# Continuity Practice Problems with Answers and Solutions

> 12 continuity practice problems with answers and full solutions: piecewise functions, finding k, types of discontinuity and the Intermediate Value Theorem.

Source: https://practicecalculus.com/practice-problems/continuity
Updated: 2026-11-06

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

## At a glance

- **Problems:** 12, with answers and full solutions
- **Level:** Calculus 1 and AP Calculus AB, Unit 1
- **You need:** One-sided limits and [evaluating limits](/practice-problems/evaluating-limits)
- **Main idea:** f is continuous at a when f(a) exists, the limit exists, and the two are equal
- **Practice with new numbers:** [Continuity lesson](/learn/calculus-1/limits/continuity)

## Key takeaways

- Check three things at $x = a$: $f(a)$ is defined, $\lim_{x\to a} f(x)$ exists, and they are equal.
- For a piecewise function, the only places to check are the break points. Compare the left limit, the right limit and the value there.
- The Intermediate Value Theorem needs continuity on a closed interval. Without it, a sign change does not guarantee a zero.

These **continuity practice problems** cover the three-part definition, piecewise functions, finding a constant that makes a function continuous, classifying discontinuities, domains of continuity and the Intermediate Value Theorem. Try each one first. The answers come right after the list, then the full solutions.

A function $f$ is continuous at $x = a$ when all three hold:

$$f(a) \text{ is defined}, \qquad \lim_{x\to a} f(x) \text{ exists}, \qquad \lim_{x\to a} f(x) = f(a)$$

For more on the idea, read the [continuity guide](/blog/continuity).

## Problems

**Is it continuous?**

1. Is $f(x) = \dfrac{x^2 - 1}{x - 1}$ continuous at $x = 1$? If not, how could you fix it?
2. Is $f(x) = \begin{cases} x^2 + 1 & x < 2 \\ 3x - 1 & x \ge 2 \end{cases}$ continuous at $x = 2$?
3. Is $f(x) = \begin{cases} 2x + 1 & x \le 1 \\ x^2 + 4 & x > 1 \end{cases}$ continuous at $x = 1$?
4. Is $f(x) = \begin{cases} \dfrac{\sin x}{x} & x \ne 0 \\ 1 & x = 0 \end{cases}$ continuous at $x = 0$?

**Make it continuous**

5. Find $k$ so that $f(x) = \begin{cases} kx + 3 & x < 2 \\ x^2 - k & x \ge 2 \end{cases}$ is continuous everywhere.
6. Find $a$ and $b$ so that $f(x) = \begin{cases} 3x + a & x < 1 \\ x^2 + bx & 1 \le x \le 2 \\ 10 & x > 2 \end{cases}$ is continuous everywhere.

**Types of discontinuity and domains**

7. Find and classify the discontinuities of $f(x) = \dfrac{x - 2}{x^2 - 5x + 6}$.
8. Where is $f(x) = \dfrac{\ln(x - 4)}{x - 6}$ continuous?
9. Where is $f(x) = \sqrt{9 - x^2}$ continuous?

**Intermediate Value Theorem**

10. Show that $x^3 + x - 1 = 0$ has a solution between $0$ and $1$.
11. Show that $\cos x = x$ has a solution in $\left(0, \frac{\pi}{2}\right)$.
12. $f(x) = \frac{1}{x}$ has $f(-1) = -1$ and $f(1) = 1$. Does the IVT say $f$ has a zero in $[-1, 1]$?

## Answers

| Problem | Answer |
| --- | --- |
| 1 | No, $f(1)$ is undefined. Removable: define $f(1) = 2$ |
| 2 | Yes, both sides give $5$ and $f(2) = 5$ |
| 3 | No, jump discontinuity (left $3$, right $5$) |
| 4 | Yes |
| 5 | $k = \dfrac{1}{3}$ |
| 6 | $a = 1$, $b = 3$ |
| 7 | Removable at $x = 2$, infinite at $x = 3$ |
| 8 | $(4, 6)\cup(6, \infty)$ |
| 9 | $[-3, 3]$ |
| 10 | $f(0) = -1 < 0$ and $f(1) = 1 > 0$, so a root exists by the IVT |
| 11 | $g(x) = \cos x - x$ changes sign on the interval, so a root exists |
| 12 | No. $f$ is not continuous on $[-1, 1]$, and it has no zero |

## Step-by-step solutions

**1.** At $x = 1$ the bottom is $0$, so $f(1)$ is undefined and $f$ is not continuous there. The limit does exist:

$$\lim_{x\to 1}\frac{(x-1)(x+1)}{x-1} = \lim_{x\to 1}(x + 1) = 2$$

This is a removable discontinuity, a hole at $(1, 2)$. Defining $f(1) = 2$ fills the hole and makes $f$ continuous.

