What you’ll learn:

• The 3 conditions that define continuity at a point
• How to classify discontinuities as removable, jump, or infinite
• Intermediate value theorem (IVT)

Defining continuity with limits

In section 1.1, you used one-sided limits to decide whether an overall limit exists. That same idea leads directly to continuity.

A function is continuous at a point $a$ when:

• the left-hand and right-hand limits agree, and
• that common limit equals the function’s actual value at $a$.

More formally, all three conditions below must be true.

Definition of continuity at point $a$:

1. $f(a)$ is defined

2. $\underset{x\to a}{\lim }f(x)$ exists, meaning

$$\underset{x\to a^{-}}{\lim }f(x)=\underset{x\to a^{+}}{\lim }f(x)$$

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

If any condition fails, the function is discontinuous at $x=a$.

Example

This type of question often shows up on the AP exam:

What value of $k$ will make the piecewise function $f(x)$ continuous everywhere?

$$f(x)=\{\begin{array}{ll}\frac{x^2-4x+3}{x-1}& \text{if }x\ne 1\\ k& \text{if }x=1\end{array}$$

Solution

For $f(x)$ to be continuous at a point $a$, both $f(a)$ and $\underset{x\to a}{\lim }f(x)$ must exist, and they must be equal.

Because $f(x)$ is piecewise, the breakpoint $x=1$ is the only place continuity could fail. So we test continuity at $a=1$.

Condition 1: $f(a)$ is defined.

At $x=1$, the piecewise definition gives $f(1)=k$.

Condition 2: $\underset{x\to 1}{\lim }f(x)$ must exist.

For $x\ne 1$, $f(x)$ follows the rational expression, so we compute

$$\underset{x\to 1}{\lim }\frac{x^2-4x+3}{x-1}$$

$$=\underset{x\to 1}{\lim }\frac{\cancel{(x-1)}(x-3)}{\cancel{(x-1)}}$$

$$=\underset{x\to 1}{\lim }(x-3)$$

$$=-2$$

So the limit exists and equals $-2$.

Condition 3: The limit must equal the function value.

Continuity at $x=1$ requires

$$f(1)=\underset{x\to 1}{\lim }f(x)$$

so

$$k=-2.$$

$$\boxed{k=-2}$$

This choice of $k$ fills the “hole” created by the rational expression.

Classifying discontinuities

The three main types of discontinuities are:

1. Removable discontinuity (hole)

• When $f(x)$passes condition 2 but fails condition 3.
  i.e.$\underset{x\to a}{\lim }f(x)$ exists but is $\ne f(a)$.
  • Removable discontinuities are often referred to as “holes.”

2. Jump discontinuity

• When $f(x)$fails condition 2
  i.e.$\underset{x\to a^{-}}{\lim }f(x)\ne \underset{x\to a^{+}}{\lim }f(x)$ because $f(x)$ “jumps” from one value to another at $x=a$.
  • Usually seen in piecewise functions at the breakpoint(s).

3. Infinite (asymptotic) discontinuity

• When $\underset{x\to a}{\lim }f(x)=\pm \infty$ (so the limit does not exist as a finite number). This means $f(x)$ approaches a vertical asymptote at $x=a$.
  • Often seen in rational functions where factoring and canceling still leaves a constant over a variable denominator, e.g. $f(x)=\frac{1}{x-4}$

Shown below is a visual of each:

If a question asks you to assess the continuity of a function and classify any discontinuities, follow these steps:

Check condition 1: Find potential discontinuities

Typically, discontinuities occur in:

• Rational functions (when the denominator $=0$)
• Logarithmic functions (since the argument must be $>0$)
• Piecewise functions

If $f(a)$ is undefined, then $f(x)$ is discontinuous at $x=a$. The next step is to classify the type.

Even if $f(a)$ is defined (as it often is in a piecewise function), continuity still isn’t guaranteed. Move on to condition 2.

Check condition 2: Classify discontinuity

Find the one-sided limits as $x\to a$.

