Continuity at a point

What you will be able to do

• test continuity at a point with the left-hand limit, the right-hand limit and the value;
• name the three ways a function can fail to be continuous;
• repair a removable break by giving the function the right value at one point.

The pencil test

Draw the graph of $y = x^2$ without lifting your pencil. Now try $y = \dfrac{1}{x}$. At $x = 0$ the graph has no point, so you must lift the pencil and start again on the other side. Continuity turns this picture into a test with numbers.

Limits from the left and from the right

The point $x = a$ can be approached from two sides. The left-hand limit, written $\displaystyle\lim_{x \to a^-} f(x)$, uses values of $x$ below $a$. The right-hand limit, written $\displaystyle\lim_{x \to a^+} f(x)$, uses values of $x$ above $a$. The limit $\displaystyle\lim_{x \to a} f(x)$ exists only when both one-sided limits exist and are equal.

The definition

A function $f$ is continuous at $x = a$ when three things hold.

1. $f(a)$ is defined.
2. The limit exists, so the left-hand limit equals the right-hand limit.
3. The limit equals the value: $\displaystyle\lim_{x \to a} f(x) = f(a)$.

If any one of the three fails, $f$ is discontinuous at $a$.

Three ways to fail

• A removable break (a hole). The limit exists, but $f(a)$ is missing or has the wrong value. Take $f(x) = \dfrac{x^2 - 1}{x - 1}$ for $x \ne 1$. For $x \ne 1$ it equals $x + 1$, so both one-sided limits at $1$ are $2$. The value $f(1)$ is not defined. Giving $f(1)$ the value $2$ repairs the break.
• A jump. Both one-sided limits exist, but they are different. Take $f(x) = \dfrac{|x|}{x}$. It is $-1$ for every $x < 0$ and $1$ for every $x > 0$. The jump is $1 - (-1) = 2$, and no choice of $f(0)$ can close it. The greatest integer function $[x]$ jumps by $1$ at every integer.
• An infinite break. One or both one-sided limits are not finite. Take $f(x) = \dfrac{1}{x - 2}$. Near $x = 2$ the values grow without bound, upwards on the right and downwards on the left.

A fourth kind exists. The function $\sin\dfrac{1}{x}$ swings between $-1$ and $1$ faster and faster near $0$, so it has no limit at all. A-Level, IB and AP texts also call the first kind a point discontinuity.

A worked example

Let $f(x) = \dfrac{x^2 - 9}{x - 3}$ for $x \ne 3$, and $f(3) = k$. Find $k$ so that $f$ is continuous at $x = 3$.

1. Simplify. For $x \ne 3$, $x^2 - 9 = (x - 3)(x + 3)$, so $f(x) = x + 3$.
2. Find the limit. As $x \to 3$ from either side, $x + 3 \to 6$. The left-hand and right-hand limits are both $6$.
3. Match the value. Continuity needs $f(3) = 6$, so $k = 6$.

Practise with new numbers

Try it in Python

The program looks at three functions from both sides of a point. It prints the values of each function a little to the left and a little to the right, with the distance $h$ getting smaller.

Common mistakes

• Checking only one side. A limit exists only when the left-hand and right-hand limits agree.
• Forgetting the value. The limit can exist and $f$ can still be discontinuous, because $f(a)$ is missing or different.
• Cancelling and stopping. After cancelling a factor, the simplified formula holds for $x \ne a$ only. The value at $a$ is a separate question.
• Treating a jump as repairable. A jump cannot be removed by changing one value.

Remember these

• Continuous at $a$ means: $f(a)$ is defined, the limit exists, and the limit equals $f(a)$.
• The limit exists when the left-hand limit equals the right-hand limit.
• The three breaks are removable (limit exists, value wrong or missing), jump (one-sided limits differ) and infinite.
• Only a removable break can be repaired.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Find the left-hand and right-hand limits of $f(x) = \dfrac{|x - 3|}{x - 3}$ at $x = 3$. Is $f$ continuous there?

2. Let $f(x) = \dfrac{x^2 - 25}{x - 5}$ for $x \ne 5$ and $f(5) = 7$. Is $f$ continuous at $5$?

3. What kind of break does $f(x) = \dfrac{1}{x + 4}$ have at $x = -4$?

4. Find the left-hand and right-hand limits of $[x]$ at $x = 3$.

5. True or false: if $\displaystyle\lim_{x \to a} f(x)$ exists, then $f$ is continuous at $a$.

References

• OpenStax, Calculus Volume 1, Chapter 2, section on continuity.
• NCERT, Mathematics Part I for Class XII, Chapter 5, section on continuity.