Squeeze Theorem

Core Idea

The Squeeze Theorem states that if g(x) <= f(x) <= h(x) near x = a, and lim g(x) = lim h(x) = L, then lim f(x) = L. The function f is "squeezed" between two functions that converge to the same limit. The most famous application is proving lim(x->0) sin(x)/x = 1, a result needed for the derivative of sin(x).

How It's Best Learned

Prove lim(x->0) sin(x)/x = 1 geometrically using the unit circle area argument. Then apply the squeeze theorem to related limits like lim(x->0) (1 - cos(x))/x = 0. Practice identifying bounding functions in other squeeze theorem problems (e.g., x² * sin(1/x) near 0).

Common Misconceptions

• Trying to apply the squeeze theorem when the bounding functions do not converge to the same limit.
• Confusing the squeeze theorem with limit comparison (they are different tools).
• Not verifying the inequality g(x) <= f(x) <= h(x) holds near the point of interest.

Explainer

From your study of limit laws, you know how to compute limits of sums, products, and quotients algebraically. But some functions resist those techniques — particularly functions like sin(x)/x near x = 0, where direct substitution gives 0/0, an indeterminate form. The Squeeze Theorem is a geometric argument that sidesteps algebraic manipulation by trapping the function between two simpler ones.

The logic is intuitive: if you know a value lies between 3.99 and 4.01, you know it's close to 4. The Squeeze Theorem makes this precise for limits. If g(x) ≤ f(x) ≤ h(x) for all x near a (but not necessarily at a), and both g and h approach the same limit L as x → a, then f is permanently trapped between values approaching L — so f must also approach L. There is no room for f to wander off to a different value.

The canonical application is proving lim(x→0) sin(x)/x = 1, which is essential for deriving d/dx[sin x] = cos x. The proof uses a geometric argument on the unit circle: comparing the areas of a small triangle, a circular sector, and a larger triangle, one obtains sin x ≤ x ≤ tan x for small positive x (in radians). Dividing through by sin x and taking reciprocals gives the bounds cos x ≤ sin(x)/x ≤ 1. Since lim(x→0) cos x = 1 and lim(x→0) 1 = 1, the Squeeze Theorem forces lim(x→0) sin(x)/x = 1. Every trigonometric derivative you will compute from this point onward rests on this single limit.

A second class of applications involves functions like f(x) = x² sin(1/x) near x = 0. Direct substitution fails (1/x is undefined at 0), and algebraic manipulation is impossible because sin(1/x) oscillates infinitely rapidly. But you know −1 ≤ sin(1/x) ≤ 1 always, so −x² ≤ x² sin(1/x) ≤ x². Both bounding functions approach 0 as x → 0, so the limit is 0. The bounding strategy — find a function that is always ≥ f and one always ≤ f, both with the same limit — is the skill to internalize. The bounds do not need to be tight everywhere, only near the point of interest.