**2.** Check all three conditions:

- Left limit: $2^2 + 1 = 5$.
- Right limit: $3(2) - 1 = 5$.
- Value: $f(2) = 3(2) - 1 = 5$.

All three agree, so $f$ is continuous at $x = 2$.

**3.** The left limit is $2(1) + 1 = 3$ and the right limit is $1^2 + 4 = 5$. The one-sided limits differ, so the limit does not exist. This is a jump discontinuity.

**4.** The value is $f(0) = 1$, and the special limit gives $\lim_{x\to 0}\frac{\sin x}{x} = 1$. They match, so $f$ is continuous at $0$.

**5.** Each piece is a polynomial, so only $x = 2$ needs checking. Set the left limit equal to the right limit:

$$2k + 3 = 4 - k \quad\Rightarrow\quad 3k = 1 \quad\Rightarrow\quad k = \frac{1}{3}$$

**6.** There are two break points, so you get two equations.

At $x = 2$: the middle piece gives $4 + 2b$ and the right piece gives $10$, so $b = 3$.

At $x = 1$: the left piece gives $3 + a$ and the middle piece gives $1 + b = 4$, so $a = 1$.

**7.** Factor the bottom: $x^2 - 5x + 6 = (x - 2)(x - 3)$, so $f$ is undefined at $x = 2$ and $x = 3$.

$$f(x) = \frac{x - 2}{(x - 2)(x - 3)} = \frac{1}{x - 3} \quad (x \ne 2)$$

At $x = 2$ the limit is $\frac{1}{2 - 3} = -1$, so that is a removable discontinuity. At $x = 3$ the simplified form still has a zero bottom, so $f\to\pm\infty$: an infinite discontinuity with a vertical asymptote.

**8.** The log needs $x - 4 > 0$, so $x > 4$. The bottom needs $x \ne 6$. Combining these, $f$ is continuous on $(4, 6)\cup(6, \infty)$.

**9.** The root needs $9 - x^2 \ge 0$, so $-3 \le x \le 3$. The function is continuous on the closed interval $[-3, 3]$, using one-sided continuity at the endpoints.

**10.** Let $f(x) = x^3 + x - 1$. It is a polynomial, so it is continuous on $[0, 1]$. Since $f(0) = -1$ and $f(1) = 1$, the value $0$ lies between them. By the [Intermediate Value Theorem](/theorems/intermediate-value-theorem) there is a $c$ in $(0, 1)$ with $f(c) = 0$.

**11.** Rewrite as $g(x) = \cos x - x = 0$. The function $g$ is continuous everywhere. Check the endpoints:

$$g(0) = 1 > 0, \qquad g\!\left(\tfrac{\pi}{2}\right) = 0 - \tfrac{\pi}{2} < 0$$

The sign changes, so by the IVT there is a $c$ in $\left(0, \frac{\pi}{2}\right)$ with $\cos c = c$.

**12.** No. The IVT needs $f$ to be continuous on the whole closed interval, and $\frac{1}{x}$ is not defined at $0$. The theorem says nothing here, and in fact $\frac{1}{x}$ is never $0$.

## Common mistakes on these problems

- **Checking only the limit.** In problem 1 the limit exists, but the function is still discontinuous because $f(1)$ is missing.
- **Using the wrong piece at the break point.** In problem 2, $f(2)$ comes from the piece with $x \ge 2$. Read the inequality signs carefully.
- **Checking only one break point.** Problem 6 has two break points, and each one gives its own equation.
- **Applying the IVT without continuity.** Problem 12 shows that a sign change alone proves nothing.

## Keep practicing

These 12 problems use fixed numbers. For unlimited practice with new functions and instant feedback, try the [continuity lesson](/learn/calculus-1/limits/continuity) and the [Intermediate Value Theorem lesson](/learn/calculus-1/limits/intermediate-value-theorem). Continuity is also the first hypothesis of the [Mean Value Theorem](/practice-problems/mean-value-theorem), which you can practice next.

## Frequently asked questions

### How do you check if a function is continuous at a point?

Use the three-part test. Make sure $f(a)$ is defined, find the left and right limits and confirm they match, then confirm that common limit equals $f(a)$. If any step fails, $f$ is discontinuous at $a$.

### How do you find k to make a piecewise function continuous?

Set the left-hand limit equal to the right-hand limit at the break point and solve for $k$. In problem 5, $2k + 3 = 4 - k$ gives $k = \frac{1}{3}$.

### What are the types of discontinuity?

A removable discontinuity (a hole) is where the limit exists but does not equal $f(a)$ or $f(a)$ is missing. A jump discontinuity is where the one-sided limits are different. An infinite discontinuity is at a vertical asymptote.