Short on time? This page is the recap. For the proofs, the graphs and worked examples, use the full study guide. This topic is shared by Calculus AB and Calculus BC.
Notation: lim (x → a) f(x) means “the limit as x approaches a of f(x)”. |x| is the absolute value of x. Angles are in radians.
Recap
- Use the squeeze theorem when the limit laws fail, usually because one factor (such as sin(1/x)) has no limit.
- Theorem. If g(x) ≤ f(x) ≤ h(x) for all x in an open interval around a (except possibly at a), and g and h both tend to L, then f tends to L.
- Three conditions to show: the inequality; that it holds on both sides of a, close to a; equal outer limits.
- If the outer limits differ, the theorem gives no conclusion. It does not prove the limit fails to exist.
- The theorem gives the limit, never the value f(a).
Key relationships
| Idea | Statement | Use |
|---|---|---|
| Starting bound | −1 ≤ sin u ≤ 1, −1 ≤ cos u ≤ 1 for any u | Build bounds for sin(1/x), cos(4/x) and so on |
| Never-negative factor | Multiply by x², x⁴ or |x|: signs stay | −x² ≤ x² sin(1/x) ≤ x² |
| Factor that can be negative | |x cos u| ≤ |x| gives −|x| ≤ x cos u ≤ |x| | Valid on both sides of 0 |
| Key limit 1 | lim (x → 0) (sin x)/x = 1 | From cos x ≤ (sin x)/x ≤ 1 near 0 |
| Key limit 2 | lim (x → 0) (1 − cos x)/x = 0 | From −|x|/2 ≤ (1 − cos x)/x ≤ |x|/2 |
| Matching form | (sin kx)/x = k × (sin kx)/(kx) → k | Angle and denominator must match |
Assumptions behind the method
- The inequality must hold for every x near a, on both sides, not just at a few points.
- Both outer limits must exist and be equal.
- The trig results need radians. In degrees, (sin x)/x → π/180 instead of 1.
Mistakes to avoid
- Using the product law when one factor has no limit.
- Writing −x ≤ x cos u ≤ x. That is wrong for x < 0. Use |x|.
- “Different bounds, so the limit does not exist.” No conclusion is the correct conclusion.
- Claiming f(a) from the squeeze. Only the limit follows.
- “(sin 5x)/x → 1”. Match the angle: the limit is 5.
- Not naming the theorem. Write “by the squeeze theorem” and show both outer limits.
Quick self-check
- Find lim (x → 0) x² cos(7/x). (0: −x² ≤ x² cos(7/x) ≤ x², and both bounds tend to 0)
- Find lim (x → 0) (sin 6x)/(2x). (3: write 3 × (sin 6x)/(6x))
- For x ≠ 1 near 1, 1 − (x − 1)² ≤ f(x) ≤ 1 + 2|x − 1|. What is lim (x → 1) f(x)? (1: both bounds tend to 1)
- Find lim (x → 0) (1 − cos x)/(4x). (0: it is (1/4) × (1 − cos x)/x)
Next: practice questions.