1.8 Determining Limits Using the Squeeze Theorem

The squeeze theorem says: if f(x) ≤ g(x) ≤ h(x) near a (except maybe at a) and lim_{x→a} f(x) = lim_{x→a} h(x) = L, then lim_{x→a} g(x) = L. To use it on a limit: 1. Find two simpler functions f and h that bound g from below and above for x near the limit point. 2. Prove the inequalities hold (often trig inequalities or algebraic manipulation). 3. Evaluate the limits of the bounds. If both equal the same L, conclude the middle function’s limit is L. Classic AP examples: use geometric/trig inequalities to show cos x ≤ 1 and −1 ≤ sin x ≤ 1 to get lim_{x→0} sin x / x = 1 and lim_{x→0} (1−cos x)/x = 0 (see CED EK LIM-1.E.2). Squeeze-theorem problems often appear on AP sections testing LIM-1.E; be ready to justify the bounds briefly. For guided practice and examples, check the Topic 1.8 study guide (https://library.fiveable.me/ap-calculus/unit-1/determining-limits-using-squeeze-theorem/study-guide/0Ax6y3Qku88ex24KGwiG) or try practice problems at Fiveable (https://library.fiveable.me/practice/ap-calculus).

The squeeze theorem says: if g(x) ≤ f(x) ≤ h(x) near a (except maybe at a) and lim_{x→a} g(x) = lim_{x→a} h(x) = L, then lim_{x→a} f(x) = L. You use it when direct substitution fails (like 0/0 or undefined) and you can bound the troublesome function between two simpler functions whose limits you know. Classic AP examples from the CED: lim_{x→0} sin x / x = 1 and lim_{x→0} (1 − cos x)/x = 0—both are shown by bounding with trig inequalities or geometric arguments. Use direct substitution whenever the function is continuous at the point (substitution gives the limit). Use the squeeze theorem when you can’t simplify algebraically but can find upper and lower bounds that squeeze to the same value. For an AP-focused study guide and worked examples, see the Topic 1.8 study guide (https://library.fiveable.me/ap-calculus/unit-1/determining-limits-using-squeeze-theorem/study-guide/0Ax6y3Qku88ex24KGwiG). For extra practice, try problems at (https://library.fiveable.me/practice/ap-calculus).

Squeeze theorem idea (step-by-step)—example: lim(x→0) (1 − cos x)/x = 0. 1) Find two functions that “trap” your function near the limit point. Use trig identity: 1 − cos x = 2 sin^2(x/2). So 0 ≤ 1 − cos x = 2 sin^2(x/2). 2) Bound sin^2 by a simpler function. For small u, |sin u| ≤ |u|, so 0 ≤ 2 sin^2(x/2) ≤ 2 (x/2)^2 = x^2/2. 3) Divide the inequality by x (for x ≠ 0), keeping sign info: for x→0 we can write 0 ≤ (1 − cos x)/x ≤ x/2 for x near 0 (taking x>0; similar for x<0 gives same limit). 4) Take limits of the two bounds as x→0: lim x→0 0 = 0 and lim x→0 x/2 = 0. 5) By the squeeze (sandwich) theorem, the middle expression has the same limit: lim x→0 (1 − cos x)/x = 0. This is exactly the kind of CED example (LIM-1.E.2). For another classic, use the unit-circle geometric bounds to squeeze sin x/x → 1 as x→0. For more worked examples and AP-style practice, see the Topic 1.8 study guide (https://library.fiveable.me/ap-calculus/unit-1/determining-limits-using-squeeze-theorem/study-guide/0Ax6y3Qku88ex24KGwiG) and try problems at (https://library.fiveable.me/practice/ap-calculus).

Think of the squeeze theorem like picking two simple “fences” that you already know squeeze your complicated function. Tips to pick them: - Look for obvious inequalities. If your function has sin, cos, or absolute values, use standard bounds: −1 ≤ sin x ≤ 1, |f(x)| ≤ g(x), or for 0https://library.fiveable.me/ap-calculus/unit-1/determining-limits-using-squeeze-theorem/study-guide/0Ax6y3Qku88ex24KGwiG). For lots more problems, try the AP practice bank (https://library.fiveable.me/practice/ap-calculus).

