Unit 1 · Topic 1.10 Beta

Types of Discontinuity

A discontinuity is a point where the graph breaks.

Practice 2: Connecting RepresentationsPractice 3: Justification

Question set for this topic

Part 1 · Hook

Why this matters

A road can be broken in different ways: a single missing paving stone, a step where one side is higher than the other, or a cliff with no bottom in sight. Graphs break in the same three ways, and calculus names each one. Knowing which break you face tells you whether it can be repaired.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. If limx→2−f(x)=3\lim_{x \to 2^-} f(x) = 3 and limx→2+f(x)=3\lim_{x \to 2^+} f(x) = 3, then

  1. limx→2f(x)=3\lim_{x \to 2} f(x) = 3
  2. f(2)=3f(2) = 3
  3. limx→2f(x)\lim_{x \to 2} f(x) does not exist
  4. f(2)f(2) is undefined
Show the answer

Equal one-sided limits give the two-sided limit. They say nothing about f(2).

  • Correct: limx→2f(x)=3\lim_{x \to 2} f(x) = 3:
  • f(2)=3f(2) = 3:
  • limx→2f(x)\lim_{x \to 2} f(x) does not exist:
  • f(2)f(2) is undefined:

2. Substituting x=4x = 4 into x+1x−4\frac{x + 1}{x - 4} gives 50\frac{5}{0}. Near x=4x = 4, the expression is

  1. unbounded
  2. close to 5
  3. close to 0
  4. close to 1
Show the answer

A nonzero number over a shrinking denominator grows without bound.

  • Correct: unbounded:
  • close to 5:
  • close to 0:
  • close to 1:

Part 4 · See it

See it first

Three graphs, each with a break at x = 2. Removable: the curve has a hole at (2, 3), an open circle, while f(2) = 1 is a closed dot below it; the limit exists. Jump: the left piece ends at a closed dot at (2, 1) and the right piece starts at an open circle at (2, 3); the one-sided limits differ. Infinite: the curve rises without bound on both sides of the dashed vertical asymptote x = 2.
The three kinds of discontinuity: removable, jump and infinite. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. limx→cf(x)\lim_{x \to c} f(x) exists, but f(c)f(c) is missing or differentthe graph has a hole: a removable discontinuity
  2. Both one-sided limits are finite but not equalthe graph steps from one height to another: a jump discontinuity
  3. ff is unbounded on at least one side of ccthe graph has a vertical asymptote x=cx = c: an infinite discontinuity
  4. A formula gives 00\frac{0}{0} at cc and the zero factor cancelsexpect a hole; a factor that stays in the denominator gives an asymptote

Part 6 · Key ideas

Key ideas

  • A function is continuous at x=cx = c when limx→cf(x)\lim_{x \to c} f(x) exists and equals f(c)f(c); its graph has no break there. A discontinuity is a point where this fails.
  • Removable: the limit exists, but the point is missing or misplaced (a hole).
  • Jump: both one-sided limits are finite and different.
  • Infinite: the function is unbounded near cc (a vertical asymptote).
  • To classify, find the one-sided limits first, then compare them with f(c)f(c).

Part 7 · Misconception

A common mistake

The wrong idea: If a function is undefined at a point, its graph has a vertical asymptote there.

What actually happens: Undefined is not the same as unbounded. x2−4x−2\frac{x^2 - 4}{x - 2} is undefined at 2 but equals x+2x + 2 nearby, so its graph has a hole at (2,4)(2, 4), a removable discontinuity. An asymptote needs the values to grow without bound.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Graph

The graph of ff

−4−3−2−112345−1123456xyf

1. What kind of discontinuity does ff have at x=−2x = -2?

  1. Removable
  2. Jump
  3. Infinite
  4. No discontinuity; ff is continuous there
Show the answer

From both sides the curve approaches height 1, so limx→−2f(x)=1\lim_{x \to -2} f(x) = 1. But f(−2)=3f(-2) = 3. The limit exists and differs from the value: a removable discontinuity.

  • Correct: Removable: The limit exists (1) but does not equal f(−2)=3f(-2) = 3; moving one point would fix it.
  • Jump: A jump needs different one-sided limits. Both sides approach 1 here.
  • Infinite: Nothing is unbounded near −2-2.
  • No discontinuity; ff is continuous there: The value f(−2)=3f(-2) = 3 is not where the curve is heading, so there is a break.

2. What kind of discontinuity does ff have at x=1x = 1?

  1. Removable
  2. Jump
  3. Infinite
  4. No discontinuity; f(1)f(1) is defined
Show the answer

limx→1−f(x)=4\lim_{x \to 1^-} f(x) = 4 and limx→1+f(x)\lim_{x \to 1^+} f(x)=12= \frac{1}{2}. Both one-sided limits exist but differ: a jump discontinuity.

  • Removable: A removable discontinuity has a two-sided limit. Here the sides disagree.
  • Correct: Jump: Left side to 4, right side to 12\frac{1}{2}: two different finite one-sided limits.
  • Infinite: Both one-sided limits are finite numbers.
  • No discontinuity; f(1)f(1) is defined: A defined value is not enough; the sides must also meet at that value.

3. What kind of discontinuity does ff have at x=3x = 3?

  1. Jump
  2. Removable
  3. Infinite
  4. No discontinuity; ff is simply not defined at 3
Show the answer

Near x=3x = 3 the graph rises without bound on both sides of the dashed vertical asymptote, so the discontinuity is infinite.

