Q.Find whether the function is continuous or discontinuous at the indicated point: $f(x) = \begin{cases} \dfrac{x^2}{2}, & 0 \le x \le 1 \\ 2x^2 - 3x + \dfrac{3}{2}, & 1 < x \le 2 \end{cases}$ at $x = 1$.

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

Step 1: Compute $f(1)$

Since $x = 1$ falls in the first piece ($0 \le x \le 1$),

$f(1) = \frac{1^2}{2} = \frac12$.

Step 2: Compute left-hand limit

For $x \to 1^-$, use the first piece:

$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} \frac{x^2}{2} = \frac12$.

Step 3: Compute right-hand limit

For $x \to 1^+$, use the second piece: …

The Core Idea: Continuity at a Point

A function is continuous at a point if there is no "break" or "jump" there. For a piecewise function like this, the danger zone is exactly where the definition changes — here, at $x=1$. The function is defined by one rule on $[0,1]$ and another on $(1,2]$. For continuity at $x=1$, three things must match perfectly:

1. The value of the function at $x=1$ (using the first piece, since $1$ is in $0 \le x \le 1$).
2. The limit as $x$ approaches $1$ from the left (using the first piece).
3. The limit as $x$ approaches $1$ from the right (using the second piece).

If all three are the same number, the function is continuous. If even one differs, it's discontinuous.

Step-by-Step Work

1. Find $f(1)$ directly.

Since $x=1$ falls in the first case ($0 \le x \le 1$), we use $f(x) = \dfrac{x^2}{2}$.

2. Compute the left-hand limit as $x \to 1^-$.

When $x$ approaches $1$ from values less than $1$, we are still in the first piece. So:

This matches $f(1)$ — so far so good.

3. Compute the right-hand limit as $x \to 1^+$.

For $x$ just greater than $1$, we use the second piece: $f(x) = 2x^2 - 3x + \frac{3}{2}$.

Substitute $x=1$ directly (the expression is a polynomial, so it's continuous everywhere):

$= 2(1)^2 - 3(1) + \frac{3}{2} = 2 - 3 + \frac{3}{2} = -1 + \frac{3}{2} = \frac{1}{2}$ …

Method: Checking Continuity at the Junction Point of a Piecewise Function

This is the general technique for testing continuity exactly where a piecewise definition switches from one formula to another — the only point where a break could occur, since each formula is continuous (being a polynomial or similarly well-behaved expression) everywhere strictly inside its own interval.

Steps

Step 1: Determine which piece actually defines the function value at the junction point.

Read the inequality signs carefully — the junction point itself belongs to exactly one of the two pieces (e.g. "$\le$" on one side means the boundary point is evaluated using that piece's formula).

Step 2: Compute the left-hand limit using the formula valid immediately to the left of the junction.

Since that formula (typically a polynomial) is itself continuous, the left-hand limit equals the value obtained by direct substitution of the junction point into that formula. …

Common Mistakes

Mistake 1: Using the wrong piece of the definition for one of the one-sided limits.

Why it's wrong: at the junction $x=1$, the left-hand limit must use the formula valid for $x<1$ (or $x\le1$) and the right-hand limit must use the formula valid for $x>1$ — swapping them, or using the same formula for both sides, produces an incorrect comparison. Correct approach: check the inequality that defines each piece carefully before substituting, and use only the formula that is actually valid on that side of the junction.

Mistake 2: Assuming that because the two formulas look different, the function must be discontinuous at the junction. …