• If either one-sided limit is $\pm \infty$, it’s an infinite discontinuity.
• If the one-sided limits are finite but don’t match, it’s a jump discontinuity.
• If they match and the (two-sided) limit exists, move on to condition 3.

Check condition 3: Limit $=$ f(a)

Compare the limit from condition 2 with the function value from condition 1.

• If $f(a)$ does not equal the limit, it’s a removable discontinuity.
• If they are equal, the function is continuous at $x=a$.

Examples

1. Classify the discontinuities on $f(x)=\frac{x^2-4x+3}{x^2-x-6}$, if any.

Solution

Answer: Infinite discontinuity at $x=2$, removable discontinuity at $x=3.$

Condition 1: Potential discontinuities

A rational function is undefined where its denominator is 0:

$$\begin{gathered} x^2-x-6=0 \\ (x+2)(x-3)=0 \\ x=-2,3 \end{gathered}$$

So the function is discontinuous at $x=-2$ and $x=3$.

Condition 2: Classify

Check each point using conditions 2 and 3.

• For $a=-2$:

Evaluate

$$\underset{x\to -2}{\lim }\frac{x^2-4x+3}{x^2-x-6}$$

Direct substitution gives $\frac{0}{0}$, so factor and simplify:

$$=\underset{x\to -2}{\lim }\frac{(x-1)\cancel{(x-3)}}{(x+2)\cancel{(x-3)}}$$

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

Now direct substitution gives $\frac{-3}{0}$, which indicates a vertical asymptote. The one-sided limits confirm this:

$$\text{Left: }\underset{x\to -2^{-}}{\lim }\frac{(x-1)}{(x+2)}=\infty$$

$$\text{Right: }\underset{x\to -2^{+}}{\lim }\frac{(x-1)}{(x+2)}=-\infty$$

Because the function grows without bound near $x=-2$, there is an infinite discontinuity at $x=-2$.

• For $a=3$:

Evaluate

$$\underset{x\to 3}{\lim }\frac{x^2-4x+3}{x^2-x-6}$$

Using the same factoring and canceling,

$$\underset{x\to 3}{\lim }\frac{(x-1)}{(x+2)}$$

$$=\frac{2}{5}$$

So the limit exists, which satisfies condition 2.

Condition 3: Matching

Even though the limit exists at $x=3$, the original function still has denominator 0 there, so $f(3)$ is undefined. That means condition 3 fails.

Therefore there is a removable discontinuity at $x=3$ - a hole that would be filled at $y=\frac{2}{5}$.

Let’s explore another example - a piecewise function:

2. Classify any discontinuities on $f(x)$, the piecewise function defined by:

$$f(x)=\{\begin{array}{ll}x+2& \text{if }x<1\\ 1& \text{if }x=1\\ -x+4& \text{if }x>1\end{array}$$

Solution

Intermediate value theorem

One important consequence of continuity is the Intermediate Value Theorem (IVT).

Intermediate value theorem:

If $f(x)$ is continuous on the interval $[a,b]$, and $L$ is any number between $f(a)$ and $f(b)$, then there exists at least one number $c$ in $[a,b]$ such that $f(c)=L$.

In plain language: if a function is continuous on $[a,b]$, it can’t “skip over” values. It must take on every $y$-value between $f(a)$ and $f(b)$ somewhere in that interval.

Examples

1. Let $f(x)=x^2-4$ on the interval $[1,3]$. Use the IVT to show that there is a root in the interval and find the value of $c$ guaranteed by the IVT.

Solution

$f(x)$ is a polynomial, so it’s continuous everywhere, including on $[1,3]$.

A root means $f(x)=0$, which corresponds to crossing the $x$-axis (so here $L=0$).

Evaluate the endpoints:

• $f(1)=1^2-4=-3$
• $f(3)=3^2-4=5$

Because $-3<0<5$, the IVT guarantees at least one $c$ in $[1,3]$ such that $f(c)=0$.

