Curriculum
General Mathematics — Second Secondary
Unit 1: Real Functions and Curve Sketching
0/10
Unit 2: Exponents, Logarithms and Applications
0/10
Unit 3: Limits
0/6
Unit 4: Trigonometry
0/4
Text lesson
Finding Limits Algebraically
Finding Limits Algebraically
Worked examples
Worked examples
Example 1: direct substitution
- Find limx → 4 (3x − 1) / (x + 2).
- Solution The denominator at 4 is 6, which is not zero, so we substitute directly: (3 × 4 − 1) / (4 + 2) = 11 / 6.
Example 2: factorising
- Find limx → 3 (x2 − 9) / (x − 3).
- Solution Substitution gives 0/0. Factorise the numerator as (x − 3)(x + 3) and cancel (x − 3), since x ≠ 3. The expression becomes x + 3, whose limit at 3 is 6.
Example 3: factorising a difference of cubes
- Find limx → 2 (x3 − 8) / (x − 2).
- Solution Use the difference of cubes: x3 − 8 = (x − 2)(x2 + 2x + 4). The fraction equals x2 + 2x + 4 for every x ≠ 2. Substituting 2 gives 4 + 4 + 4 = 12.
Example 4: multiplying by the conjugate
- Find limx → 0 (√(x + 4) − 2) / x.
- Solution Substitution gives 0/0. Multiply the numerator and the denominator by (√(x + 4) + 2):
- (√(x + 4) − 2)(√(x + 4) + 2) = x + 4 − 4 = x. So the fraction becomes x / (x(√(x + 4) + 2)) = 1 / (√(x + 4) + 2).
- At x = 0 the limit is 1 / (2 + 2) = 1/4.
Example 5: factorising a trinomial
- Find limx → −1 (2x2 + 3x + 1) / (x + 1).
- Solution Factorise the numerator: 2x2 + 3x + 1 = (2x + 1)(x + 1). The fraction equals 2x + 1 for every x ≠ −1. At x = −1 this gives 2(−1) + 1 = −1.
Example 6: a demanding problem
- Find limx → 5 (1/x − 1/5) / (x − 5).
- Solution Combine the numerator: 1/x − 1/5 = (5 − x) / (5x). The fraction becomes (5 − x) / (5x(x − 5)). Since 5 − x = −(x − 5), the fraction equals −1 / (5x) for every x ≠ 5. At x = 5 we get −1/25.
Check→At x = 5.01 the value of the fraction is about −0.04, which agrees with −1/25 = −0.04.
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