Unit 1 · Topic 1.8

1.8 Determining Limits Using the Squeeze Theorem

The Squeeze Theorem finds a limit by trapping a tricky function between two simpler ones that have the same limit. It is how you prove lim (x→0) sin x / x = 1 and how you handle a function like x²·sin(1/x) near 0.

Key terms

squeeze theorem
sandwich theorem
upper and lower bounds
lim (x→0) sin(x)/x = 1

The theorem, with its conditions

Suppose g(x) ≤ f(x) ≤ h(x) for all x in an open interval containing c (except possibly at c itself). If lim (x→c) g(x) = L and lim (x→c) h(x) = L, then lim (x→c) f(x) = L.

Both conditions matter: f must be trapped between the two functions near c, and the two bounding functions must have the same limit. If the bounds approach different numbers, you can't conclude anything about f. It is also called the Sandwich Theorem.

The classic use: something bounded times something going to 0

Sine and cosine are always between −1 and 1. So for x²·sin(1/x), multiply −1 ≤ sin(1/x) ≤ 1 by x², which is never negative: −x² ≤ x² sin(1/x) ≤ x².

Both −x² and x² approach 0 as x→0, so x² sin(1/x) is squeezed to 0. The same pattern works for x·cos(1/x) using −|x| and |x|.

Where sin x / x → 1 comes from

Draw the unit circle with a small positive angle x. Comparing the areas of a small triangle, the circular sector, and a larger triangle gives, for small x ≠ 0, cos x ≤ sin x / x ≤ 1. Both cos x and 1 approach 1 as x→0, so the Squeeze Theorem gives lim (x→0) sin x / x = 1.

This limit matters later: it is the key step in showing the derivative of sin x is cos x.

Choosing the bounds

The hard part of a squeeze is finding the two bounding functions. The usual move is to spot a factor that stays bounded, like sin(anything) or cos(anything) between −1 and 1, multiplied by a factor that goes to 0.

A compact version: if |f(x)| ≤ g(x) near c and g(x) → 0, then f(x) → 0. That's because −g(x) ≤ f(x) ≤ g(x), and both bounds approach 0. You'll see this with expressions like x·sin(1/x), (x − 2)·cos(1/(x − 2)) or sin(x)/x as x→∞.

If the problem hands you the inequality, your job is just to check that the two bounds have the same limit at c and then name the theorem.

Writing it up

A full squeeze argument has three parts: the inequality and where it holds, the limit of each bound, and the conclusion naming the Squeeze Theorem. Writing “by the Squeeze Theorem” is part of the justification.

Worked examples

Try each one yourself first, then open the solution.

Example 1
Trap: the product rule for limits fails here
Find lim (x→0) x² sin(1/x). Justify your answer.
Show the solutionHide the solution
1. Step 1: Direct substitution fails because sin(1/0) is undefined, and the limit of sin(1/x) doesn't exist, so the product property doesn't apply.
2. Step 2: For all x ≠ 0, −1 ≤ sin(1/x) ≤ 1. Multiply through by x², which is positive for x ≠ 0: −x² ≤ x² sin(1/x) ≤ x².
3. Step 3: lim (x→0) (−x²) = 0 and lim (x→0) x² = 0.
4. Step 4: The function is trapped between two functions that both approach 0.
Answer: By the Squeeze Theorem, lim (x→0) x² sin(1/x) = 0.
Example 2
Squeeze with given bounds
For all x, 4x − 9 ≤ f(x) ≤ x² − 4x + 7. Find lim (x→4) f(x).
Show the solutionHide the solution
1. Step 1: Check that the bounds make sense: (x² − 4x + 7) − (4x − 9) = x² − 8x + 16 = (x − 4)², which is never negative, so the lower bound really is below the upper bound.
2. Step 2: Lower bound: lim (x→4) (4x − 9) = 16 − 9 = 7.
3. Step 3: Upper bound: lim (x→4) (x² − 4x + 7) = 16 − 16 + 7 = 7.
4. Step 4: Both bounds approach 7 and f is between them.
Answer: By the Squeeze Theorem, lim (x→4) f(x) = 7.
Example 3
Squeeze at infinity
Find lim (x→∞) cos x / x.
Show the solutionHide the solution
1. Step 1: For x > 0, −1 ≤ cos x ≤ 1. Divide by x, which is positive: −1/x ≤ cos x / x ≤ 1/x.
2. Step 2: As x→∞, −1/x → 0 and 1/x → 0.
3. Step 3: The function is trapped between two functions that both approach 0.
Answer: By the Squeeze Theorem, lim (x→∞) cos x / x = 0, so y = 0 is a horizontal asymptote (the graph crosses it infinitely often).

Common mistakes

Splitting x² sin(1/x) into (lim x²)(lim sin(1/x)) = 0 × (something). The product property needs both limits to exist, and lim sin(1/x) doesn't.
Multiplying an inequality by a quantity that might be negative without flipping the signs. Use x² or |x|, which are never negative.
Using bounds whose limits are different. If g → 0 and h → 2, the Squeeze Theorem tells you nothing.

On the exam

Multiple-choice questions may give inequalities like the second example and ask for the limit, or ask which condition makes the Squeeze Theorem apply.
In free response, name the theorem and show both bounding limits. The limit sin x / x = 1 is something you should simply know.

Connected topics

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