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Calculus BC · Unit 1 · 1.10
AP Calculus · Limits and Continuity
Exploring Types of Discontinuities
Four kinds, told apart by one question: does the two-sided limit exist?
key idea$\text{Classify by WHY the three-part test fails}$
The categories
Continuity at $x=c$ needs three things at once: $f(c)$ exists, $\lim_{x\to c}f(x)$ exists, and the two are equal. A discontinuity is named by which of those fails — so classify by cause, not by how the graph looks at a glance.
[Figure: four panels showing removable, jump, infinite and oscillating discontinuities]
The four kinds side by side. Only the first is removable, because it is the only one where the two-sided limit exists.Telling them apart
| Type | Does $\lim_{x\to c}f$ exist? | One-sided limits | Why it fails | Removable? |
|---|
| Removable | yes | agree, both finite | $f(c)$ is missing or is the wrong value | yes — redefine $f(c)$ |
| Jump | no | both finite but differ | the two sides disagree | no |
| Infinite | no | one or both are $\pm\infty$ | the function blows up — vertical asymptote | no |
| Oscillating | no | neither settles | values keep cycling without approaching anything | no |
The second column does the work: if the two-sided limit exists, the discontinuity is removable; if it does not, the reason it does not gives you the name.
| Spotted in the algebra | Almost always means |
|---|
| a factor cancels, e.g. $\dfrac{x^{2}-4}{x-2}$ | removable at the cancelled root |
| a factor stays in the denominator, e.g. $\dfrac{1}{x-2}$ | infinite at that root |
| a piecewise definition with different one-sided values | jump at the seam |
| $\sin\!\left(\tfrac1x\right)$ or $\cos\!\left(\tfrac1x\right)$ near $0$ | oscillating at $0$ |
A quick triage before you compute anything.
Worked examples
EX 1
Classify from the formula
$\dfrac{x^{2}-9}{x-3}$Classify the discontinuity of $f(x)=\dfrac{x^{2}-9}{x-3}$ at $x=3$.
1
Check $f(3)$
Substituting gives $\tfrac00$, so $f(3)$ is undefined. Part 1 of the test already fails.
2
Does the limit exist?
Factor: $\tfrac{(x-3)(x+3)}{x-3}=x+3$ for $x\neq3$. So $\lim_{x\to3}f(x)=6$ — the limit exists and is finite.
3
Name it
The limit exists but $f(3)$ does not, so the discontinuity is removable.
4
Remove it
Define $f(3)=6$. The extended function is continuous everywhere.
EX 2
Same-looking expression, different answer
$\dfrac{x+3}{x-3}$Classify the discontinuity of $g(x)=\dfrac{x+3}{x-3}$ at $x=3$.
1
Check $g(3)$
Substituting gives $\tfrac{6}{0}$ — undefined, and not the indeterminate $\tfrac00$.
2
Does the limit exist?
Nothing cancels. As $x\to3^{+}$ the denominator is a small positive, so $g\to+\infty$; as $x\to3^{-}$ it is a small negative, so $g\to-\infty$.
3
Name it
The magnitude is unbounded, so this is an infinite discontinuity, with a vertical asymptote at $x=3$.
4
Contrast with Example 1
The two expressions differ only in the numerator. Whether the offending factor cancels is what separates removable from infinite.
EX 3
A piecewise seam
jumpLet $h(x)=\begin{cases} x+1, & x\lt 2 \\ x^{2}-1, & x\ge 2\end{cases}$. Classify the behaviour at $x=2$.
1
Left limit
Using the top piece: $\lim_{x\to2^{-}}h(x)=2+1=3$.
2
Right limit
Using the bottom piece: $\lim_{x\to2^{+}}h(x)=2^{2}-1=3$.
3
Compare
Both one-sided limits equal $3$, so the two-sided limit exists and equals $3$.
4
Check the value
$h(2)=2^{2}-1=3$ from the second piece, which matches the limit.
5
Conclude
All three conditions hold — $h$ is continuous at $x=2$. Not every piecewise seam is a jump; you have to check.
EX 4
Make it removable on purpose
solve for $k$For which $k$ does $p(x)=\begin{cases}\dfrac{x^{2}-x-6}{x-3}, & x\neq3 \\ k, & x=3\end{cases}$ become continuous at $x=3$?
1
Recognise the type
For $x\neq3$ the numerator factors as $(x-3)(x+2)$, so the factor cancels — a removable discontinuity.
2
Compute the limit
$\lim_{x\to3}\tfrac{(x-3)(x+2)}{x-3}=\lim_{x\to3}(x+2)=5$.
3
Match the value to the limit
Continuity requires $p(3)=\lim_{x\to3}p(x)$, so $k=5$.
4
Why only removable types allow this
You can only patch a hole when the limit exists. No choice of $k$ fixes a jump, an infinite or an oscillating discontinuity.
Watch out
“Undefined” does not name the type
All four types can leave $f(c)$ undefined. What distinguishes them is the behaviour of the limit, so always evaluate the one-sided limits before naming anything.
A hole in the graph is not always removable-looking
A removable discontinuity can also occur where $f(c)$ is defined but equals the wrong value — the graph shows a filled point sitting off the curve, plus an open circle on it.
Infinite discontinuity vs. vertical asymptote
They coincide here, but the terms differ: the asymptote is the line $x=c$; the discontinuity is the behaviour of $f$ at $c$. Answer the question that was asked.
Oscillating is rare but examinable
$\sin\!\left(\tfrac1x\right)$ at $x=0$ has bounded values yet no limit. Bounded does not imply convergent.