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Limits and Their Properties

1.4 Continuity and One-Sided Limits

Focus on continuity at a point, one-sided limits, intermediate value theorem.

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Ideas in this lesson

01Continuity at a pointWorked steps · five questions02One-sided limitsWorked steps · five questions03Intermediate Value TheoremWorked steps · five questions
Idea 1 of 3

Continuity at a point

The value exists, the two-sided limit exists, and the two are equal. All three conditions are necessary.

Worked example 1

For x≠2x \ne 2, define f(x)=x2−4x−2f(x) = \dfrac{x^{2} - 4}{x - 2}. Set f(2)=kf(2) = k. Which kk makes ff continuous at 22?
  1. x2−4x−2=x+2(x≠2)\dfrac{x^{2} - 4}{x - 2} = x + 2 (x \ne 2)
  2. k=lim⁡x→2(x+2)\displaystyle k = \lim\limits_{x\to 2} (x + 2)
  3. k=4k = 4

Answer: 44

Worked example 2

For x≠3x \ne 3, define f(x)=x2f(x) = x^{2}. Set f(3)=kf(3) = k. Which kk makes ff continuous at 33?
  1. f(x)=x2f(x) = x^{2}
  2. k=lim⁡x→3x2\displaystyle k = \lim\limits_{x\to 3} x^{2}
  3. k=9k = 9

Answer: 99

Idea 2 of 3

One-sided limits

Use the rule valid on the requested side. A two-sided limit exists only if the one-sided limits agree.

Worked example 1

Find the left-hand limit at x=2x = 2.
f(x)={x+2if x<26if x≥2f(x) = \begin{cases}x + 2 & \text{if } x < 2 \\ 6 & \text{if } x \ge 2\end{cases}
  1. lim⁡x→2−f(x)=2+2=4\displaystyle \lim\limits_{x\to 2^{-}} f(x) = 2 + 2 = 4
  2. lim⁡x→2+f(x)=6\displaystyle \lim\limits_{x\to 2^{+}} f(x) = 6
  3. Requested one-sided limit =4= 4

Answer: 44

Worked example 2

Find the right-hand limit at x=3x = 3.
f(x)={x+2if x<37if x≥3f(x) = \begin{cases}x + 2 & \text{if } x < 3 \\ 7 & \text{if } x \ge 3\end{cases}
  1. lim⁡x→3−f(x)=3+2=5\displaystyle \lim\limits_{x\to 3^{-}} f(x) = 3 + 2 = 5
  2. lim⁡x→3+f(x)=7\displaystyle \lim\limits_{x\to 3^{+}} f(x) = 7
  3. Requested one-sided limit =7= 7

Answer: 77

Idea 3 of 3

Intermediate Value Theorem

A continuous function on a closed interval takes every value between its endpoint outputs. The theorem guarantees existence, not uniqueness.

Worked example 1

For f(x)=x3−2f(x) = x^{3} - 2 on [0,2][0, 2], does the Intermediate Value Theorem guarantee a zero?
  1. f(0)=−2<0f(0) = -2 < 0
  2. f(2)=23−2=6>0f(2) = 2^{3} - 2 = 6 > 0
  3. 0∈(f(0),f(2))0 \in (f(0), f(2)); ∃c∈(0,2)c \in (0, 2) with f(c)=0f(c) = 0

Answer: Yes

Worked example 2

For f(x)=x3−3f(x) = x^{3} - 3 on [0,3][0, 3], does the Intermediate Value Theorem guarantee a zero?
  1. f(0)=−3<0f(0) = -3 < 0
  2. f(3)=33−3=24>0f(3) = 3^{3} - 3 = 24 > 0
  3. 0∈(f(0),f(3))0 \in (f(0), f(3)); ∃c∈(0,3)c \in (0, 3) with f(c)=0f(c) = 0

Answer: Yes

Resource alignment: Larson & Edwards, Calculus of a Single Variable, Tenth Edition, §1.4,p1.4, p. 7070. Explanations, diagrams and numerical practice variants are original MathLab content aligned to this section.