Topic 1.10 Notes – Exploring Types of Discontinuities

What continuity at a point requires

A function $f$ is continuous at $x = a$ only if all three of these are true:

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

That’s the definition. On FRQs, you justify continuity by checking these explicitly.

A few reminders about limits:

• The limit exists only if $$\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x)$$ and both are finite numbers.
• If the function heads toward $\pm\infty$, the limit does not exist as a finite value.

If even one of the three conditions fails, the function is discontinuous at $x=a$. The type of failure tells you the type of discontinuity.

Removable discontinuity

This happens when the limit exists, but the function value doesn’t match it (or isn’t defined).

So:

• $\lim_{x \to a} f(x)$ exists
• But either $f(a)$ is undefined or $f(a) \ne \lim_{x \to a} f(x)$

Here’s what that looks like on a graph:

Removable discontinuity at $x=2$

At $x=2$, the limit is 4, but the function value is 1. The graph has a hole at the limit value.

Common causes:

• A rational function where a factor cancels, like $$\frac{(x-3)(x+1)}{x-3}$$ (after canceling, it behaves like $x+1$, except at $x=3$)
• A piecewise function that assigns a different value at one point

It’s called removable because you could redefine $f(a)$ to equal the limit and make the function continuous.

Jump discontinuity

This happens when the left- and right-hand limits are different.

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

That means the limit does not exist.

Look at the piecewise example below.

Jump discontinuity at $x = 1$

As $x \to 1^-$, the graph approaches 1. As $x \to 1^+$, it approaches 3. Since those one-sided limits are different, the overall limit does not exist.

There’s a visible vertical gap between the two sides. Even if one of the points is filled in, it doesn’t matter. If the one-sided limits don’t agree, continuity fails at condition 2.

These almost always come from piecewise functions where the formulas don’t line up at the breakpoint.

Discontinuity due to a vertical asymptote

This occurs when at least one one-sided limit approaches infinity.

Examples:

• $\lim_{x \to a^-} f(x) = \infty$
• Both sides approach $-\infty$
• One side → $\infty$, the other → $-\infty$

Here’s a classic example with a vertical asymptote at $x = 2$:

$y = \frac{1}{x - 2}$

As $x$ approaches 2, the graph shoots upward on one side and downward on the other without bound. Because the limit is not finite, condition 2 fails.

Common sources:

• Denominator equals zero and does not cancel
• Logarithmic functions at domain edges

On AP problems, you’ll often detect this algebraically before even graphing.

How to justify continuity on a test

When a question says “justify your answer,” use the definition language.

For continuity:

• State that $f(a)$ exists.
• Show left- and right-hand limits are equal.
• Conclude that the limit equals $f(a)$.

For discontinuity:

• Clearly identify which condition fails.
  • “The left- and right-hand limits are not equal, so the limit does not exist.”
  • “The limit exists, but it does not equal $f(a)$.”

Just saying “there’s a hole” won’t earn full credit. The rubric looks for reasoning from the definition.