Limits at infinity: only the biggest powers survive

The problem

A shopkeeper earns Rs 1,000,000 plus Rs 5 per customer, and spends Rs 2 per customer. With a million customers, does the million in the bank matter much? With a billion customers, hardly at all. For huge x, only the biggest terms decide the answer.

Limits at infinity make this exact. Divide everything by the highest power of x in the denominator; every smaller term becomes a/xᵖ, and those all shrink to zero.

For a positive rational p with xᵖ defined, lim(x→±∞) a/xᵖ = 0. To find a limit at infinity, divide the numerator and denominator by the highest power of x in the denominator, then use this theorem.

Key terms

1/x → 0

As x grows without end, 1/x gets as close to zero as we please. The same happens as x → −∞.

a/xᵖ → 0

Any constant divided by a positive power of x goes to zero, e.g. 3/x, −5/x², 6/x^(1/5).

Divide by the highest power

Divide each term of the numerator and denominator by the highest power of x in the denominator. Every leftover a/xᵖ vanishes, leaving the answer.

√x² = |x|

Inside a square root, x² comes out as |x|. For x → −∞, |x| = −x, so divide by −x. This changes the sign of the answer.

Short questions with model answers

1. Q1. Evaluate lim(x→−∞) (4x⁴ − 5x³)/(3x⁵ + 2x² + 1).

   • ÷ x⁵: (4/x − 5/x²)/(3 + 2/x³ + 1/x⁵)
   • (0 − 0)/(3 + 0 + 0) = 0

   0

2. Q2. Evaluate lim(x→−∞) (2 − 3x)/√(3 + 4x²) and lim(x→+∞) (2 − 3x)/√(3 + 4x²).

   • x < 0: √x² = −x ⇒ ÷ (−x): (−2/x + 3)/√(3/x² + 4) → 3/√4 = 3/2
   • x > 0: √x² = x ⇒ ÷ x: (2/x − 3)/√(3/x² + 4) → −3/2

   x → −∞: 3/2 · x → +∞: −3/2

3. Q3. Evaluate lim(x→+∞) (2x² − 3)/(5x² + 4).

   • ÷ x²: (2 − 3/x²)/(5 + 4/x²)
   • (2 − 0)/(5 + 0) = 2/5

   2/5

Common mistakes

• ✗ Writing √x² = x when x → −∞.

  ✓ √x² = |x|, and for negative x that is −x. Missing this flips the sign of the answer: 3/2 becomes −3/2.

• ✗ Saying ∞/∞ = 1.

  ✓ ∞/∞ is indeterminate. (2x² − 3)/(5x² + 4) has top and bottom both growing, yet its limit is 2/5, not 1.

• ✗ Dividing only the denominator by the highest power of x.

  ✓ Divide every term of both numerator and denominator by the same power, so the fraction's value is unchanged.

MCQs

1. 1. lim(x→+∞) 1/x equals:

   1. (a) 1
   2. (b) 0
   3. (c) ∞
   4. (d) does not exist
   Show answer

   (b) 1/x gets as close to 0 as we please when x is large enough.

2. 2. lim(x→−∞) (4x⁴ − 5x³)/(3x⁵ + 2x² + 1) equals:

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

   (b) The top's degree (4) is below the bottom's (5), so the limit is 0.

3. 3. lim(x→−∞) (2 − 3x)/√(3 + 4x²) equals:

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

   (a) Divide by −x (since √x² = −x for x < 0): 3/√4 = 3/2.

4. 4. lim(x→+∞) (2x² − 3)/(5x² + 4) equals:

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

   (c) Equal degrees: the ratio of leading coefficients, 2/5.

5. 5. For positive rational p, lim(x→∞) 6/xᵖ equals:

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

   (c) By the theorem, a constant over a positive power of x tends to 0.

Quick revision

• lim(x→±∞) a/xᵖ = 0 for positive rational p.
• Divide top and bottom by the highest power of x in the denominator; smaller degree on top gives 0, equal degrees give the ratio of leading coefficients.
• With roots, √x² = |x|: for x → −∞ divide by −x, which can change the sign of the limit.