Continuity at a point
Draw a graph without lifting your pencil: that is a continuous function. If you must lift the pencil, there is a break (a discontinuity).
A function f is continuous at x = c if f(c) is defined, the limit of f(x) as x approaches c exists, and the limit equals f(c). In short: left-hand limit = right-hand limit = f(c).
Formula: lim (x → c) f(x) = f(c)
Rule 1: Test continuity — Find f(c), the left limit and the right limit.
- Easy: f(x) = x² + 1 at x = 2: the limit is 5 and f(2) = 5, so f is continuous.
- Medium: f(x) = 2x + 1 for x ≤ 3 and x + 4 for x > 3: both sides give 7 at x = 3 and f(3) = 7, so f is continuous at 3.
- Exam-type: f(x) = kx + 1 for x ≤ 2 and 3x + 1 for x > 2 is continuous at 2 when 2k + 1 = 7, so k = 3.
Rule 2: Recognise the types of break — Look at what fails.
- Easy: f(x) = 1 / (x − 2) is not defined at x = 2.
- Medium: f(x) = x for x < 0 and x + 1 for x ≥ 0 has a jump at 0 (left limit 0, f(0) = 1).
- Exam-type: f(x) = (x² − 4) / (x − 2) for x ≠ 2 with f(2) = 5: the limit is 4, not 5; redefining f(2) = 4 repairs it (removable).
Three things must hold: the value exists, the limit exists, and they are equal.
Checking only that f(c) exists.
Pencil never lifts: left meets right meets the value.
Practice questions
- 1Is f(x) = 3x − 5 continuous at x = 2?
- 2Find the limit of (x² − 4) / (x − 2) as x approaches 2.
- 3Is f(x) = 1 / (x − 3) continuous at x = 3?
- 4Is f(x) = x + 2 for x < 1 and 4 − x for x ≥ 1 continuous at x = 1?
- 5Find k so that f(x) = kx + 1 for x ≤ 2 and 3x − 1 for x > 2 is continuous at x = 2.
- 6Find k so that f(x) = (x² − 9) / (x − 3) for x ≠ 3 and f(3) = k is continuous at x = 3.
- 7Show that f(x) = |x| is continuous at x = 0.
- 8Is the greatest integer function [x] continuous at x = 2?
Show answers (try them yourself first)
- 1Yes, it is a polynomial; the limit 1 equals f(2) = 1
- 2(x − 2)(x + 2) / (x − 2) = x + 2, so the limit is 4
- 3No, f(3) is not defined
- 4Yes: left limit 3, right limit 3 and f(1) = 3
- 52k + 1 = 5, so k = 2
- 6The limit is x + 3 = 6, so k = 6
- 7Left limit 0, right limit 0 and f(0) = 0
- 8No: the left limit is 1 but the right limit and f(2) are 2
Standard continuous functions and algebra of continuity
If two roads are unbroken, then putting them end to end or adding their heights gives an unbroken road again. Continuity survives the usual operations.
If f and g are continuous at c, so are f + g, f − g, f g and k f; and f/g too when g(c) ≠ 0. A composition of continuous functions is continuous. Polynomials, sine, cosine, exponentials and |x| are continuous everywhere; a rational function is continuous except where the denominator is zero; tan x is discontinuous at odd multiples of π/2.
Formula: f, g continuous ⇒ f ± g, f g, f/g (g ≠ 0), f∘g continuous
Rule 1: Find where a function is continuous — Remove the points where it is not defined.
- Easy: f(x) = x³ − 2x + 1 is continuous on all real numbers.
- Medium: f(x) = (x + 1) / (x − 4) is continuous for all x except x = 4.
- Exam-type: f(x) = tan x is continuous except at x = π/2, 3π/2, −π/2 and so on.
Rule 2: List the points of discontinuity — Solve denominator = 0.
- Easy: f(x) = 1 / (x² − 9) is discontinuous at x = 3 and x = −3.
- Medium: f(x) = 1 / ((x − 1)(x + 2)) is discontinuous at x = 1 and x = −2.
- Exam-type: f(x) = sin(x²) and f(x) = |x − 3| are continuous everywhere, being compositions of continuous functions.
Continuity is a property of a point; "continuous function" means continuous at every point of its domain.
Saying 1/x is discontinuous at 0 and forgetting that 0 is outside its domain.
Build from continuous blocks and the result stays continuous.
Practice questions
- 1Is f(x) = x² + 5x continuous everywhere?
- 2Find the point of discontinuity of f(x) = 1 / (x − 7).
- 3Find the points of discontinuity of f(x) = 1 / (x² − 16).
- 4Is f(x) = |x + 2| continuous at x = −2?
- 5Is f(x) = cos x + x² continuous on the real numbers?
- 6Find the points of discontinuity of f(x) = (x + 3) / (x² − 5x + 6).
- 7Is f(x) = x / |x| for x ≠ 0 and f(0) = 0 continuous at 0?
- 8Is the sum of two discontinuous functions always discontinuous?
Show answers (try them yourself first)
- 1Yes, it is a polynomial
- 2x = 7
- 3x = 4 and x = −4
- 4Yes, it is the composition of |·| with a polynomial
- 5Yes, the sum of two continuous functions
- 6x² − 5x + 6 = (x − 2)(x − 3), so x = 2 and x = 3
- 7No: the left limit is −1 and the right limit is 1
- 8No: [x] + (−[x]) = 0 is continuous even though each part is not