Continuity: no breaks, no jumps, no holes

The problem

Draw a graph without lifting your pen and it is continuous. But the pen test fails in an exam hall: you need a check you can write down. Three questions do the job.

First we look at each side of c separately: the left-hand limit (x < c) and the right-hand limit (x > c). The limit exists only when both sides agree.

f is continuous at c if and only if (i) f(c) is defined, (ii) lim(x→c) f(x) exists, and (iii) lim(x→c) f(x) = f(c). If any condition fails, f is discontinuous at c.

Key terms

Left-hand limit

The value f(x) approaches as x comes towards c from numbers smaller than c.

Right-hand limit

The value f(x) approaches as x comes towards c from numbers bigger than c.

Existence of a limit

lim(x→c) f(x) = L if and only if both one-sided limits equal L. Different sides mean no limit.

Continuous at c

f(c) is defined, the limit at c exists, and the two are equal. The graph then has no break at c.

Short questions with model answers

1. Q1. f(x) = 2x + 1 for 0 ≤ x < 2, 7 − x for 2 ≤ x < 4, and x for 4 ≤ x ≤ 6. Do lim(x→2) f(x) and lim(x→4) f(x) exist?

   • x → 2⁻: 2(2) + 1 = 5; x → 2⁺: 7 − 2 = 5
   • x → 4⁻: 7 − 4 = 3; x → 4⁺: 4

   At 2: exists, 5 · At 4: does not exist

2. Q2. f(x) = (x² − 9)/(x − 3) for x ≠ 3, with f(3) = 6. g(x) is the same formula for x ≠ 3 only. Discuss the continuity of f and g at x = 3.

   • lim(x→3) (x + 3)(x − 3)/(x − 3) = lim(x→3) (x + 3) = 6
   • f(3) = 6 = lim ⇒ f continuous; g(3) undefined ⇒ g discontinuous

   f is continuous at 3; g is discontinuous at 3

3. Q3. Discuss continuity at 3 of f(x) = x − 1 for x < 3 and 2x + 1 for 3 ≤ x.

   • f(3) = 2(3) + 1 = 7
   • x → 3⁻: 3 − 1 = 2; x → 3⁺: 2(3) + 1 = 7

   Not continuous at 3

Common mistakes

• ✗ Checking only one side of c and calling that the limit.

  ✓ For a piecewise function, find both one-sided limits with the formula valid on each side. In Example 1 at 4, the left gives 3 and the right gives 4.

• ✗ Saying f is continuous because the limit exists.

  ✓ The limit must also equal f(c). For 3x − 1 (x < 1), 4 (x = 1), 2x (x > 1), the limit at 1 is 2 but f(1) = 4, so f is discontinuous.

• ✗ Simplifying (x² − 9)/(x − 3) to x + 3 and calling it continuous at 3.

  ✓ Cancelling is allowed only for x ≠ 3. At x = 3 the original formula gives 0/0, so the function is undefined there unless f(3) is given.

MCQs

1. 1. For f(x) = 2x + 1 (0 ≤ x < 2), 7 − x (2 ≤ x < 4), lim(x→2⁻) f(x) equals:

   1. (a) 3
   2. (b) 5
   3. (c) 7
   4. (d) does not exist
   Show answer

   (b) Left of 2 the formula is 2x + 1, and 2(2) + 1 = 5.

2. 2. For f(x) = x − 1 (x < 3), 2x + 1 (3 ≤ x), lim(x→3) f(x) is:

   1. (a) 2
   2. (b) 7
   3. (c) 9/2
   4. (d) does not exist
   Show answer

   (d) Left limit 2, right limit 7. They differ, so there is no limit.

3. 3. For f(x) = 3x − 1 (x < 1), 4 (x = 1), 2x (x > 1), lim(x→1) f(x) and f(1) are:

   1. (a) 2 and 2
   2. (b) 2 and 4
   3. (c) 4 and 4
   4. (d) 3 and 4
   Show answer

   (b) Both sides give 2, but f(1) = 4. Condition (iii) fails, so f is discontinuous at 1.

4. 4. g(x) = (x² − 9)/(x − 3) for x ≠ 3, with no value given at 3. At x = 3, g is:

   1. (a) continuous, g(3) = 6
   2. (b) continuous, g(3) = 0
   3. (c) discontinuous
   4. (d) continuous, g(3) = 3
   Show answer

   (c) The limit is 6, but g(3) is not defined, so condition (i) fails.

5. 5. f(x) = (√(2x + 5) − √(x + 7))/(x − 2) for x ≠ 2, and f(2) = k. For f to be continuous at 2, k =

   1. (a) 0
   2. (b) 1/6
   3. (c) 1/3
   4. (d) 6
   Show answer

   (b) Rationalise: the top becomes x − 2, leaving 1/(√(2x + 5) + √(x + 7)) → 1/(3 + 3) = 1/6. k must equal the limit.

Quick revision

• The limit at c exists only when the left-hand and right-hand limits are equal.
• Continuous at c means three things: f(c) is defined, the limit exists, and the limit equals f(c).
• A hole, a jump, or a misplaced point each breaks a different condition; name the one that fails.