Q.Find whether the function is continuous or discontinuous at the indicated point: $f(x) = \begin{cases} 3x + 5, & x \ge 2 \\ x^2, & x < 2 \end{cases}$ at $x = 2$.

Concept: Continuity at a point — check whether $f(2)$, the left-hand limit, and the right-hand limit all agree.

Function value: since $x\ge 2$ uses $3x+5$, $f(2)=3(2)+5=11$.

Left-hand limit ($x\to 2^-$, use $x^2$): $\lim_{x\to 2^-}x^2=4$.

Right-hand limit ($x\to 2^+$, use $3x+5$): $\lim_{x\to 2^+}(3x+5)=11$.

Since $4\ne 11$, the two one-sided limits differ, so $\lim_{x\to 2}f(x)$ does not exist and continuity fails.

The idea

For $f$ to be continuous at $x=2$ we need three things to match: the value $f(2)$, the limit coming from the left, and the limit coming from the right. The rule changes exactly at $x=2$, so that boundary is the only place trouble can appear.

Step-by-step

1. Function value at $x=2$. The condition $x\ge 2$ selects the piece $3x+5$:

2. Left-hand limit. For $x<2$ the function is $x^2$:

3. Right-hand limit. For $x\ge 2$ the function is $3x+5$:

4. Compare. The left limit is $4$ and the right limit is $11$. Because

the two-sided limit $\lim_{x\to 2}f(x)$ does not exist. With no limit, the continuity test fails no matter what $f(2)$ is — the graph jumps from height $4$ up to $11$ at $x=2$.

Method: Testing Continuity of a Piecewise Function at the Boundary Point

Use this method whenever a function is defined by different formulas on either side of a specific point, and you must decide whether it is continuous there.

Steps

Step 1: Determine which piece defines $f(a)$

Check which inequality includes the equality sign at the boundary point — that piece is the one used to compute $f(a)$ itself.

Step 2: Compute the left-hand limit

Using the formula that applies for $x$ values strictly less than $a$, evaluate $\lim_{x\to a^-} f(x)$ by direct substitution (assuming that piece is itself a continuous function like a polynomial).

Step 3: Compute the right-hand limit

Using the formula that applies for $x$ values greater than (or equal to, depending on how the pieces are split) $a$, evaluate $\lim_{x\to a^+} f(x)$ similarly.

Step 4: Compare all three values

If the two one-sided limits disagree, stop there — the two-sided limit does not exist, so the function is discontinuous regardless of what $f(a)$ equals (a jump discontinuity). If they agree with each other but not with $f(a)$, that's a different kind of discontinuity (removable/misplaced-point). Only if all three agree is the function continuous.

Common Mistakes

Mistake 1: Assuming continuity just because $f(a)$ is defined

A student might see that $f(2)=11$ is a perfectly valid number and stop there, concluding the function must be continuous simply because it has a value at that point. Why it's wrong: being defined at a point is only one of the three continuity conditions — the one-sided limits must also exist and agree with that value. Correct approach: always compute both one-sided limits explicitly before drawing any conclusion, even when $f(a)$ looks unremarkable.

Mistake 2: Using the wrong piece for one of the one-sided limits

Because the boundary condition is $x \ge 2$ (not $x>2$), it's tempting to also use the $3x+5$ piece when computing the left-hand limit as $x\to 2^-$. Why it's wrong: the left-hand limit must use only the formula valid for $x$ strictly less than $2$, which here is $x^2$ — mixing up which piece belongs to which side directly causes a wrong (and possibly misleadingly "matching") answer. Correct approach: re-read the domain conditions carefully before choosing which formula to use for each one-sided limit.