The Squeeze Theorem

When a function is too wild to evaluate directly, trap it: pin it between two functions you can handle. If both of those converge to the same value, the one in the middle has nowhere else to go. That's the Squeeze Theorem — and it's the tool that proves the cornerstone limit \(\lim_{x \to 0} \frac{\sin x}{x} = 1\).

By the end you'll be able to evaluate limits of bounded oscillating functions by squeezing them between two bounds with a common limit, and explain why \(\lim_{x \to 0} \sin(x)/x = 1\).

Predict: ignore the green wiggle for a second and look only at the upper bound and the lower bound. As \(x \to 0\), what value do they BOTH close in on? That — before you've even looked at the middle function — is your prediction for its limit. Now drag the zoom slider and watch the envelope pinch the wiggle to exactly that value.

The middle function \(g(x) = x^2\sin(1/x)\) oscillates infinitely fast near 0 (the drawing under-samples there, but the true curve never leaves the envelope). Switch cases to see the same squeeze prove \(\sin(x)/x \to 1\), pinned between \(\cos x\) and \(1\). The open circle marks the hole at \(x = 0\): the function is undefined there, yet the limit exists. Hover the chart or the circle for exact values.

Window |x| ≤ 1.00 · squeeze: -1.0000 ≤ x²·sin(1/x) ≤ 1.0000 · both bounds → 0, so the trapped limit is 0

−x² ≤ x²·sin(1/x) ≤ x² — the envelope pinches to 0, dragging the wiggle with it
upper bound y = x² trapped middle x²·sin(1/x) lower bound y = −x² hole at x = 0, limit = 0

The Squeeze Theorem bounds a function between two others that share a common limit, forcing the middle function to that same limit — and it establishes the special trigonometric limit \(\lim_{x \to 0} \frac{\sin x}{x} = 1\).

Intuitive

If \(f\) is caught between two functions that both squeeze down to the same value \(L\), then \(f\) has nowhere else to go — it must go to \(L\) too. This is your tool of choice when \(f\) itself is hard to evaluate directly but easy to bound. The classic setup: something oscillating like \(\sin(1/x)\) is hopeless to track, but it never escapes \([-1, 1]\), so multiplying it by something shrinking hands you two bounds that pinch together. You watched exactly this in the visualizer: the upper and lower parabolas collapse onto 0, dragging the wiggle with them.

Formal

Suppose that near \(x = a\) — except possibly at \(a\) itself — we have \(g(x) \le f(x) \le h(x)\), and the outer functions share a common limit: \(\lim_{x \to a} g(x) = L = \lim_{x \to a} h(x)\). Then the middle function is trapped and forced to the same value: \(\lim_{x \to a} f(x) = L\). Two fine-print details do real work here: the bound only needs to hold near \(a\), not everywhere; and both outer limits must be the same \(L\) — if \(g\) and \(h\) converge to different values, the squeeze concludes nothing.

The special trigonometric limit

The theorem's most famous job is establishing \[ \lim_{x \to 0} \frac{\sin x}{x} = 1. \] A geometric argument on the unit circle gives, for small positive \(x\), \(\cos x \le \frac{\sin x}{x} \le 1\). Since \(\cos x \to 1\) and \(1 \to 1\) as \(x \to 0\), the middle term is squeezed to 1 — the second case in the visualizer above. This limit underlies the derivatives of sine and cosine (coming up in the next section) and the related result \(\lim_{x \to 0} \frac{1 - \cos x}{x} = 0\).

Which rescue tool, when?

You now have two rescues for limits that direct substitution can't do. When substitution gives 0/0 because the numerator and denominator share a vanishing factor, you factor-and-cancel or multiply by a conjugate — algebra removes the obstruction. When the obstruction is a bounded factor oscillating too fast to track, no algebra will simplify it: instead you bound it and squeeze. Ask "what's actually blocking substitution?" and pick the tool that matches.

Worked example

Evaluate \(\lim_{x \to 0} x^2 \sin(1/x)\). Substitution fails because \(\sin(1/x)\) oscillates wildly near 0. But sine is always bounded: \(-1 \le \sin(1/x) \le 1\). Multiplying through by \(x^2 \ge 0\) (which can't flip the inequalities): \[ -x^2 \le x^2\sin(1/x) \le x^2. \] Both bounds \(\to 0\) as \(x \to 0\), so by the Squeeze Theorem the limit is 0.

Your turn

Same playbook, a different wiggle: evaluate \(\lim_{x \to 0} x^4 \cos(1/x)\). Cosine is bounded: \(-1 \le \cos(1/x) \le 1\). Multiplying through by \(x^4 \ge 0\) gives \(-x^4 \le x^4\cos(1/x) \le x^4\). Both bounds \(\to\) ____ as \(x \to 0\), so the limit is ____.

Reveal the answer

Both \(-x^4\) and \(x^4\) go to \(0\) as \(x \to 0\), so the Squeeze Theorem forces \(\lim_{x \to 0} x^4\cos(1/x) = \) 0. Zoom the slider all the way in on the first case above — a fourth-power envelope pinches even faster than the squared one you're watching.

More info — the pitfalls that break a squeeze

Three ways a squeeze attempt goes wrong. First, using bounds that do not share a common limit — if the outer functions converge to different values, the theorem doesn't apply (that's why the quiz below refuses to conclude anything from \(-1 \le f(x) \le 1\)). Second, flipping direction: multiplying an inequality by a negative quantity reverses it, so track signs when you build the bounds — in the worked example we multiplied by \(x^2 \ge 0\) precisely so the inequalities kept their direction. Third, demanding too much: the bound only needs to hold near the point, not everywhere. For the full unit-circle derivation of \(\cos x \le \sin(x)/x \le 1\), read the OpenStax section linked in Dive deeper below.

Check your understanding

Question 1 of 4

Evaluate \(\lim_{x \to 0} x^2 \sin(5/x)\).

Question 2 of 4

What is \(\lim_{x \to 0} \dfrac{\sin x}{x}\), and why?

Question 3 of 4

Near \(x = 0\) you know only that \(-1 \le f(x) \le 1\). What does the Squeeze Theorem let you conclude about \(\lim_{x \to 0} f(x)\)?

Question 4 of 4

You face two limits: (a) \(\lim_{x \to 3} \dfrac{x^2 - 9}{x - 3}\) and (b) \(\lim_{x \to 0} x \sin(1/x)\). Which technique fits each?

Recap

  • If \(g(x) \le f(x) \le h(x)\) near \(a\) (except possibly at \(a\)) and \(\lim_{x \to a} g(x) = L = \lim_{x \to a} h(x)\), then \(\lim_{x \to a} f(x) = L\).
  • Both bounds must converge to the same limit — bounds with different limits conclude nothing.
  • Classic use: a bounded oscillation times a shrinking factor, like \(-x^2 \le x^2\sin(1/x) \le x^2 \Rightarrow\) limit 0.
  • The squeeze \(\cos x \le \frac{\sin x}{x} \le 1\) establishes \(\lim_{x \to 0} \frac{\sin x}{x} = 1\), which underlies the derivatives of sine and cosine.
  • Tool choice: factor/rationalize a 0/0 with a common factor; bound-and-squeeze a wild-but-bounded oscillation.

Dive deeper

Sources

  • The Squeeze Theorem