Squeeze theorem (sandwich theorem) formula: if f(x) ≤ g(x) ≤ h(x) for all x near a (except maybe at a) and lim_{x→a} f(x) = lim_{x→a} h(x) = L, then lim_{x→a} g(x) = L. How to apply (step-by-step): 1. Find two simpler functions f and h that bound your function g: f(x) ≤ g(x) ≤ h(x) for x near the limit point. 2. Compute the limits of f and h as x→a. 3. If those two limits are equal (call it L), conclude lim_{x→a} g(x) = L. Two classic AP examples (appear in the CED for Topic 1.8): - Show lim_{x→0} (sin x)/x = 1 by using geometric trig inequalities from the unit circle to get cos x ≤ (sin x)/x ≤ 1 for small |x|, then both bounds → 1. - Show lim_{x→0} (1 − cos x)/x = 0 by noting 0 ≤ (1 − cos x)/x ≤ x/2 (or another suitable bound) and both bounds → 0. On the AP exam you may be asked to justify bounds or use trig inequalities; be ready to state the bounding inequalities and compute the two outer limits. For a quick topic review, check the Fiveable study guide (https://library.fiveable.me/ap-calculus/unit-1/determining-limits-using-squeeze-theorem/study-guide/0Ax6y3Qku88ex24KGwiG). For more practice, use https://library.fiveable.me/practice/ap-calculus.

Use the squeeze theorem when you can find two simpler functions that “trap” your function and both have the same limit at the point. Practically, that often happens when: - The function oscillates but is bounded (like f(x)=sin x) and is multiplied by something that → 0 (e.g., x·sin(1/x) or sin x / x). - You recognize a standard trig limit (sin x / x → 1 or (1 − cos x)/x → 0) but can’t simplify algebraically, so you bound using trig inequalities or a unit-circle argument. - The expression is hard to algebraically simplify but you can produce obvious upper/lower bounds that converge to the same value. Before using squeeze, try algebraic factoring, limit laws, substitution, or L’Hôpital’s Rule (when applicable and conditions hold). If those fail or the function is naturally trapped between two simpler limits, squeeze is the right tool. On the AP exam you should state the inequalities, show both bounding limits equal the same value, and cite the squeeze theorem (see examples in the CED: sin x/x and (1−cos x)/x). For a quick review and examples, see the Topic 1.8 study guide (https://library.fiveable.me/ap-calculus/unit-1/determining-limits-using-squeeze-theorem/study-guide/0Ax6y3Qku88ex24KGwiG). For extra practice, try problems at (https://library.fiveable.me/practice/ap-calculus).

Use the unit-circle inequalities and the Squeeze Theorem. For 0 < x < π/2 the geometric facts give sin x < x < tan x = sin x / cos x. Divide the chain by x (positive) and rearrange to get cos x < sin x / x < 1. As x → 0+, cos x → 1 and the right-hand bound is 1, so by the Squeeze Theorem sin x / x → 1. For x → 0− use oddness: sin(−x)/(−x) = sin x / x, so the same limit holds from the left. Therefore lim_{x→0} (sin x)/x = 1. This is the classic CED illustrative example for Topic 1.8 (LIM-1.E.2). If you want a step-by-step geometric proof or an ε–δ justification, check the Topic 1.8 study guide on Fiveable (https://library.fiveable.me/ap-calculus/unit-1/determining-limits-using-squeeze-theorem/study-guide/0Ax6y3Qku88ex24KGwiG) and practice more problems at (https://library.fiveable.me/practice/ap-calculus).

Think of the squeeze theorem like a sandwich: if f(x) is always between two functions g(x) ≤ f(x) ≤ h(x), and both g and h get arbitrarily close to the same number L as x → a, then f has no room but to be close to L too. Intuitively, once g and h are within, say, ±ε of L for x near a, f—being stuck between them—must also lie within ±ε of L. A compact epsilon–delta sketch: for any ε>0 pick δ so both |g(x)−L|<ε and |h(x)−L|<ε when 0<|x−a|<δ. Then g(x)≥f(x)≥h(x) implies |f(x)−L|<ε, proving the limit is L. That’s exactly how we justify limits like lim_{x→0} (sin x)/x = 1 using trig bounds from the unit circle. For step-by-step examples and practice aligned to the AP CED (LIM-1.E), see the Topic 1.8 study guide (https://library.fiveable.me/ap-calculus/unit-1/determining-limits-using-squeeze-theorem/study-guide/0Ax6y3Qku88ex24KGwiG). For more problems, check Fiveable’s practice set (https://library.fiveable.me/practice/ap-calculus).

