Q.Find all points of discontinuity of $f$, where $f$ is defined by $f(x) = \begin{cases} x^3-3, & \text{if } x \leq 2 \\ x^2+1, & \text{if } x > 2 \end{cases}$

Concept: Continuity At A Point — A function is continuous at $x = a$ if $\lim_{x \to a} f(x) = f(a)$. For a piecewise function, the critical point is the boundary where the definition changes.

Here the boundary is $x = 2$. We check the left-hand limit, right-hand limit, and the function value at $x = 2$.

Step 1: Compute $f(2)$. Since $x \leq 2$ uses the first piece,

$f(2) = 2^3 - 3 = 8 - 3 = 5$.

Step 2: Find the left-hand limit as $x \to 2^-$ (using $x^3 - 3$):

$\lim_{x \to 2^-} f(x) = 2^3 - 3 = 5$.

Step 3: Find the right-hand limit as $x \to 2^+$ (using $x^2 + 1$): …

The key idea: a piecewise function can only be discontinuous at the "break" points where the definition changes. Here, the only candidate is $x=2$. Everywhere else, $f$ is a polynomial — and polynomials are continuous on their entire domain. So the entire question reduces to checking what happens at $x=2$.

Why check limits? Continuity at a point $x=a$ means three things must hold simultaneously:

1. $f(a)$ is defined.
2. $\lim_{x\to a} f(x)$ exists.
3. That limit equals $f(a)$.

For a piecewise function, the limit exists only if the left-hand limit and right-hand limit are equal. So we compute both sides at $x=2$.

1. Find $f(2)$. Since $x=2$ falls in the first piece ($x \leq 2$), we use $f(x)=x^3-3$.

2. Left-hand limit as $x\to 2^-$. For $x<2$, the function is $x^3-3$, a polynomial. Polynomials are continuous, so the limit is just the value at $x=2$:

3. Right-hand limit as $x\to 2^+$. For $x>2$, the function is $x^2+1$, also a polynomial. So:

4. Compare. Left-hand limit = $5$, right-hand limit = $5$, and $f(2)=5$. All three are equal. Therefore the limit exists and equals $f(2)$. …

Method: Comparing Two Formulas at Their Shared Boundary

This is the standard approach for any function built from two different expressions meeting at a single switch-over point.

Steps

Step 1: Identify which formula defines the boundary point itself

Look at the inequality signs in the definition — whichever piece's inequality includes "$=$" is the one that determines $f(a)$ at the boundary; the other piece is only used for the limit approaching from its side.

Step 2: Compute $f(a)$ from the correct piece

Substitute $a$ into the formula identified in Step 1.

Step 3: Compute the left-hand and right-hand limits separately …

Common Mistakes

Mistake 1: Using $x^2+1$ instead of $x^3-3$ to compute $f(2)$

Why it's wrong: the boundary condition is $x \leq 2$, so $x=2$ belongs to the first piece ($x^3-3$), not the second. Correct approach: identify exactly which inequality includes the equals sign before evaluating $f(a)$, and use only that piece.

Mistake 2: Stopping once the left-hand limit matches $f(2)$, without also checking the right-hand limit …