Learning Objectives
Choose among squeeze, trigonometric, leading-term, and conjugate methods.
Justify each limit using conditions that actually hold.
Explain when a proposed method gives insufficient information.
Prerequisites: both introductory AP Calculus sections. Trigonometric inputs are in radians. This review uses algebra and established limit facts, without derivative-based methods.
1. Match the Expression to the Method
Feature | Likely method |
|---|---|
Bounded oscillation times a shrinking factor | Squeeze using an absolute-value bound |
sin(kx) divided by x near zero | Match the angle with its denominator |
1 − cos(kx) near zero | Conjugate or known cosine limit |
Rational expression at infinity | Compare leading terms |
Difference of radical expressions | Multiply by a conjugate |
A method is useful only when its hypotheses hold. The product law needs individual finite limits; the Squeeze Theorem needs matching outer limits; the fundamental sine limit needs radian inputs. State the relevant condition before calculating.
2. Combine Two Familiar Patterns
Worked Example: Trigonometry and Algebra
Evaluate lim(x → 0) [sin(2x)/x] × [(√(1 + x) − 1)/x].
The first factor is 2[sin(2x)/(2x)], so it approaches 2.
Rationalizing the second factor gives 1/[√(1 + x) + 1], which approaches 1/2.
Both finite limits exist, so the product law applies and the result is 2 × 1/2 = 1. Each original denominator is nonzero near, but not at, the target.
3. Identify an Invalid Argument
A learner writes lim(x → 0) x sin(1/x) = 0 × lim(x → 0) sin(1/x) = 0. The conclusion happens to be correct, but the displayed product-law argument is invalid because sin(1/x) has no limit at zero. A valid justification is |x sin(1/x)| ≤ |x|, followed by the Squeeze Theorem.
Common Mistakes
A correct numerical answer does not repair an invalid limit-law step.
At negative infinity, replacing |x|/x by 1 gives the wrong sign.
Bounds with different limits provide no squeeze conclusion, even if the middle function independently has a limit.
4. Mixed Practice
1. Find lim(x → 0) x² cos(7/x). State the bounds.
2. Find lim(x → 0) sin(6x)/sin(2x).
3. Find lim(x → 0) (1 − cos(4x))/x².
4. Find lim(x → −∞) (7x² − 3)/(2x² + x).
5. Find lim(x → −∞) √(16x² + 1)/x.
6. Find lim(x → +∞) [√(x² + 8x + 3) − x].
7. Suppose −|x| ≤ f(x) − 3 ≤ x² near zero. Find the limit of f(x) at zero.
Worked Solutions
1. The bounds are −x² and x², both approaching zero. The limit is 0.
2. Write the expression as 3[sin(6x)/(6x)][2x/sin(2x)]. Both bracketed factors approach 1, giving 3.
3. Rewrite as 16[1 − cos(4x)]/(4x)². The result is 16(1/2) = 8.
4. Equal polynomial degrees give 7/2. Dividing every term by x² verifies the result.
5. Write the quotient as (|x|/x)√(16 + 1/x²). At negative infinity this tends to −4.
6. Rationalization gives (8x + 3)/[√(x² + 8x + 3) + x]. Divide by positive x to obtain (8 + 3/x)/[√(1 + 8/x + 3/x²) + 1]. The limit is 8/2 = 4.
7. Both bounds for f(x) − 3 approach zero. Hence f(x) − 3 approaches zero and f(x) approaches 3.
5. Summary
- Look for structure before choosing a limit technique.
- Use bounds for oscillation, identities for trigonometry, and leading terms for rational end behavior.
- Track signs and scale factors explicitly.
- Justify the method as well as the final number.