Determining Limits Using the Squeeze Theorem: Study Guide (Calculus AB 1.8)

Course scope. This topic is shared by Calculus AB and Calculus BC. Everything here is examinable for both.

A note on notation

This page has no equation renderer, so limits are written in a compact form:

lim (x → 0) f(x) means “the limit as x approaches 0 of f(x)”.

On paper, write the usual form with “x → 0” under “lim”. The vertical bars in |x| mean the absolute value of x, so |x| is never negative. All angles are in radians.

When rewriting is not enough

In Topic 1.6 you turned 0/0 limits into easier ones by factoring, using conjugates or using identities. Some limits do not give in to any of these.

Example A. lim (x → 0) (sin x)/x. Substitution gives 0/0. But sin x has no factor of x to cancel, and no identity removes the problem.

Example B. lim (x → 0) x² sin(1/x). You might try the product law: “limit of x² times limit of sin(1/x)”. That fails, because sin(1/x) has no limit at 0. As x → 0, 1/x grows without bound, so sin(1/x) swings between −1 and 1 infinitely often. The product law only applies when both limits exist.

Both limits do exist. To find them you need a different tool: trap the awkward function between two simpler ones.

The squeeze theorem

Squeeze theorem. Suppose that, for all x in an open interval containing a (except possibly at x = a),

g(x) ≤ f(x) ≤ h(x)

and that lim (x → a) g(x) = L and lim (x → a) h(x) = L. Then lim (x → a) f(x) = L.

Think of g as a floor and h as a ceiling. Near a, the floor and the ceiling both move towards the same height L. The graph of f is stuck between them, so it has nowhere else to go.

Before you use the theorem, check three conditions. In a written answer, show each one.

1. The inequality. You have two functions g and h with g(x) ≤ f(x) ≤ h(x). The order matters: the lower bound must really be below f, and the upper bound must really be above it.
2. Where it holds. The inequality is true for every x close to a on both sides. It need not hold far away from a, and it need not hold at a itself.
3. Equal outer limits. The limits of g and h at a exist and are the same number L.

If condition 3 fails (say g → 1 and h → 4), the theorem says nothing. It does not say the limit of f fails to exist. It simply gives no answer.

Notice what the theorem does not tell you. It gives the limit of f at a, not the value f(a). Because the inequality only needs to hold for x ≠ a, f(a) could be anything, or f might not be defined at a at all.

How to build the bounds

Most squeeze questions start from one fact you already know:

−1 ≤ sin u ≤ 1 and −1 ≤ cos u ≤ 1, for every real number u.

Here u can be anything: 1/x, 4/x, 1/(x − 3). Then build up to f in small, legal steps.

• Multiply by a quantity that is never negative, such as x², x⁴ or |x|. The inequality signs stay the same. Example: −x² ≤ x² sin(1/x) ≤ x².
• Add or subtract a constant on all three parts. The signs stay the same.
• Watch out for factors that can be negative, such as x or x³. If you multiply −1 ≤ cos u ≤ 1 by x, the signs flip when x < 0. The safe route is absolute values: |x cos u| = |x| |cos u| ≤ |x|, so −|x| ≤ x cos u ≤ |x|. This holds for every x, on both sides of 0.

Then check that both bounds have the same limit. With bounds like ±x² or ±|x| at a = 0, both limits are 0, so the trapped function tends to 0 as well.

Worked example 1: a wiggle around a nonzero value

Question. Find lim (x → 0) [3 + x² sin(4/x)], justifying your answer.

1. Try the limit laws. The limit of 3 is 3. But lim (x → 0) sin(4/x) does not exist (it oscillates between −1 and 1), so the product law cannot be used on x² sin(4/x).
2. Start from a known inequality. For every x ≠ 0: −1 ≤ sin(4/x) ≤ 1.
3. Multiply by x². For x ≠ 0, x² > 0, so the signs stay the same: −x² ≤ x² sin(4/x) ≤ x².
4. Add 3. For every x ≠ 0: 3 − x² ≤ 3 + x² sin(4/x) ≤ 3 + x²
5. Check the conditions. The inequality holds for all x ≠ 0, so it holds on an open interval around 0, except at 0. The bounds are polynomials, so lim (x → 0) (3 − x²) = 3 and lim (x → 0) (3 + x²) = 3. The outer limits are equal.
6. Conclude. By the squeeze theorem, lim (x → 0) [3 + x² sin(4/x)] = 3.

Check. At x = 0.1 the function equals about 3.0075, and the bounds are 2.99 and 3.01. The value sits between them, as it must. At x = 0.01 it is about 2.99991.

Interpretation. Figure 1 shows this function. The function is not defined at x = 0, so its graph has a hole at (0, 3), yet the limit is 3.

Worked example 2: given bounds, and what they do not tell you

Question. A function f is not given by a formula. You are told that for 0 < x < 3, x ≠ 1,

x + 1 ≤ f(x) ≤ x² − x + 2.

(a) Find lim (x → 1) f(x). (b) What can you say about f(1)? (c) Can you find lim (x → 2) f(x)?

(a)

1. Is the inequality possible? The upper bound minus the lower bound is x² − 2x + 1 = (x − 1)², which is never negative. So the floor really is below the ceiling. (The two touch only at x = 1.)
2. Where does it hold? On 0 < x < 3 except x = 1. That is an open interval around 1, with only x = 1 removed. Condition 2 is met.
3. Outer limits. Both bounds are polynomials, so substitute: 1 + 1 = 2 and 1 − 1 + 2 = 2. They are equal.
4. Conclude. By the squeeze theorem, lim (x → 1) f(x) = 2.

(b) Nothing. The inequality was not claimed at x = 1, so f(1) could be 2, 7 or undefined. The limit and the function value are separate questions. (Topic 1.10 onwards looks at when they match.)

