§ 01The Intuition — Trapped Between Two Friends
Imagine three people walking towards the same doorway, shoulder to shoulder. The person in the middle cannot go anywhere except through the same door as the two on either side. That is the Squeeze Theorem.
Some limits cannot be computed by substitution, factoring, or L'Hôpital's Rule, particularly those involving functions that oscillate wildly near the limit point (like sin(1/x) near 0) or products that create 0 · (something unbounded). In these situations, direct analysis of the target function breaks down entirely.
The Squeeze Theorem sidesteps the problem. Instead of evaluating the troublesome function f directly, we find two "bounding" functions g and h that are simpler and whose limits we already know. If f is always sandwiched between g and h near the point of interest, and g and h converge to the same value L, then f is logically forced to converge to L as well.
The key insight is that we never need to understand f's behaviour in full detail. All we need to know is that it is bounded between two functions whose limits are known and equal.
§ 02The Formal Statement
Here is the theorem in full. Note that it works for two-sided limits, one-sided limits, and limits at infinity, the setup is identical in all cases.
Three conditions must all hold:
Checklist for Applying the Squeeze Theorem
- The inequality g(x) ≤ f(x) ≤ h(x) holds in some open interval containing a (it need not hold at a itself).
- Both outer limits exist and are equal: limx→a g(x) = limx→a h(x) = L.
- Conclude: limx→a f(x) = L. The middle function has no choice.
§ 03Geometric Intuition
On a graph, the Squeeze Theorem says that if f(x) is always between the graphs of g(x) and h(x), and both g and h funnel into the same point (a, L), then f must also pass through (a, L): or at minimum approach L as x → a.
Notice that f oscillates unpredictably away from a, we could never compute its limit by substitution or by any algebraic manipulation. Yet because it is pinned between g and h, and both g and h arrive at L, f has no alternative destination. The geometry makes the conclusion feel inevitable.
§ 04The Strategy — How to Find the Bounds
The hardest part of applying the Squeeze Theorem is not invoking it, it is finding the right bounding functions g and h in the first place. The following strategy covers the vast majority of exam and textbook problems.
4-Step Strategy for Squeeze Theorem Problems
- Identify the oscillating or unbounded part. Usually this is a trig function (sin, cos, tan) whose argument blows up, or a factor that oscillates. The key property is that it is bounded: |sin(anything)| ≤ 1 and |cos(anything)| ≤ 1 always.
- Write the bounding inequality for the oscillating part. For example: −1 ≤ sin(1/x) ≤ 1, or −1 ≤ cos(x²) ≤ 1.
- Multiply through by the remaining non-negative factor to produce bounds on f(x) as a whole. Be careful: if you multiply by a factor that can be negative, the inequalities flip and you need to handle cases.
- Take the limit of both outer bounds. If they both equal L, invoke the Squeeze Theorem and conclude lim f(x) = L.
This single fact, combined with multiplication by x², |x|, or any non-negative expression, generates the bounding inequalities for nearly every standard squeeze problem.
§ 05Worked Examples
Direct substitution is impossible: at x = 0, sin(1/x) is undefined and oscillates infinitely. Factoring or L'Hôpital don't help. Use the squeeze.
- Bound the oscillating part: −1 ≤ sin(1/x) ≤ 1 for all x ≠ 0.
- Multiply by x² ≥ 0: −x² ≤ x²·sin(1/x) ≤ x².
- Take limits of bounds: limx→0 (−x²) = 0 and limx→0 x² = 0.
- Both bounds → 0, so by the Squeeze Theorem: limx→0 x²·sin(1/x) = 0.
- −1 ≤ cos(1/x) ≤ 1 for all x ≠ 0.
- Multiply by |x| (we need the factor to be non-negative for safe multiplication): −|x| ≤ x·cos(1/x) ≤ |x|.
- limx→0 (−|x|) = 0 and limx→0 |x| = 0.
- Both → 0, so limx→0 x·cos(1/x) = 0.
- −1 ≤ cos(2/x) ≤ 1.
- Multiply by x⁴ ≥ 0: −x⁴ ≤ x⁴·cos(2/x) ≤ x⁴.
- Both bounds → 0 as x → 0.
