Unit 1 · Topic 1.11 Beta

Continuity at a Point

f is continuous at c when f(c) is defined, lim as x → c, f(x) exists, and the two are equal.

Practice 3: Justification

Question set for this topic

Part 1 · Hook

Why this matters

An engineer signing off a bridge checks every joint, not just most of them. Continuity at a point is a three-joint inspection: the point must be there, the curve must arrive from both sides at one height, and that height must be where the point is. Skip one joint and the conclusion can fail.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. A removable discontinuity at x=cx = c means

  1. the limit at cc exists, but the value is missing or different
  2. the one-sided limits at cc differ
  3. the function is unbounded near cc
  4. the function is defined at cc
Show the answer

Removable: a hole, possibly with the point elsewhere.

  • Correct: the limit at cc exists, but the value is missing or different:
  • the one-sided limits at cc differ:
  • the function is unbounded near cc:
  • the function is defined at cc:

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

  1. equals 2
  2. does not exist
  3. equals f(1)f(1)
  4. equals 4
Show the answer

Equal one-sided limits give the two-sided limit.

  • Correct: equals 2:
  • does not exist:
  • equals f(1)f(1):
  • equals 4:

Part 4 · See it

See it first

Four graphs near x = c. First: an open circle and no closed dot, so f(c) is not defined and condition 1 fails. Second: a jump, so the limit at c does not exist and condition 2 fails. Third: a hole with the closed dot f(c) placed above it, so the limit exists but does not equal f(c) and condition 3 fails. Fourth: an unbroken curve through a closed dot, where all three conditions hold.
Each of the three conditions can fail on its own. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. f(c)f(c) is undefinedcondition 1 fails: there is no point on the graph at x=cx = c
  2. f(c)f(c) is defined, but the one-sided limits differ or are unboundedcondition 2 fails: limx→cf(x)\lim_{x \to c} f(x) does not exist
  3. The value and the limit both exist but are different numberscondition 3 fails: the point is not where the curve arrives
  4. All three conditions holdff is continuous at cc, and limx→cf(x)\lim_{x \to c} f(x)=f(c)= f(c), so the limit can be found by substitution

Part 6 · Key ideas

Key ideas

  • Continuous at cc means: (1) f(c)f(c) is defined, (2) limx→cf(x)\lim_{x \to c} f(x) exists, (3) limx→cf(x)\lim_{x \to c} f(x)=f(c)= f(c).
  • Check the conditions in order. The first one that fails tells you the reason, and the kind of discontinuity.
  • At the break point of a piecewise function, find both one-sided limits from the pieces, then compare with the value the function is given there.
  • A justification shows all three conditions with numbers: "f(2)=3f(2) = 3, limx→2f(x)=3\lim_{x \to 2} f(x) = 3, so limx→2f(x)\lim_{x \to 2} f(x)=f(2)= f(2)."
  • If ff is continuous at cc, its limit at cc is just f(c)f(c). That is why direct substitution works.

Part 7 · Misconception

A common mistake

The wrong idea: f(2)f(2) is defined, so ff is continuous at x=2x = 2.

What actually happens: A defined value is only the first condition. The limit must exist and equal it. A function can be defined at a jump (the value sits on one side) or at a misplaced point above a hole.

Part 8 · Check yourself

Check yourself

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

Table

A function defined in three parts

The function g is given by a different rule left of 2, at 2, and right of 2
WhereRule for g(x)g(x)
x<2x < 2x2−1x^2 - 1
x=2x = 233
x>2x > 24x−54x - 5

1. Which is a complete justification that gg is continuous at x=2x = 2?

  1. g(2)=3g(2) = 3, and both one-sided limits are 3, so limx→2g(x)\lim_{x \to 2} g(x)=g(2)= g(2)
  2. Both one-sided limits at 2 equal 3, so the limit limx→2g(x)\lim_{x \to 2} g(x) exists
  3. g(2)=3g(2) = 3 is defined, and each piece of gg is a polynomial on its interval
  4. gg is defined at x=2x = 2, so the graph of gg can have no break at x=2x = 2
Show the answer

All three conditions must be shown: g(2)g(2) exists, limx→2g(x)\lim_{x \to 2} g(x) exists, and they are equal.

  • Correct: g(2)=3g(2) = 3, and both one-sided limits are 3, so limx→2g(x)\lim_{x \to 2} g(x)=g(2)= g(2): Defined value, existing limit, and the two equal: all three conditions.
  • Both one-sided limits at 2 equal 3, so the limit limx→2g(x)\lim_{x \to 2} g(x) exists: This shows the limit exists but never compares it with g(2)g(2).
  • g(2)=3g(2) = 3 is defined, and each piece of gg is a polynomial on its interval: Polynomial pieces can still fail to meet at the break point; the limits must be checked.
  • gg is defined at x=2x = 2, so the graph of gg can have no break at x=2x = 2: A defined value is the first condition only; a function can be defined at a jump.

2. Suppose the rule at x=2x = 2 were changed to g(2)=5g(2) = 5. Which condition for continuity at 2 would fail?

  1. g(2)g(2) is defined; changing it leaves g(2)g(2) undefined
  2. limx→2g(x)\lim_{x \to 2} g(x)=g(2)= g(2); the limit is 3 but the value is 5
  3. limx→2g(x)\lim_{x \to 2} g(x) exists; the one-sided limits would differ
  4. No condition; gg would still be continuous at x=2x = 2
Show the answer

The one-sided limits come from the rules left and right of 2, which do not change. The limit is still 3, but now g(2)=5g(2) = 5, so the third condition fails: a removable discontinuity.