(c) No, not from this information. At x = 2 the bounds tend to 2 + 1 = 3 and 4 − 2 + 2 = 4. The outer limits are different, so condition 3 fails. All you know is that f(x) stays between about 3 and 4 near x = 2. The limit might be 3.5, or it might not exist. The squeeze theorem is silent.

The two trig limits you must know

The squeeze theorem proves two results that the rest of the course depends on. Both need x in radians.

Result 1: lim (x → 0) (sin x)/x = 1

Take 0 < x < π/2 and look at the unit circle in Figure 2. Three shapes share the corner O and the side OA of length 1.

1. Compare the areas. Triangle OAP has base 1 and height sin x, so area ½ sin x. The sector has area ½ × 1² × x = ½ x (this is where radians matter). Triangle OAT has base 1 and height tan x, so area ½ tan x. The shapes are nested, so

   sin x < x < tan x.

2. Divide by sin x, which is positive here: 1 < x / sin x < 1/cos x.

3. Take reciprocals (all parts positive, so the signs reverse):

   cos x < (sin x)/x < 1.

4. Negative x. Both cos x and (sin x)/x are even functions (replacing x by −x gives the same value). So the same inequality holds for −π/2 < x < 0.

5. Squeeze. The inequality holds for all x near 0 except 0. lim (x → 0) cos x = 1 and lim (x → 0) 1 = 1. So lim (x → 0) (sin x)/x = 1.

Check. sin(0.1)/0.1 ≈ 0.99833, and cos 0.1 ≈ 0.99500. The value sits between cos x and 1.

Result 2: lim (x → 0) (1 − cos x)/x = 0

Step 1 above gives sin t < t for 0 < t < π/2. For t ≥ π/2, sin t ≤ 1 < t. Since sin is odd, |sin t| ≤ |t| for every t. Now use the identity 1 − cos x = 2 sin²(x/2), which comes from cos 2θ = 1 − 2 sin²θ:

1 − cos x = 2 sin²(x/2) ≤ 2 (x/2)² = x²/2.

Also 1 − cos x ≥ 0. Divide by |x| (positive for x ≠ 0): 0 ≤ (1 − cos x)/|x| ≤ |x|/2. Since (1 − cos x)/x is either that value or its negative,

−|x|/2 ≤ (1 − cos x)/x ≤ |x|/2 for x ≠ 0.

Both bounds tend to 0, so lim (x → 0) (1 − cos x)/x = 0. (At x = 0.1 the value is about 0.04996, below the bound 0.05.)

Degrees warning. If x were measured in degrees, the first limit would be π/180 ≈ 0.01745, not 1. That is one reason calculus always uses radians.

Worked example 3: using the result for sin x / x

Question. Find lim (x → 0) (sin 3x)/(x + x cos x).

1. Substitution gives 0/0, so rewrite.
2. Factor the bottom: x + x cos x = x(1 + cos x).
3. Create the form sin u / u. Multiply and divide by 3: (sin 3x)/(x(1 + cos x)) = 3 × (sin 3x)/(3x) × 1/(1 + cos x), for x ≠ 0 near 0.
4. Take each limit. Let u = 3x. As x → 0, u → 0, so (sin 3x)/(3x) → 1. Also 1/(1 + cos x) → 1/2.
5. Product law (now every limit exists): 3 × 1 × 1/2 = 3/2.

Check. At x = 0.01 the expression is about 1.49981, close to 1.5.

A useful shortcut: bounded times something that tends to 0

Worked example 1 is an instance of a general pattern. Suppose |b(x)| ≤ M near a, for some fixed number M, and lim (x → a) z(x) = 0. Then

−M |z(x)| ≤ b(x) z(x) ≤ M |z(x)|,

and both bounds tend to 0. So b(x) z(x) → 0. On a free-response answer, show the inequality and name the squeeze theorem. “Bounded times zero” on its own is a slogan, not a justification.

The word bounded matters. The factor 1/x is not bounded near 0, and x × (1/x) = 1 for every x ≠ 0, so that limit is 1, not 0.

Common misconceptions

• “Use the product law on x² sin(1/x).” The product law needs both limits to exist. lim (x → 0) sin(1/x) does not, so the law does not apply. Use the squeeze theorem.
• Multiplying by x without care. From −1 ≤ cos u ≤ 1, writing −x ≤ x cos u ≤ x is wrong for x < 0. Use −|x| ≤ x cos u ≤ |x|.
• Different outer limits, so “the limit does not exist”. If the bounds tend to different values, the theorem gives no conclusion at all.
• Checking the inequality at a single point. It must hold for every x in an interval around a, on both sides (except possibly at a).
• “The squeeze gives f(a).” It gives the limit only. f(a) can differ, or be undefined.
• Cancelling the x in sin x / x. sin x is not x times anything. The value 1 comes from the squeeze argument, not from cancelling.
• Forgetting radians. sin x / x → 1 only when x is in radians.
• “lim (x → 0) (sin 5x)/x = 1.” The angle and the denominator must match. Rewrite as 5 × (sin 5x)/(5x), so the limit is 5.

Where this leads

Topic 1.9 brings graphs, tables, equations and words together, so you can find the same limit from whichever representation you are given. In Unit 2 the two trig limits proved here give the derivatives of sin x and cos x. Squeeze arguments also return when you meet limits at infinity later in Unit 1, for example to show that (sin x)/x → 0 as x grows without bound. Continue with Topic 1.9, Connecting Multiple Representations of Limits, or go back to Topic 1.7, Selecting Procedures for Determining Limits. Students in either course can return to the Calculus AB roadmap or the Calculus BC roadmap.

Try the practice questions now, then use the revision notes and the checklist to consolidate.