This is the most important limit in calculus. It cannot be proved by L'Hôpital (that would be circular, L'Hôpital requires the derivative of sin, which itself requires this limit). The Squeeze Theorem provides the rigorous foundation.
We use a geometric argument on the unit circle. For 0 < x < π/2, compare the areas of three regions: triangle OAP, circular sector OAP, and triangle OAT (where O is the origin, A = (1,0), P = (cos x, sin x), T = (1, tan x)).
- Compute each area: Area(△OAP) = ½·1·sin x = sin x / 2. Area(sector OAP) = x / 2. Area(△OAT) = ½·1·tan x = tan x / 2.
- The area inequality gives: sin x / 2 ≤ x / 2 ≤ tan x / 2.
- Multiply through by 2/sin x (positive for 0 < x < π/2): 1 ≤ x/sin x ≤ 1/cos x.
- Take reciprocals (reverse inequalities): cos x ≤ sin x / x ≤ 1.
- As x → 0⁺: lim cos x = 1 and lim 1 = 1. By the Squeeze Theorem: lim sin x / x = 1.
- The result for x → 0⁻ follows by symmetry (sin and x are both odd functions, so sin x / x is even).
This result also follows from the Squeeze-derived sin(x)/x = 1, via an algebraic identity trick.
- Multiply numerator and denominator by (1 + cos x):
- (1 − cos x)(1 + cos x) / [x(1 + cos x)] = sin²x / [x(1 + cos x)]
- Rewrite: [sin x / x] · [sin x / (1 + cos x)]
- As x → 0: [1] · [0 / 2] = 0
Now x → ∞. The numerator oscillates between −1 and 1 but the denominator grows without bound.
- −1 ≤ sin(x) ≤ 1 for all x.
- Divide by x > 0 (for large x): −1/x ≤ sin(x)/x ≤ 1/x.
- limx→∞ (−1/x) = 0 and limx→∞ 1/x = 0.
- −1 ≤ sin(1/x) ≤ 1 for x ≠ 0.
- Multiply by √x ≥ 0: −√x ≤ √x·sin(1/x) ≤ √x.
- limx→0⁺ (−√x) = 0 and limx→0⁺ √x = 0.
The expression inside can range from 2 + (−1) = 1 to 2 + 1 = 3.
- −1 ≤ sin(1/x²) ≤ 1, so 1 ≤ 2 + sin(1/x²) ≤ 3.
- Multiply by x² ≥ 0: x² ≤ x²·(2 + sin(1/x²)) ≤ 3x².
- limx→0 x² = 0 and limx→0 3x² = 0.
- Both bounds → 0, so the middle → 0.
§ 06The Squeeze Theorem for Sequences
The Squeeze Theorem applies equally to sequences. If aₙ ≤ bₙ ≤ cₙ for all sufficiently large n, and limn→∞ aₙ = limn→∞ cₙ = L, then limn→∞ bₙ = L.
- −1 ≤ sin(n) ≤ 1 for all n.
- Divide by n > 0: −1/n ≤ sin(n)/n ≤ 1/n.
- limn→∞ (−1/n) = 0 and limn→∞ 1/n = 0.
Rewrite: n·sin(1/n) = sin(1/n)/(1/n). Let t = 1/n. As n → ∞, t → 0⁺.