  • g(2)g(2) is defined; changing it leaves g(2)g(2) undefined: g(2)=5g(2) = 5 is defined.
  • Correct: limx→2g(x)\lim_{x \to 2} g(x)=g(2)= g(2); the limit is 3 but the value is 5: Value 5, limit 3: the third condition fails.
  • limx→2g(x)\lim_{x \to 2} g(x) exists; the one-sided limits would differ: The pieces left and right of 2 are unchanged, so both one-sided limits are still 3.
  • No condition; gg would still be continuous at x=2x = 2: A value different from the limit breaks the graph.

Graph

The graph of hh

12345612345xyh

3. Why is hh not continuous at x=1x = 1?

  1. limx→1h(x)\lim_{x \to 1} h(x) does not exist
  2. limx→1h(x)\lim_{x \to 1} h(x) is not equal to h(1)=1.5h(1) = 1.5
  3. hh is unbounded near x=1x = 1
  4. h(1)h(1) is not defined
Show the answer

The graph has an open circle at (1,1.5)(1, 1.5) and no closed dot above x=1x = 1, so h(1)h(1) does not exist. The first condition fails, even though limx→1h(x)=1.5\lim_{x \to 1} h(x) = 1.5.

  • limx→1h(x)\lim_{x \to 1} h(x) does not exist: From both sides the line heads to 1.5, so the limit exists.
  • limx→1h(x)\lim_{x \to 1} h(x) is not equal to h(1)=1.5h(1) = 1.5: 1.5 is the limit, not a value: the point is missing.
  • hh is unbounded near x=1x = 1: The line stays near 1.5; nothing grows without bound.
  • Correct: h(1)h(1) is not defined: An open circle with no closed dot means there is no value at 1.

4. h(3)=2.5h(3) = 2.5 is defined. Why is hh still not continuous at x=3x = 3?

  1. limx→3h(x)\lim_{x \to 3} h(x) does not exist: the sides approach 2.5 and 3.5
  2. h(3)h(3) is not defined because there is an open circle at x=3x = 3
  3. limx→3h(x)\lim_{x \to 3} h(x) exists but equals 3.5, which is not h(3)h(3)
  4. hh is unbounded near x=3x = 3, so there is a vertical asymptote
Show the answer

limx→3−h(x)\lim_{x \to 3^-} h(x)=2.5= 2.5 and limx→3+h(x)\lim_{x \to 3^+} h(x)=3.5= 3.5. Unequal one-sided limits mean the limit does not exist, so the second condition fails: a jump.

  • Correct: limx→3h(x)\lim_{x \to 3} h(x) does not exist: the sides approach 2.5 and 3.5: The second condition fails: the one-sided limits differ.
  • h(3)h(3) is not defined because there is an open circle at x=3x = 3: The closed dot at (3,2.5)(3, 2.5) gives h(3)=2.5h(3) = 2.5.
  • limx→3h(x)\lim_{x \to 3} h(x) exists but equals 3.5, which is not h(3)h(3): 3.5 is only the right-hand limit; the left side heads to 2.5.
  • hh is unbounded near x=3x = 3, so there is a vertical asymptote: Both one-sided limits are finite.

5. Let f(x)=x2−9x−3f(x) = \frac{x^2 - 9}{x - 3} for x≠3x \ne 3 and f(3)=3f(3) = 3. Which describes ff at x=3x = 3?

  1. Not continuous: f(3)f(3) is not defined
  2. Not continuous: the limit at 3 does not exist
  3. Not continuous: the limit is 6 but f(3)=3f(3) = 3
  4. Continuous: f(3)f(3) and the limit both exist
Show the answer

limx→3f(x)\lim_{x \to 3} f(x)=limx→3(x+3)= \lim_{x \to 3} (x + 3)=6= 6, which is not f(3)=3f(3) = 3. The third condition fails.

  • Not continuous: f(3)f(3) is not defined: The second rule defines f(3)=3f(3) = 3.
  • Not continuous: the limit at 3 does not exist: f(x)=x+3f(x) = x + 3 near 3, so the limit is 6.
  • Correct: Not continuous: the limit is 6 but f(3)=3f(3) = 3: Value 3, limit 6: they must be equal.
  • Continuous: f(3)f(3) and the limit both exist: Both exist, but they are not equal.

6. ff is continuous at x=4x = 4, and limx→4(f(x)+2x)\lim_{x \to 4} \left(f(x) + 2x\right)=13= 13. Find f(4)f(4).

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

By the limit laws, limx→4f(x)\lim_{x \to 4} f(x)=13−limx→42x= 13 - \lim_{x \to 4} 2x=13−8= 13 - 8=5= 5. Since ff is continuous at 4, f(4)f(4)=limx→4f(x)= \lim_{x \to 4} f(x)=5= 5.

  • Answer: 5

Part 9 · Summary

Summary

ff is continuous at cc when f(c)f(c) is defined, limx→cf(x)\lim_{x \to c} f(x) exists, and the two are equal. Check the three conditions in order and state each one with its number. Continuity at cc is exactly what makes limx→cf(x)\lim_{x \to c} f(x)=f(c)= f(c).

Part 10 · Up next

What comes next

Part 11 · Connections

Connections