  • Jump: A jump needs two finite one-sided limits. These values grow without bound.
  • Removable: There is no finite limit at 3 to fill a hole with.
  • Correct: Infinite: ff is unbounded near 3: an infinite discontinuity, with a vertical asymptote.
  • No discontinuity; ff is simply not defined at 3: Not being defined at 3 already breaks continuity there, and ff is unbounded nearby: an infinite discontinuity.

Table

Four functions with breaks

Four functions defined by formulas
FunctionFormula
aaa(x)=x2−1x−1a(x) = \frac{x^2 - 1}{x - 1}
bbb(x)=x+1x−1b(x) = \frac{x + 1}{x - 1}
ccc(x)=|x−1|x−1c(x) = \frac{|x - 1|}{x - 1}
ddd(x)d(x)=x2+x−2x2−1= \frac{x^2 + x - 2}{x^2 - 1}

4. Which describes all the discontinuities of dd?

  1. Removable at x=1x = 1 and infinite at x=−1x = -1
  2. Infinite at x=1x = 1 and infinite at x=−1x = -1
  3. Removable at x=1x = 1 and removable at x=−1x = -1
  4. Removable at x=−2x = -2 and infinite at x=1x = 1
Show the answer

d(x)d(x)=(x+2)(x−1)(x+1)(x−1)= \frac{(x + 2)(x - 1)}{(x + 1)(x - 1)}. The factor x−1x - 1 cancels, leaving a hole at x=1x = 1. The factor x+1x + 1 does not cancel, so dd is unbounded near x=−1x = -1.

  • Correct: Removable at x=1x = 1 and infinite at x=−1x = -1: A canceled factor gives a hole; a factor left in the denominator gives a vertical asymptote.
  • Infinite at x=1x = 1 and infinite at x=−1x = -1: At x=1x = 1 the zero factor cancels, so the limit is 32\frac{3}{2}, a finite number.
  • Removable at x=1x = 1 and removable at x=−1x = -1: At x=−1x = -1 the numerator is −2≠0-2 \ne 0 and the denominator is 0, so dd is unbounded there.
  • Removable at x=−2x = -2 and infinite at x=1x = 1: x=−2x = -2 makes the numerator 0, which gives a zero of dd, not a break.

5. A function has limx→5−g(x)=2\lim_{x \to 5^-} g(x) = 2, limx→5+g(x)=2\lim_{x \to 5^+} g(x) = 2, and g(5)=6g(5) = 6. What kind of discontinuity is at x=5x = 5?

  1. Jump
  2. Infinite
  3. There is no discontinuity
  4. Removable
Show the answer

The one-sided limits agree, so limx→5g(x)=2\lim_{x \to 5} g(x) = 2 exists. It does not equal g(5)=6g(5) = 6: a removable discontinuity.

  • Jump: A jump needs unequal one-sided limits; these are both 2.
  • Infinite: All the limits are finite.
  • There is no discontinuity: g(5)=6g(5) = 6 is not the limit 2, so the graph has a break.
  • Correct: Removable: The limit exists but the point is in the wrong place, like a hole with a dot above it.

6. A function has limx→5−g(x)=2\lim_{x \to 5^-} g(x) = 2, limx→5+g(x)=6\lim_{x \to 5^+} g(x) = 6, and g(5)=6g(5) = 6. What kind of discontinuity is at x=5x = 5?

  1. Jump
  2. Removable
  3. Infinite
  4. There is no discontinuity
Show the answer

The one-sided limits, 2 and 6, are finite and different, so the limit does not exist: a jump. The value matching one side does not change that.

  • Correct: Jump: Two different finite one-sided limits make a jump.
  • Removable: A removable discontinuity needs the two-sided limit to exist.
  • Infinite: Both one-sided limits are finite.
  • There is no discontinuity: The left side heads to 2, not 6, so the graph breaks.

7. A student says f(x)=x2−4x−2f(x) = \frac{x^2 - 4}{x - 2} has a vertical asymptote at x=2x = 2 because f(2)f(2) is undefined. What is the best response?

  1. Yes; any point where ff is undefined has a vertical asymptote
  2. No; f(2)=0f(2) = 0, so ff has no discontinuity at x=2x = 2
  3. Yes; the denominator is 0 at x=2x = 2, so ff is unbounded
  4. No; f(x)=x+2f(x) = x + 2 for x≠2x \ne 2, so there is a hole at (2,4)(2, 4)
Show the answer

Undefined is not the same as unbounded. Here the zero factor cancels: (x−2)(x+2)x−2\frac{(x - 2)(x + 2)}{x - 2}=x+2= x + 2, so the limit at 2 is 4 and the discontinuity is removable.

  • Yes; any point where ff is undefined has a vertical asymptote: A hole is also a point where the function is undefined.
  • No; f(2)=0f(2) = 0, so ff has no discontinuity at x=2x = 2: f(2)f(2) is 00\frac{0}{0}, which is undefined, not 0.
  • Yes; the denominator is 0 at x=2x = 2, so ff is unbounded: The numerator is also 0 at 2, so the denominator alone does not decide. The limit is 4.
  • Correct: No; f(x)=x+2f(x) = x + 2 for x≠2x \ne 2, so there is a hole at (2,4)(2, 4): The canceled factor leaves a hole at height 4, not an asymptote.

Part 9 · Summary

Summary

A discontinuity is a point where the graph breaks. Removable: the limit exists but the value is missing or different (a hole). Jump: finite one-sided limits that differ. Infinite: unbounded near the point (a vertical asymptote). Classify by finding the one-sided limits and comparing them with f(c)f(c).

Part 10 · Up next

What comes next

Part 11 · Connections

Connections