Exploring Types of Discontinuities
Introduced| English |
|---|
| discontinuity/dɪskɒntɪˈnjuːɪti/ |
| removable discontinuity/rɪˈmuːvəbl dɪskɒntɪˈnjuːɪti/ |
| jump discontinuity/dʒʌmp dɪskɒntɪˈnjuːɪti/ |
| infinite discontinuity/ˈɪnfɪnət dɪskɒntɪˈnjuːɪti/ |
| vertical asymptote/ˈvɜːtɪkl ˈæsɪmptəʊt/ |
Where a graph "breaks"
- A continuous curve can be drawn without lifting your pen. A discontinuity 间断点 is a break in that flow.
- Not all breaks are the same — there are three distinct types, each with its own graph signature.
- Naming the type tells you instantly whether the break can be repaired.
- Let's meet all three: removable, jump, and infinite.
Removable: a single missing point
- A removable discontinuity 可去间断点 is just a hole in an otherwise smooth curve.
- The limit exists (both sides meet at one height), but $f(c)$ is missing or plotted elsewhere.
- $f(x)=\dfrac{x^2-4}{x-2}$ has a hole at $x=2$: the limit is $4$, yet $f(2)$ is undefined.
- Because the limit exists, you could "fill the hole" — hence removable (lesson 1.13).
At $x=3$, $f(x)=\dfrac{x^2-9}{x-3}$ has a limit of $6$ but $f(3)$ undefined. The discontinuity is...
A limit exists but the value is missing — a hole, i.e. removable.
Jump: the sides disagree
- A jump discontinuity 跳跃间断点 happens when the left and right limits are both finite but different.
- The graph leaps from one height to another — think of a step or a tax bracket.
- Here $\displaystyle\lim_{x\to c}f(x)$ does not exist, because the one-sided limits disagree.
- No single point value can bridge a gap of nonzero width, so a jump is not removable.
A curve that blows up
y = a / (x − b)
A reciprocal shoots to $\pm\infty$ near where its denominator hits zero — that is an infinite discontinuity and a vertical asymptote.
A jump discontinuity can be removed by redefining the function at the single point.
The one-sided limits differ, so no single value bridges the gap — jumps are not removable.
A step function is $1$ for $x<0$ and $4$ for $x\ge0$. What is the size of the jump at $x=0$?
Right minus left: $4-1=3$.
Infinite: off to $\pm\infty$
- An infinite discontinuity 无穷间断点 is where the function grows without bound near $c$.
- The graph rockets up or down along a vertical asymptote — as in $f(x)=\dfrac1{x-2}$ at $x=2$.
- The limit is infinite (or DNE), so this break is also not removable.
- Signature: a denominator heading to zero while the numerator does not.
An infinite discontinuity appears on a graph as a vertical ____.
The function grows without bound along the vertical asymptote.
Which discontinuity type is the only one where the two-sided limit exists?
Removable = a hole where both sides agree, so the limit exists.
Select all functions with an infinite discontinuity at the marked point.
The third is removable (limit $=2$). The others blow up to $\pm\infty$.
Only the removable type has a genuine two-sided limit. For jump and infinite discontinuities the limit does not exist, so no clever redefinition at the single point can make the function continuous. Don't try to "fill" a jump — the gap has real width.
Classify the discontinuity of each at the marked point:
- $\dfrac{x^2-9}{x-3}$ at $x=3$: limit $=6$, value undefined → removable (hole).
- $f(x)=\begin{cases}1,&x<0\\2,&x\ge0\end{cases}$ at $x=0$: left $\to1$, right $\to2$ → jump.
- $\dfrac{1}{x^2}$ at $x=0$: grows to $+\infty$ → infinite.
Three discontinuity types: removable (a hole — the limit exists but $f(c)$ doesn't match), jump (finite one-sided limits that disagree), and infinite (unbounded near $c$, a vertical asymptote). Only the removable type has a two-sided limit, so only it can be repaired.