The Squeeze Theorem

One function caught between two others

Some limits cannot be found by substituting or by rewriting. Take $f\left(x\right)=x^2\sin \left(\frac{1}{x}\right)$f(x) = x² sin(1/x) as $x\to 0$x → 0. At x = 0 the expression has no value, because $\frac{1}{0}$1/0 is undefined, and $\sin \left(\frac{1}{x}\right)$sin(1/x) cannot be factored away.

What can be said is that sin of any number lies between −1 and 1. So $-1\le \sin \left(\frac{1}{x}\right)\le 1$−1 ≤ sin(1/x) ≤ 1 for every $x\ne 0$x ≠ 0. Multiply through by $x^2$x², which is positive for $x\ne 0$x ≠ 0, so the inequality signs stay as they are: $-x^2\le x^2\sin \left(\frac{1}{x}\right)\le x^2$−x² ≤ x² sin(1/x) ≤ x².

So the graph of f lies between the parabola $y=x^2$y = x² above and the parabola $y=-x^2$y = −x² below, touching each wherever $\sin \left(\frac{1}{x}\right)$sin(1/x) is 1 or −1.

The gold curve is $y=x^2\sin \left(\frac{1}{x}\right)$y = x² sin(1/x), between the dashed parabolas $y=x^2$y = x² and $y=-x^2$y = −x². Away from 0 it swings slowly; close to 0 it swings faster and faster, always inside the two parabolas.

The middle has nowhere else to go

The squeeze theorem: if $g\left(x\right)\le f\left(x\right)\le h\left(x\right)$g(x) ≤ f(x) ≤ h(x) for every x near c, except perhaps at c itself, and g(x) and h(x) both tend to the same limit L as $x\to c$x → c, then f(x) tends to L as well.

Here $g\left(x\right)=-x^2$g(x) = −x² and $h\left(x\right)=x^2$h(x) = x², and both tend to 0 as $x\to 0$x → 0. Every value of f lies between them, so f(x) is never further from 0 than $x^2$x² is. At x = 0.1 the bounds are $\pm 0.01$±0.01, and f(0.1) = 0.01 × sin 10 = −0.00544. At x = 0.01 the bounds are $\pm 0.0001$±0.0001, and f(0.01) = −0.0000506. Taking x close enough to 0 makes $x^2$x², and with it every value of f, as close to 0 as you like.

So $\lim x^2\sin \left(\frac{1}{x}\right)$lim x² sin(1/x) as $x\to 0$x → 0 is 0. The theorem says nothing about f at 0 itself, and needs nothing: f has no value there, and the limit is still 0.

$x^2\sin \left(\frac{1}{x}\right)$x² sin(1/x) in gold between the dashed parabolas $y=x^2$y = x² and $y=-x^2$y = −x². The two upright bars clamp the strip $|x|\le \delta$|x| ≤ δ, and inside it every value of f lies between $-{\delta }^2$−δ² and ${\delta }^2$δ². Close the clamp to $\delta =0.1$δ = 0.1: the jaws are then $2{\delta }^2=0.02$2δ² = 0.02 apart, and they close to nothing as $\delta$δ does.

Endless swings, one limit

$\sin \left(\frac{1}{x}\right)$sin(1/x) is 0 whenever $\frac{1}{x}$1/x is a multiple of $\pi$π, that is at $x=\frac{1}{\pi }$x = 1/π, $\frac{1}{2\pi }$1/(2π), $\frac{1}{3\pi }$1/(3π) and so on: 0.318, 0.159, 0.106, 0.0796, and on toward 0. Between each pair f swings up to the upper parabola or down to the lower one. There are infinitely many of these swings between 0 and any positive x, so no drawing can show all of them.

The swinging never stops, but its size does: the swing at x is at most $x^2$x² either way. A limit is about how far the values are from L, not about whether they stay still, so the swings do not prevent the limit 0.

The same three curves close to the origin, from x = −0.4 to 0.4. The dashed parabolas pinch together at (0, 0), and the gold curve between them is pinched with them.

Both conditions are needed

The bounds must hold for every x near c. A bound that holds at a few points, or only far from c, says nothing about the limit.

The two bounds must also have the same limit. $\sin \left(\frac{1}{x}\right)$sin(1/x) itself lies between −1 and 1, but those bounds tend to −1 and 1, two different numbers, so they leave a gap of 2 for $\sin \left(\frac{1}{x}\right)$sin(1/x) to move in. It does: it is 1 at $x=\frac{1}{\frac{\pi }{2}+2k\pi }$x = 1/(π/2 + 2kπ) and −1 at $x=\frac{1}{\frac{3\pi }{2}+2k\pi }$x = 1/(3π/2 + 2kπ) for every whole number k, as close to 0 as you like, so $\sin \left(\frac{1}{x}\right)$sin(1/x) has no limit at 0.

The product law cannot be used for $x^2\sin \left(\frac{1}{x}\right)$x² sin(1/x) either. It needs both factors to have limits, and $\sin \left(\frac{1}{x}\right)$sin(1/x) has none. The squeeze avoids that factor’s limit altogether by bounding it.

A squeeze far out

The theorem works for $x\to \infty$x → ∞ too, with "near c" read as "for all large x". For x > 0, dividing $-1\le \sin x\le 1$−1 ≤ sin x ≤ 1 by the positive number x gives $-\frac{1}{x}\le \frac{\sin x}{x}\le \frac{1}{x}$−1/x ≤ (sin x)/x ≤ 1/x. Both bounds tend to 0 as $x\to \infty$x → ∞, so $\frac{\sin x}{x}\to 0$(sin x)/x → 0. At x = 100 it is −0.00506, inside the bounds $\pm 0.01$±0.01.

The usual mistakes

Adding the two limits. If both bounds tend to 5, f tends to 5, not 10: f sits between them, and between 5 and 5 there is only 5.

Multiplying an inequality by a negative number without reversing it. Multiplying by $x^2$x² is safe because $x^2$x² is positive; multiplying $-1\le \sin \left(\frac{1}{x}\right)\le 1$−1 ≤ sin(1/x) ≤ 1 by a negative x would turn the signs round.

Concluding that there is no limit because the function oscillates. $x^2\sin \left(\frac{1}{x}\right)$x² sin(1/x) swings forever near 0 and still tends to 0.

Using bounds with different limits, such as −1 and 1 for $x^2\sin \left(\frac{1}{x}\right)$x² sin(1/x). They are true bounds but they prove nothing; $-x^2$−x² and $x^2$x² are the bounds that close together.

A lathe and a tapering pin

The application below has exactly this function as the error in a machined surface, $x^2\sin \left(\frac{1}{x}\right)$x² sin(1/x) mm at a distance x mm from the tip, and uses the same bound to say how far out the surface can be near the tip.