Topic 1.13 Notes – Removing Discontinuities

Removable Discontinuities

A removable discontinuity is a hole in the graph at $x = a$.

It happens when:

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

Continuity at $x=a$ requires all three:

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

If #2 fails, you cannot fix it.

Here’s what a removable discontinuity looks like. This graph shows a function that follows the line $y = x + 1$ everywhere except at one missing point.

Removable discontinuity at $x = 1$

Notice the graph behaves like a line, but there’s an open circle at $x=1$. That’s the hole. The limit as $x \to 1$ is 2, but the function is not defined at that point.

How to Remove a Hole in a Rational Function

Most removable discontinuities in AB come from canceling factors.

Suppose:

$$f(x) = \frac{x^2 - 1}{x - 1}$$

First factor:

$$x^2 - 1 = (x-1)(x+1)$$

So for $x \neq 1$,

$$f(x) = x + 1$$

Now the limit:

$$\lim_{x \to 1} f(x) = 2$$

The hole exists because the original function was undefined at $x=1$.
To remove it, define:

$$f(1) = 2$$

That’s it. You don’t change the formula everywhere. You only redefine the single point.

Step-by-step process

1. Factor completely.
2. Cancel common factors.
3. Find the x-value that made the denominator zero.
4. Plug that value into the simplified expression.
5. Redefine the function at that point to equal the limit.

When You Can’t Remove It

• If nothing cancels → vertical asymptote
• If left and right limits differ → jump discontinuity

Only holes can be patched.

This shows up on free-response questions where you must justify continuity using the definition. You’ll need to explicitly state that the limit exists and equals the function value.

Making a Piecewise Function Continuous

Now let’s switch to piecewise functions. These often ask you to solve for a constant that makes the function continuous at a boundary.

Suppose:

$$f(x)= \begin{cases} 3x - 2 & x < 4 \\ k & x = 4 \\ x^2 - 5 & x > 4 \end{cases}$$

Continuity at $x=4$ requires:

$$\text{Left limit} = \text{Right limit} = f(4)$$

Compute the one-sided limits

Left-hand limit:

$$3(4) - 2 = 10$$

Right-hand limit:

$$4^2 - 5 = 11$$

These are not equal, so this function cannot be made continuous by choosing $k$. The limit does not exist.

That’s a subtle AP trap. If the two expressions don’t agree, no constant will fix it.

Now suppose instead the right side were $2x + 2$.

Right-hand limit:

$$2(4) + 2 = 10$$

Now both sides approach 10. So define:

$$k = 10$$

Then the function is continuous.

Graphically, both pieces approach the same y-value at $x=4$, and we choose $k$ to fill in that point:

Piecewise function made continuous at $x=4$ by choosing $k=10$

Step-by-step for solving parameters

1. Identify the boundary point.
2. Compute left-hand limit.
3. Compute right-hand limit.
4. Set them equal.
5. Make that value equal to $f(a)$.
6. Solve for the parameter.

On tests, they often skip directly to “Find the value of $k$ that makes the function continuous.” That’s your cue to set the expressions equal at the boundary.