DWG NO. 10.4 — Lesson 4 of 5

Continuity

Unit 10: Limits and an Introduction to Calculus · ~20 min

Objective Determine whether a function is continuous at a point using the three-part definition, and classify discontinuities as removable, jump, or infinite.

Lessons 10.1 and 10.2 kept finding limits that matched the function value, and limits that didn't — sometimes because the function wasn't even defined there. Continuity is the word for when a function's graph has no break, hole, or jump at a point: intuitively, you could trace it through that point without lifting your pencil.

The three-part definition

A function f is continuous at x = a if all three conditions hold:

1. f(a) is defined — the function actually has a value at a.
2. li\(m_{x→a}\) f(x) exists — the left-hand and right-hand limits agree.
3. li\(m_{x→a}\) f(x) = f(a) — the value the function approaches matches the value it actually takes.

If any one of these three fails, f is discontinuous at a. Which condition fails tells you what kind of discontinuity it is.

Three types of discontinuity

Type	What fails	Fixable?
Removable	The limit exists, but f(a) is undefined or doesn't match it — a single-point hole.	Yes — redefine f(a) to equal the limit.
Jump	The left-hand and right-hand limits both exist but disagree.	No — the graph genuinely steps to a different level.
Infinite	One or both one-sided limits are ±∞ (a vertical asymptote).	No — the function is unbounded near a.

A jump discontinuity at x = 2: li\(m_{x→2^{-}}\) f(x) = 3 (open circle) but f(2) = 1 (filled circle) and li\(m_{x→2^{+}}\) f(x) = 1 — the one-sided limits disagree

Worked Example 1 · Removable discontinuity

ProblemCheck whether f(x) = (x²−4)/(x−2) is continuous at x = 2.

1Condition 1 fails immediately: f(2) is undefined (0/0), so f cannot be continuous at x=2.

2But li\(m_{x→2}\) f(x) = li\(m_{x→2}\) (x+2) = 4 does exist — only condition 1 fails, not condition 2.

Discontinuous at x=2 — removable (redefining f(2)=4 would fix it)

Worked Example 2 · Jump discontinuity

ProblemLet g(x) = x² for x < 0, and g(x) = x+1 for x ≥ 0. Check continuity at x = 0.

1g(0) = 0+1 = 1, so condition 1 is fine. Now check the limit: left-hand li\(m_{x→0^{-}}\) x² = 0; right-hand li\(m_{x→0^{+}}\) (x+1) = 1.

2The one-sided limits disagree (0 ≠ 1), so li\(m_{x→0}\) g(x) does not exist — condition 2 fails.

Discontinuous at x=0 — jump discontinuity

Worked Example 3 · Choosing a constant for continuity

ProblemFind k so that f(x) = kx+1 for x < 3, and f(x) = x²−2 for x ≥ 3, is continuous at x = 3.

1For continuity, the left-hand and right-hand limits at x=3 must be equal (and match f(3)).

2Left-hand: li\(m_{x→3^{-}}\) (kx+1) = 3k+1. Right-hand and f(3): 3²−2 = 7.

3Set them equal: 3k+1 = 7, so 3k = 6.

k = 2

Guided practice

1.Classify the discontinuity of f(x) = 1/(x−4) at x = 4.
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As x→4, f(x) grows without bound in magnitude — a vertical asymptote. Infinite discontinuity.
2.Is f(x) = x²−1 continuous at every real number? Explain briefly.
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Yes. Polynomials are defined everywhere, and direct substitution always gives li\(m_{x→a}\) f(x) = f(a), so all three conditions hold at every a.
3.Let g(x) = x² for x < 0, and g(x) = x+1 for x ≥ 0 (as in Worked Example 2, but re-examine the type). Confirm this is a jump discontinuity and not removable.
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Both one-sided limits exist (0 from the left, 1 from the right) but disagree — that mismatch can't be fixed by redefining a single point, since the two pieces genuinely approach different heights. Jump discontinuity, not removable.
4.f(x) = (x²+x−2)/(x−1). Determine whether the discontinuity at x=1 is removable, and if so, what value would make f continuous there.
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Factor: (x−1)(x+2)/(x−1) = x+2 for x≠1, so li\(m_{x→1}\) f(x) = 3, but f(1) is undefined. Removable — defining f(1) = 3 would make it continuous.
5.Find k so that f(x) = 2x+k for x ≤ 1, and f(x) = x²+3 for x > 1, is continuous at x = 1.
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f(1) = 2(1)+k = 2+k. Right-hand limit: li\(m_{x→1^{+}}\) (x²+3) = 4. Set 2+k = 4. k = 2.

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