condition 2 of the continuity test fails.

Therefore $f$ is discontinuous at $x = 2$. Since both one-sided limits are finite but unequal, this is a jump (finite) discontinuity; the size of the jump is $|5 - 3| = 2$.

Check (dual-solve): note that $f(2)=3$ matches the left-hand limit, so $f$ is left-continuous at $2$ but not right-continuous (RHL $=5 \neq 3$). A function continuous at a point must be both; failing right-continuity alone already forces discontinuity — consistent with the jump classification.

A common error is to substitute $x=2$ into both pieces and average them, or to use only the piece $x\le 2$ and wrongly conclude continuity because $f(2)=3$ matches the left side. Continuity needs BOTH one-sided limits to equal $f(2)$; here the right-hand limit is $5$, so the function jumps and is discontinuous regardless of the left-side match.