Is $f(x) = \frac{x^2 - 4}{x - 2}$ continuous at $x = 2$? If not, name the discontinuity.

Not continuous: a removable discontinuity at $(2, 4)$.

Where is $f(x) = \frac{x + 1}{x^2 - x - 6}$ discontinuous, and what kind of discontinuity is each?

Infinite discontinuities at $x = 3$ and $x = -2$.

Is $f(x) = \begin{cases} 2x + 1, & x < 2 \\ x^2 - 1, & x \ge 2 \end{cases}$ continuous at $x = 2$?

No: a jump discontinuity at $x = 2$, from $5$ down to $3$.

Grade 12 · MCV4U · Calculus & Vectors

MCV4U: Continuity

A function is continuous at $x = a$ if you can draw its graph through that point without lifting your pencil. Calculus needs a version you can check with algebra, and limits give it.

How It Works

$f$ is continuous at $x = a$ when all three of these hold:

1. $f(a)$ is defined.

2. $\lim_{x \to a} f(x)$ exists, which means the left-hand and right-hand limits are equal.

3. $\lim_{x \to a} f(x) = f(a)$ .

Each condition that fails gives a different kind of break.

Removable discontinuity (a hole). The limit exists but $f(a)$ is missing or sits somewhere else. This is what happens when a factor cancels, like $\frac{x^2 - 4}{x - 2}$ at $x = 2$ .

Jump discontinuity. The left and right limits are both finite but different, so the graph steps up or down. Piecewise functions are where you meet these.

Infinite discontinuity. The function grows without bound near $a$ , which is a vertical asymptote. You get one where a denominator is zero and the numerator is not.

You only need to test the suspicious points. Polynomials are continuous everywhere, and a rational function is continuous everywhere except where its denominator is zero. For a piecewise function, each piece is usually a polynomial, so the only places to check are the break points where the rule changes.

At a break point, use the piece for $x < a$ to find the left-hand limit and the piece for $x > a$ for the right-hand limit. Then check which piece owns $x = a$ itself, from the $\le$ or $\ge$ sign, to find $f(a)$ .

When a question asks for the constant that makes a piecewise function continuous, set the two one-sided limits equal and solve. With two unknowns you need two break points, which gives two equations.

5 Worked Examples

Easy first, test level last. Try each one on paper, then open the solution.

Easy
Is $f(x) = \frac{x^2 - 4}{x - 2}$ continuous at $x = 2$ ? If not, name the discontinuity.
Show solution
1. Condition 1 fails: $f(2) = \frac{0}{0}$ is undefined.
2. The limit still exists: $\lim_{x \to 2}\frac{(x - 2)(x + 2)}{x - 2} = \lim_{x \to 2}(x + 2) = 4$ .
3. So the graph is the line $y = x + 2$ with a hole at $(2, 4)$ .
Answer: Not continuous: a removable discontinuity at $(2, 4)$ .
Easy
Where is $f(x) = \frac{x + 1}{x^2 - x - 6}$ discontinuous, and what kind of discontinuity is each?
Show solution
1. A rational function is continuous except where its denominator is zero: $x^2 - x - 6 = (x - 3)(x + 2) = 0$ at $x = 3$ and $x = -2$ .
2. The numerator is $4$ at $x = 3$ and $-1$ at $x = -2$ . Neither is zero, so nothing cancels.
3. A nonzero number over zero means the function is unbounded: vertical asymptotes.
Answer: Infinite discontinuities at $x = 3$ and $x = -2$ .
Medium
Is $f(x) = \begin{cases} 2x + 1, & x < 2 \\ x^2 - 1, & x \ge 2 \end{cases}$ continuous at $x = 2$ ?
Show solution
1. Left-hand limit, from the $x < 2$ piece: $2(2) + 1 = 5$ .
2. Right-hand limit, from the $x \ge 2$ piece: $2^2 - 1 = 3$ .
3. $5 \ne 3$ , so $\lim_{x \to 2} f(x)$ does not exist. $f(2) = 3$ is defined, but condition 2 fails.
Answer: No: a jump discontinuity at $x = 2$ , from $5$ down to $3$ .
Medium
Find $k$ so that $f(x) = \begin{cases} kx + 3, & x < 2 \\ x^2 - k, & x \ge 2 \end{cases}$ is continuous everywhere.
Show solution
1. Each piece is a polynomial, so the only point to check is $x = 2$ .
2. Left-hand limit: $2k + 3$ . Right-hand limit and $f(2)$ : $4 - k$ .
3. Set them equal: $2k + 3 = 4 - k$ , so $3k = 1$ and $k = \frac{1}{3}$ .
4. Check: both sides give $\frac{11}{3}$ .
Answer: $k = \frac{1}{3}$
Test level
Find $a$ and $b$ so that $f(x) = \begin{cases} ax - 1, & x < 1 \\ x^2 + b, & 1 \le x \le 2 \\ bx + 3, & x > 2 \end{cases}$ is continuous everywhere.
Show solution
1. Two break points, so two equations.
2. At $x = 1$ : the left piece gives $a - 1$ and the middle piece gives $1 + b$ , so $a - 1 = 1 + b$ .
3. At $x = 2$ : the middle piece gives $4 + b$ and the right piece gives $2b + 3$ , so $4 + b = 2b + 3$ and $b = 1$ .
4. Substitute into the first equation: $a - 1 = 2$ , so $a = 3$ .
5. Check: at $x = 1$ both sides are $2$ , and at $x = 2$ both sides are $5$ .
Answer: $a = 3$ , $b = 1$

Learn It on Martian Lab

Martian Lab teaches this in the Piecewise Functions lesson of its MCV4U Limits unit. Watch it, then take the Limits practice tests to see if it stuck.

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MCV4U: Continuity

How It Works

5 Worked Examples

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