AP Calculus AB and BC

When the Squeeze Theorem Fails to Apply

Q: What is the classic use of the Squeeze Theorem?

Showing x^2 \sin(1/x) \to 0 as x \to 0. The bounds -x^2 \le x^2 \sin(1/x) \le x^2 both tend to 0, so the common limit is 0. Notice the difference from the counterexample above: here the bounds close in, and there they do not.

Q: Can the Squeeze Theorem prove a limit does not exist?

No. It only ever concludes that a limit exists and equals L. To show a limit does not exist you need a different argument, usually two paths giving different values, or an oscillation that never settles.

The Squeeze Theorem needs two things: the inequality has to hold near the point, and the two bounding functions have to share the same limit there. Bounds that merely exist are worthless: sin(1/x) sits between -1 and 1 everywhere and has no limit at 0 at all.

Squeeze Theorem

g(x) \le f(x) \le h(x) \ \text{ and } \lim_{x \to a} g = \lim_{x \to a} h = L \implies \lim_{x \to a} f = L

If f is trapped between g and h near a, and g and h both approach the same value L at a, then f is forced to approach L as well.

The hypotheses, and what each one buys

A theorem is only as strong as its conditions. Below, each hypothesis is dropped on its own while every other one is kept, so you can see precisely what it was holding up.

1
The two bounds share a common limit at a
$\lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L$
The squeeze is only as tight as the gap between the bounds. If that gap does not close, the trapped function has room to do whatever it likes, and no conclusion follows. This is the hypothesis students skip, because finding some pair of bounds feels like progress.
Drop it and the theorem fails
$-1 \le \sin\!\left(\frac{1}{x}\right) \le 1$
sin(1/x) trapped between -1 and 1 near x = 0
The inequality is perfectly true, and both bounds are constants with limits of their own. But the limits are $-1$ and 1, and they do not agree. The trapped function has no limit at 0 whatever: however close you look, it still swings across the full range from $-1$ to 1. A valid squeeze would have proved a limit exists, and here none does.
2
The inequality holds on an interval around a, except possibly at a itself
$g(x) \le f(x) \le h(x) \text{ for } 0 < |x - a| < \delta$
The bounds have to hold on a whole punctured neighbourhood, not at a scattering of convenient points. What happens at $a$ itself is irrelevant, which is the usual situation for limits, but what happens on every side of it is not. A pair of bounds that only holds where you happened to check gives you nothing, and no elementary formula displays that failure cleanly, so none is claimed here.
No counterexample built from an elementary formula breaks this one on its own, so none is claimed here.

The counterexample above is checked numerically on every build: each function is evaluated and the conclusion is confirmed to fail. A counterexample that stopped working would fail the build rather than sit here misleading you.

Why it is true

Fix a tolerance $\varepsilon > 0$ . Since $g \to L$ , there is a radius inside which $g(x) > L - \varepsilon$ .
Since $h \to L$ , there is a radius inside which $h(x) < L + \varepsilon$ . Take the smaller of the two radii, and shrink again if needed so the inequality $g \le f \le h$ holds there too.
Inside that radius, $L - \varepsilon < g(x) \le f(x) \le h(x) < L + \varepsilon$ , so $|f(x) - L| < \varepsilon$ .
Since $\varepsilon$ was arbitrary, $f \to L$ . Notice that the argument used the common value L in both directions; two different limits leave the chain broken in the middle.

What it does not say

Any pair of bounds will do.

Only bounds whose limits agree. Bounding a function between $-1$ and 1 is always possible for a bounded function and always proves nothing.

It works for limits at infinity only, or at finite points only.

It works for both, and for one-sided limits, and for sequences. What matters is that the same limiting process is applied to all three functions.

It gives the limit of f when f has no limit.

If the hypotheses hold, $f$ genuinely has the limit L. If $f$ has no limit, some hypothesis failed, and the usual culprit is that the bounds did not converge to a common value.

Frequently asked questions

What is the classic use of the Squeeze Theorem?

Showing $x^2 \sin(1/x) \to 0$ as $x \to 0$ . The bounds $-x^2 \le x^2 \sin(1/x) \le x^2$ both tend to 0, so the common limit is 0. Notice the difference from the counterexample above: here the bounds close in, and there they do not.

Can the Squeeze Theorem prove a limit does not exist?