Q.Find all the points of discontinuity of the function $f$ defined by $f(x) = \begin{cases} x + 2, & \text{if } x < 1 \\ 0, & \text{if } x = 1 \\ x - 2, & \text{if } x > 1 \end{cases}$.

Concept: Continuity At A Point — A function is continuous at $x = a$ if $\lim_{x \to a} f(x) = f(a)$.

Step 1: For $x < 1$, $f(x) = x+2$ is a polynomial, hence continuous. For $x > 1$, $f(x) = x-2$ is also continuous. Only $x = 1$ needs checking.

Step 2: Compute left-hand limit:

$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x+2) = 3$.

Step 3: Compute right-hand limit: …

The key idea here is continuity at a point. A function $f$ is continuous at $x = a$ if three things match perfectly: the left-hand limit, the right-hand limit, and the actual function value at $a$. If even one of these is different, the function jumps or breaks — that’s a discontinuity.

For a piecewise function like this, the only place where a break can happen is at the boundary where the definition changes. Here, that’s $x = 1$. Everywhere else — for $x < 1$ and $x > 1$ — the function is given by a simple polynomial ($x+2$ or $x-2$), and polynomials are continuous everywhere. So the only candidate for discontinuity is $x = 1$.

Let’s check $x = 1$ step by step.

1. Find the left-hand limit as $x \to 1^-$ When $x$ is just less than 1, the function uses the rule $f(x) = x + 2$. So

2. Find the right-hand limit as $x \to 1^+$ When $x$ is just greater than 1, the rule is $f(x) = x - 2$. So

3. Find the function value at $x = 1$ The definition says directly: $f(1) = 0$.

Now compare:

• Left-hand limit = 3
• Right-hand limit = -1
• $f(1) = 0$

All three are different numbers. For continuity, they must all be equal. Since they are not, the function is discontinuous at $x = 1$. …

Method: Testing a Piecewise Function for Continuity at a Boundary Point

This method applies whenever a function is defined by different formulas on different intervals, and you must determine where (if anywhere) it breaks.

Steps

Step 1: Identify the only candidate points.

Each individual piece (here, linear expressions like $x+2$ and $x-2$) is continuous on the open interval where it applies — polynomials never break on their own. The only places a discontinuity can occur are the boundary points where the rule for $f$ changes.

Step 2: At each boundary point $x=a$, compute the left-hand limit and right-hand limit separately.

Use the formula that applies just to the left of $a$ for the left-hand limit, and the formula that applies just to the right of $a$ for the right-hand limit:

A very common slip is plugging the wrong piece into the wrong side — always double-check which formula governs $x<a$ versus $x>a$. …

Common Mistakes

Mistake 1: Substituting into the wrong branch when computing a one-sided limit.

Why it's wrong: for $\lim_{x\to 1^-}$, values of $x$ are slightly less than 1, so only the piece defined for $x<1$ applies — using the $x>1$ formula (or vice versa) gives a completely wrong limit. Correct approach: always re-read which inequality each piece corresponds to before substituting, and mentally confirm "$x$ is approaching from below/above this piece's domain."

Mistake 2: Stopping once the one-sided limits disagree without stating the full conclusion.

Why it's wrong: once $\lim_{x\to 1^-} f(x) \neq \lim_{x\to 1^+} f(x)$, the two-sided limit fails to exist, so $f$ is discontinuous at that point — a full answer should still say so explicitly rather than leaving the comparison implicit, since board marking schemes award marks for stating the reasoning, not just the numbers. …