Good question—they’re different tools for different situations. - L’Hôpital’s Rule: use when a limit gives an indeterminate form 0/0 or ∞/∞ and both numerator and denominator are differentiable near the point. You can replace the original limit with the limit of their derivatives (provided that new limit exists). Good for algebraic or analytic expressions like (e^x−1)/x as x→0. - Squeeze (Sandwich) Theorem: use when you can bound your function between two simpler functions that have the same limit at the point. This is essential for trig limits and geometric arguments—e.g., sin x / x → 1 as x→0 is usually shown by trig inequalities and the unit-circle argument (CED LIM-1.E.2). Squeeze works even if derivatives don’t exist or L’Hôpital’s hypotheses fail. AP tip: know both. On the exam, use squeeze for trig/inequality arguments (Topic 1.8), and use L’Hôpital only when you can justify differentiability and an indeterminate form. For the Topic 1.8 study guide see (https://library.fiveable.me/ap-calculus/unit-1/determining-limits-using-squeeze-theorem/study-guide/0Ax6y3Qku88ex24KGwiG). For more practice, try the problems at (https://library.fiveable.me/practice/ap-calculus).

Think: you need two simpler functions that squeeze your function from below and above and both have the same limit at the point. How to get those bounds: 1. Look for known inequalities or simple comparisons. Common tools: -1 ≤ sin x ≤ 1, |sin x| ≤ |x| (near 0), 1−cos x ≥ 0 and 1−cos x ≤ x^2/2 (from Taylor/geometry), monotonicity, or algebraic manipulation (factor, multiply by conjugate). 2. Rewrite the function so you can apply a known bound. Example: for 0 < x < π/2, cos x ≤ (sin x)/x ≤ 1. So cos x ≤ sin x/x ≤ 1 and both cos x and 1 → 1 as x → 0, so sin x/x → 1 (this is the squeeze used in the CED example). 3. Use absolute-value bounds when sign isn’t fixed. If |f(x)| ≤ g(x) and g(x) → 0, then f(x) → 0. 4. Check endpoints and one-sided limits: make sure the inequalities hold on the interval you’re approaching from. If you want more worked examples and practice (including the two CED examples sin x/x and (1−cos x)/x), see the Topic 1.8 study guide (https://library.fiveable.me/ap-calculus/unit-1/determining-limits-using-squeeze-theorem/study-guide/0Ax6y3Qku88ex24KGwiG) and try problems at Fiveable practice (https://library.fiveable.me/practice/ap-calculus).

Take 0 < x < π/2 (same idea works for small negative x by symmetry). On the unit circle one gets the standard inequalities sin x < x < tan x = sin x / cos x. Divide through by x·(>0) to get (sin x)/x < 1 < (sin x)/(x·cos x). Equivalently, dividing the middle chain by (x/ sin x) or taking reciprocals carefully gives cos x < (sin x)/x < 1. As x → 0+, cos x → 1, so by the squeeze theorem (cos x ≤ (sin x)/x ≤ 1) we conclude lim_{x→0} (sin x)/x = 1. For x < 0 note (sin x)/x is even ((sin(−x))/(−x) = (sin x)/x), so the same limit holds from the left; hence the two-sided limit is 1. This is the classic squeeze-theorem proof used on the AP CED (Topic 1.8). For a quick review, see the Topic 1.8 study guide (https://library.fiveable.me/ap-calculus/unit-1/determining-limits-using-squeeze-theorem/study-guide/0Ax6y3Qku88ex24KGwiG). For extra practice, try related problems at (https://library.fiveable.me/practice/ap-calculus).

