Topic 1.13 Notes – Removing Discontinuities

1. Removable Discontinuities and How to Fix Them

A removable discontinuity happens at $x = a$ when:

• $\lim_{x \to a} f(x)$ exists,
• but either $f(a)$ is undefined or $f(a) \neq \lim_{x \to a} f(x)$.

If the limit exists, you can “fix” the function by defining

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

That’s it. You fill the hole with the limit value.

Quick reminder about continuity at $x = a$:

1. $f(a)$ is defined
2. $\lim_{x \to a} f(x)$ exists
3. They are equal

If any one fails, the function is not continuous there.

If the limit does not exist (like a jump or vertical asymptote), redefining one point will not help.

Here’s what a removable discontinuity looks like:

Removable discontinuity (hole) at $x = 1$

The graph follows a straight line, but there’s an open circle at $x = 1$. If we define $f(1) = 2$, the graph becomes continuous.

2. Algebraic Holes in Rational Functions

Most removable discontinuities on tests come from rational functions.

They usually show up when:

• Plugging in gives $\frac{0}{0}$
• The numerator and denominator share a factor like $(x - a)$

That shared factor causes the hole.

Example

$$f(x) = \frac{x^2 - 9}{x - 3}$$

Factor the numerator:

$$x^2 - 9 = (x - 3)(x + 3)$$

So for $x \ne 3$,

$$f(x) = x + 3$$

The simplified function tells you the limit.

$$\lim_{x \to 3} f(x) = 6$$

To remove the discontinuity, define $f(3) = 6$.

What They Like to Ask

• “Find the value of $k$ that makes the function continuous at $x = a$.”
• “Determine whether the discontinuity is removable.”
• “Redefine the function so it is continuous.”

If you see $\frac{0}{0}$, think factor and cancel.

Be careful:

• Only cancel factors, not individual terms.
• A non-canceling zero in the denominator means vertical asymptote, not a hole.

3. Making Piecewise Functions Continuous at a Boundary

Now think about piecewise functions.

The trouble spot is where the formula changes, say at $x = a$.

For continuity there, you must have:

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

All three values must match.

What That Means Practically

• Evaluate the left expression at $a$
• Evaluate the right expression at $a$
• Set them equal
• Make sure that equals the defined value at $a$

Example

$$f(x) = \begin{cases} 2x + 1 & x < 4 \\ kx - 3 & x \ge 4 \end{cases}$$

Left-hand limit at 4:

$$2(4) + 1 = 9$$

Right-hand expression at 4:

$$4k - 3$$

Set them equal:

$$4k - 3 = 9$$

$$4k = 12 \Rightarrow k = 3$$

Now the left limit, right limit, and $f(4)$ all equal 9.

Common mistake on quizzes: students only match one side to $f(a)$ and forget to match both sides to each other.

4. When You Cannot Remove a Discontinuity

You cannot remove:

Jump Discontinuities

Left and right limits exist but are different.

Infinite Discontinuities

The function approaches $\pm \infty$.
This usually happens when a denominator is zero and nothing cancels.

If the limit is not a finite number, redefining one point won’t fix it.

5. Graph Thinking and Recognition

Fast mental checklist:

• Rational function + $0/0$ → likely removable.
• Denominator zero but no cancellation → vertical asymptote.
• Piecewise boundary → compare left, right, and function value.
• Parameter in the problem → you’re solving for continuity.

On no-calculator multiple choice, factoring quickly is key. On FRQs, show the limit reasoning clearly. The graders want to see that you understand why redefining the function works.