- limt→0⁺ sin(t)/t = 1 (the classic result from §05)
§ 07Common Mistakes
| Mistake | What Goes Wrong | Fix |
|---|---|---|
| Multiplying an inequality by a factor that could be negative | If −1 ≤ sin(1/x) ≤ 1 is multiplied by x, and x can be negative, the inequalities flip and you get the wrong bounds. | Multiply by |x| or x² instead, or split into cases x > 0 and x < 0. Always check the sign of the multiplier. |
| Using bounds that don't converge to the same value | If lim g ≠ lim h, the theorem gives no conclusion at all — not even that the limit doesn't exist. | Both outer limits must equal the same L. If they differ, find tighter bounds or use a different method. |
| Applying the theorem to the wrong variable | Writing bounds in terms of a different variable than the limit variable, creating an invalid chain of inequalities. | All three functions — g, f, and h — must be functions of the same variable approaching the same limit point. |
| Claiming lim f = lim g when f ≤ g (without an upper bound) | Knowing only f ≤ g is not enough — you need both a lower and an upper bound that share the same limit. | Always provide two bounds: one from below and one from above. |
| Using the theorem when f is not actually between g and h | The bounds must hold in a full (deleted) neighbourhood of a, not just at isolated points. | Verify the inequality holds for all x sufficiently close to a (or for all large n for sequences). |
§ 08Quick Reference — Key Results via Squeeze
| Limit | Bounds Used | Result |
|---|---|---|
| limx→0 x²·sin(1/x) | −x² ≤ f ≤ x² | 0 |
| limx→0 x·cos(1/x) | −|x| ≤ f ≤ |x| | 0 |
| limx→∞ sin(x)/x | −1/x ≤ f ≤ 1/x | 0 |
| limx→∞ cos(x)/x² | −1/x² ≤ f ≤ 1/x² | 0 |
| limx→0 sin(x)/x | cos x ≤ f ≤ 1 (geometric proof) | 1 |
| limx→0 (1 − cos x)/x | Derived from sin(x)/x | 0 |
| limn→∞ sin(n)/n | −1/n ≤ f ≤ 1/n | 0 |
| limn→∞ n·sin(1/n) | Substitution t=1/n → sin(t)/t | 1 |
§ 09Practice Quiz — The Squeeze Theorem
10 questions covering the theorem statement, bounding strategy, and standard applications. Immediate feedback on every answer.
The Squeeze Theorem concludes lim f(x) = L when:
You need both a lower bound g and an upper bound h, with lim g = lim h = L. A single bound is not sufficient, the theorem requires a sandwich.
Evaluate: limx→0 x²·cos(5/x)
−1 ≤ cos(5/x) ≤ 1. Multiply by x² ≥ 0: −x² ≤ x²cos(5/x) ≤ x². Both bounds → 0. Answer: 0.
Why can we not evaluate limx→0 x·sin(1/x) by direct substitution?
sin(1/x) oscillates between −1 and 1 infinitely rapidly as x → 0, and is undefined at x = 0 itself. There is no single value it approaches: the Squeeze Theorem bypasses this by bounding it.
Evaluate: limx→∞ cos(x²)/x
−1 ≤ cos(x²) ≤ 1. Divide by x > 0: −1/x ≤ cos(x²)/x ≤ 1/x. Both bounds → 0 as x → ∞. Answer: 0.
In the geometric proof of limx→0 sin(x)/x = 1, which area inequality is used?
The inner triangle △OAP has the smallest area, the circular sector OAP sits in the middle, and the outer triangle △OAT (with the tangent point) has the largest area. This gives sin x / 2 ≤ x / 2 ≤ tan x / 2, which leads to the squeeze.
Evaluate: limx→0 x⁴·sin(1/x³)
−1 ≤ sin(1/x³) ≤ 1. Multiply by x⁴ ≥ 0: −x⁴ ≤ x⁴sin(1/x³) ≤ x⁴. Both bounds → 0. Answer: 0.
When multiplying −1 ≤ sin(1/x) ≤ 1 by a factor that contains x, which factor is safe to use without flipping the inequalities?
Multiplying an inequality by a non-negative quantity preserves the direction. x² ≥ 0 always, so it is safe. x can be negative (for x < 0), which would flip the inequalities. Always check the sign of the multiplier.
What is limx→0 sin(x)/x?
limx→0 sin(x)/x = 1. This is proved rigorously using the Squeeze Theorem via a geometric area argument on the unit circle. It is the foundation for the derivative of sin(x).
Using limx→0 sin(x)/x = 1, evaluate: limx→0 sin(4x)/(2x)
Rewrite: sin(4x)/(2x) = (4x/2x)·(sin(4x)/4x) = 2·[sin(4x)/4x]. Let u = 4x; as x→0, u→0. So = 2·limu→0 sin(u)/u = 2·1 = 2.
If g(x) ≤ f(x) ≤ h(x) near a but lim g(x) = 3 and lim h(x) = 5, what can you conclude about lim f(x)?
The Squeeze Theorem requires lim g = lim h = L (the same value). If the outer limits differ, the theorem gives no information about lim f. The limit of f may or may not exist: we simply cannot conclude anything from this setup alone.
Quiz complete!