Think of the squeeze as “which one’s bigger?”—put the larger function on top and the smaller on the bottom. The goal is to find two simpler functions g(x) ≤ f(x) ≤ h(x) near the limit point, where both g and h have the same limit L; then f must also tend to L by the squeeze theorem (CED EK LIM-1.E.2). Quick checklist: - Identify simple bounds you already know (e.g., −1 ≤ sin x ≤ 1, cos x ≤ 1, or the geometric inequality sin x ≤ x ≤ tan x for 0https://library.fiveable.me/ap-calculus/unit-1/determining-limits-using-squeeze-theorem/study-guide/0Ax6y3Qku88ex24KGwiG) and try problems on Fiveable’s AP Calc practice page (https://library.fiveable.me/practice/ap-calculus).

If a limit gives 0/0, that just tells you it’s indeterminate—it doesn’t force one method. Pick the tool that fits the function. Quick checklist: - Can you algebraically simplify (factor, cancel, rationalize)? Do that first—it’s often enough. - Is the expression a trig limit like sin x/x or (1 − cos x)/x near 0? Those are classic squeeze-theorem cases (use inequalities or the unit-circle argument). See Topic 1.8 in the Fiveable study guide (https://library.fiveable.me/ap-calculus/unit-1/determining-limits-using-squeeze-theorem/study-guide/0Ax6y3Qku88ex24KGwiG). - Are numerator and denominator differentiable near the point and the derivative limit easier to find? Then L’Hôpital’s Rule is convenient: if lim f = lim g = 0 (or ±∞) and lim f′/g′ exists, then lim f/g = lim f′/g′. - If neither applies, try series expansion or bounding (squeeze). Remember AP expectations: use the squeeze theorem where bounding/inequalities are natural (trig examples in the CED), and use L’Hôpital when differentiation simplifies the limit. For more practice problems and examples, check the Unit 1 overview (https://library.fiveable.me/ap-calculus/unit-1) and the AP practice bank (https://library.fiveable.me/practice/ap-calculus).

You can check your squeeze setup quickly with three things: 1. Inequality valid on a punctured neighborhood: show g(x) ≤ f(x) ≤ h(x) for all x near a (except maybe at a). Don’t just test one x—either derive the inequality algebraically or argue from known facts (e.g., |sin x| ≤ 1, or 1 − cos x ≥ 0). 2. Both bounds have the same limit at a: compute lim_{x→a} g(x) and lim_{x→a} h(x). If they’re equal to L, you’ve met the second required hypothesis of the squeeze theorem (CED EK LIM-1.E.2). 3. Check endpoints / one-sided behavior: if you need a one-sided limit, make sure the inequalities hold from that side only. Also verify you didn’t accidentally reverse an inequality when multiplying/dividing by something that can change sign. Quick sanity checks: plug in values close to a to see the inequality numerically, and compare to standard squeezes (e.g., −1 ≤ sin x ≤ 1 gives −1/|x| ≤ sin x/x ≤ 1/|x| for x≠0) or the sin x / x and (1−cos x)/x examples in the AP study guide (see the Topic 1.8 study guide: https://library.fiveable.me/ap-calculus/unit-1/determining-limits-using-squeeze-theorem/study-guide/0Ax6y3Qku88ex24KGwiG). For more practice problems, use the AP practice bank (https://library.fiveable.me/practice/ap-calculus).

Good question—the squeeze theorem shows up when a function oscillates or is hard to simplify but you can tightly bound it. Common examples you should memorize and recognize on the AP exam: - sin x / x as x → 0 (limit = 1) and (1 − cos x)/x as x → 0 (limit = 0)—classic trig uses of trigonometric inequalities and the unit-circle geometric argument (CED: sin(x)/x limit, (1−cos x)/x limit). - x·sin(1/x) or x^2·sin(1/x) as x → 0: use −|x| ≤ x·sin(1/x) ≤ |x| to get limit 0. - Functions of the form f(x)·g(x) where g(x) oscillates between −1 and 1 (e.g., sin(1/x)) and f(x) → 0. - Sequence analogues a_n = b_n·sin(c_n) where b_n → 0—squeeze for sequences. On the AP exam you’ll need to justify bounds (inequalities) and connect them to the limit (LIM-1.E). For a focused review, see the Topic 1.8 study guide (https://library.fiveable.me/ap-calculus/unit-1/determining-limits-using-squeeze-theorem/study-guide/0Ax6y3Qku88ex24KGwiG). For extra practice, try problems at (https://library.fiveable.me/practice/ap-calculus).