Limits by algebra: 0/0 is a clue, not an answer

The problem

Put x = 1 into (x³ − x)/(x² − x) and you get 0/0. Many students write “undefined” and move on. But the table of values near x = 1 clearly heads to 2. The function is hiding a common factor that makes both top and bottom zero.

Remove that factor and the answer appears. Because x → 1 means x ≠ 1, cancelling (x − 1) is completely allowed.

If substituting gives 0/0, simplify first: factorise and cancel the common factor, or rationalise a square root. Then substitute. Key results: lim(x→a) (xⁿ − aⁿ)/(x − a) = naⁿ⁻¹ and lim(x→0) (√(x + a) − √a)/x = 1/(2√a).

Key terms

0/0 form

0/0 is an indeterminate form, not a value. It tells you that numerator and denominator share a factor that vanishes at x = a.

Factorise and cancel

Factor both parts, cancel the common factor (x − a), then substitute. Allowed because x ≠ a while x → a.

Rationalise

With a square root, multiply top and bottom by the conjugate, e.g. √(x + a) + √a. The difference of squares removes the root and exposes the factor to cancel.

The xⁿ − aⁿ rule

xⁿ − aⁿ = (x − a)(xⁿ⁻¹ + axⁿ⁻² + … + aⁿ⁻¹). Cancelling (x − a) leaves n terms, each equal to aⁿ⁻¹ at x = a.

Short questions with model answers

1. Q1. Evaluate lim(x→1) (x³ − x)/(x² − x).

   • (1 − 1)/(1 − 1) = 0/0
   • x(x − 1)(x + 1) / x(x − 1) = x + 1
   • lim(x→1) (x + 1) = 2

   2

2. Q2. Evaluate lim(x→3) (x − 3)/(√x − √3).

   • x − 3 = (√x + √3)(√x − √3)
   • lim(x→3) (√x + √3) = 2√3 ≈ 3.464

   2√3

3. Q3. Use the xⁿ − aⁿ rule to find lim(x→2) (x⁵ − 32)/(x − 2).

   • 32 = 2⁵, so n = 5, a = 2
   • naⁿ⁻¹ = 5 × 2⁴ = 80

   80

Common mistakes

• ✗ Writing 0/0 = 0, or 0/0 = 1, as the answer.

  ✓ 0/0 has no value. It is a signal to simplify; the limit can be any number, like 2 or 2√3 above.

• ✗ Writing lim (xⁿ − aⁿ)/(x − a) = naⁿ.

  ✓ The power drops by one: naⁿ⁻¹. Check with n = 2: (x² − a²)/(x − a) = x + a → 2a = 2a¹.

• ✗ Rationalising by multiplying only the numerator by the conjugate.

  ✓ Multiply top and bottom by the same conjugate, so the value of the fraction does not change.

MCQs

1. 1. lim(x→−1) (x³ − x)/(x + 1) equals:

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

   (c) x³ − x = x(x − 1)(x + 1); cancel (x + 1) to get x(x − 1) → (−1)(−2) = 2.

2. 2. lim(x→1) (x⁴ − 1)/(x − 1) equals:

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

   (c) naⁿ⁻¹ with n = 4, a = 1: 4 × 1³ = 4.

3. 3. lim(x→2) (x³ − 8)/(x² + x − 6) equals:

   1. (a) 12/5
   2. (b) 0
   3. (c) 4/3
   4. (d) undefined
   Show answer

   (a) Cancel (x − 2): (x² + 2x + 4)/(x + 3) → 12/5.

4. 4. lim(x→0) (√(x + 4) − 2)/x equals:

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

   (b) 1/(2√a) with a = 4: 1/(2 × 2) = 1/4.

5. 5. Why may we cancel (x − a) when finding lim(x→a)?

   1. (a) because x − a = 0
   2. (b) because x ≠ a while x → a, so x − a ≠ 0
   3. (c) because a = 0
   4. (d) it is never allowed
   Show answer

   (b) x approaches a but never equals it, so x − a is never zero and can be cancelled.

Quick revision

• 0/0 means simplify: factorise and cancel, or rationalise, then substitute.
• lim(x→a) (xⁿ − aⁿ)/(x − a) = naⁿ⁻¹.
• lim(x→0) (√(x + a) − √a)/x = 1/(2√a).