To find $c$:

Solve $f(c)=0$:

$$\begin{gathered} c^2-4=0 \\ c=\pm 2 \end{gathered}$$

Only $\boxed{c=2}$ lies in the interval $[1,3]$.

AP tip:

If a free-response question requires the use of the IVT, explicitly state:

• $f(x)$ is continuous on $[a,b]$, and
• the target value $L$ is between $f(a)$ and $f(b)$,

then conclude that there exists a $c\in [a,b]$ such that $f(c)=L$.

2. Let $f(x)$ be a continuous function defined on the interval $[1,4]$. The values of $f(x)$ at various points are given in the table: | $x$ | $f(x)$ | |:—:|:—:| | $1$ | $2$ | | $2$ | $5$ | | $3$ | $3$ | | $4$ | $7$ |

What is the minimum number of times $f(x)=4$ on the interval $[1,4]$?

Solution

(spoiler)

Answer: $\boxed{3\text{times}}$

Because $f$ is continuous, every time the function goes from below $4$ to above $4$ (or above to below), it must cross the line $y=4$ at least once.

You can apply the IVT on each sub-interval, but it’s often easiest to visualize the situation by sketching a possible graph through the points:

From the table:

• On $[1,2]$, $f(1)=2$ and $f(2)=5$, so $4$ is between them $\Rightarrow$ at least 1 solution.
• On $[2,3]$, $f(2)=5$ and $f(3)=3$, so $4$ is between them $\Rightarrow$ at least 1 solution.
• On $[3,4]$, $f(3)=3$ and $f(4)=7$, so $4$ is between them $\Rightarrow$ at least 1 solution.

So the minimum number of times $f(x)=4$ is 3.

Challenge problem

1. For what values of $m$ and $b$ will $f(x)$ be continuous over all real numbers?

$$f(x)=\{\begin{array}{ll}m{x}^2+b& \text{if }x\le -2\\ 3x-m& \text{if }-2\le x<2\\ x^3-b& \text{if }x\ge 2\end{array}$$

Solution

(spoiler)

Answer: $\boxed{m=-\frac{4}{3}\text{ and }b=\frac{2}{3}}$

Each piece is a polynomial, so it’s continuous on its own interval. The only places continuity can fail are the breakpoints $x=-2$ and $x=2$.

Breakpoint $x=-2$:

Continuity at $x=-2$ requires

$$f(-2)=\underset{x\to -2^{-}}{\lim }f(x)=\underset{x\to -2^{+}}{\lim }f(x)$$

Compute each expression:

First,

$$\begin{array}{rl}f(-2)& =m(-2{)}^2+b\\ & =4m+b\end{array}$$

One-sided limits:

• Left: $\underset{x\to -2^{-}}{\lim }f(x)$

$$\begin{array}{rl}& =m(-2{)}^2+b\\ & =4m+b\end{array}$$

• Right: $\underset{x\to -2^{+}}{\lim }f(x)$

$$\begin{array}{rl}& =3(-2)-m\\ & =-6-m\end{array}$$

Set left and right equal:

$$4m+b=-6-m$$

which simplifies to

$$5m+b=-6.$$

Breakpoint $x=2$:

Continuity at $x=2$ requires

$$f(2)=\underset{x\to 2^{-}}{\lim }f(x)=\underset{x\to 2^{+}}{\lim }f(x)$$

Compute each expression:

First,

$$\begin{array}{rl}f(2)& =(2{)}^3-b\\ & =8-b\end{array}$$

One-sided limits:

$$\underset{x\to 2^{-}}{\lim }f(x)=3(2)-m=6-m$$

$$\underset{x\to 2^{+}}{\lim }f(x)=(2{)}^3-b=8-b$$

Set them equal:

$$6-m=8-b$$

which simplifies to

$$m-b=-2.$$

Now solve the system of equations

$$\begin{array}{rl}5m+b& =-6\\ m-b& =-2\end{array}$$

This gives $m=-\frac{4}{3}$ and $b=\frac{2}{3}$.