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Calculus Limits and continuity

Squeeze theorem

20 practice questions 0 video lessons Theory + worked examples

The Squeeze Theorem

California Calculus • Standard 1.3 • Limits and Continuity

The Squeeze Theorem is a topic in Limits and Continuity in the California Calculus Standards. It is aligned to Standard 1.3, which requires students to prove and use special limits, such as \(\dfrac{\sin x}{x}\) and \(\dfrac{1-\cos x}{x}\) as \(x\to 0\), whose standard proof is the squeeze theorem.

The squeeze theorem finds a limit by trapping a function between two others that share the same limit. It is the tool behind the special limits \(\lim_{x\to 0}\dfrac{\sin x}{x}=1\) and \(\lim_{x\to 0}\dfrac{1-\cos x}{x}=0\) (angles in radians).

California Calculus › Limits and Continuity › The Squeeze Theorem  —  Standard 1.3

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Theory

The squeeze theorem finds a limit by trapping a function between two others that share the same limit. It is the tool behind the special limits \(\lim_{x\to 0}\dfrac{\sin x}{x}=1\) and \(\lim_{x\to 0}\dfrac{1-\cos x}{x}=0\). Angles are always in radians.

The squeeze theorem (also called the sandwich theorem) says: if \(g(x)\le f(x)\le h(x)\) for all \(x\) near \(a\), and

\[\lim_{x\to a} g(x)=\lim_{x\to a} h(x)=L,\]

then \(\lim_{x\to a} f(x)=L\) as well. If the outer curves are forced to the same value, the trapped curve has nowhere else to go.

It is especially useful when \(f\) oscillates or has no simple formula for its limit — you bound it instead of computing it directly.

Special limits (radians): the standard proof of \(\dfrac{\sin x}{x}\to 1\) is a squeeze between \(\cos x\) and \(1\). These special limits are the calculus building blocks for the derivatives of \(\sin x\) and \(\cos x\).
A function squeezed between two others that meet at L The oscillating curve f is trapped between an upper curve h and a lower curve g; both outer curves pass through L at x equals a, forcing f to L as well. x y a L h g f
\(g\le f\le h\); since \(g\) and \(h\) meet at \(L\), so does \(f\).
The graph of sin x over x approaching 1 at the origin The curve sin x over x rises to an open circle at height 1 above the origin, showing the special limit equals 1. x y → 1
\(\dfrac{\sin x}{x}\to 1\) as \(x\to 0\) (a hole at the origin).

The squeeze theorem:

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

if g is at most f is at most h and g and h go to L then f goes to L

The two special trigonometric limits (angles in radians):

\[\lim_{x\to 0}\dfrac{\sin x}{x}=1,\qquad\lim_{x\to 0}\dfrac{1-\cos x}{x}=0\]

sin x over x approaches 1 and one minus cos x over x approaches 0

Radians only: these special limits are false in degrees. Always work in radians in calculus.

How to use the squeeze theorem

  1. Bound the tricky factor. For an oscillation, use \(-1\le\sin(\cdots)\le 1\) (or \(\cos\)).
  2. Multiply through by the remaining factor to get \(g\le f\le h\).
  3. Take limits of \(g\) and \(h\). If they agree, that value is the limit of \(f\).
For \(\dfrac{\sin(\text{angle})}{\text{something}}\): rewrite it as \(\dfrac{\sin\theta}{\theta}\) times a constant, then use \(\dfrac{\sin\theta}{\theta}\to 1\).
Example 1 — Squeeze a bounded oscillation
Evaluate \(\displaystyle\lim_{x\to 0} x^2\sin\!\left(\dfrac{1}{x}\right)\).
Solution

Since \(-1\le \sin\!\left(\dfrac{1}{x}\right)\le 1\), multiply by \(x^2\ge 0\):

\(-x^2\)\(\le\)\(x^2\sin\!\left(\dfrac{1}{x}\right)\le x^2\)

Both \(-x^2\) and \(x^2\) approach \(0\), so by the squeeze theorem the middle does too.

\(\lim_{x\to 0} x^2\sin\!\left(\dfrac{1}{x}\right)\)\(=\)\(0\)

limit equals 0 by the squeeze theorem

Example 2 — Use \(\sin x/x\to 1\)
Evaluate \(\displaystyle\lim_{x\to 0}\dfrac{\sin 3x}{x}\).
Solution

Match the angle inside the sine with the denominator by inserting a \(3\).

\(\dfrac{\sin 3x}{x}\)\(=\)\(3\cdot\dfrac{\sin 3x}{3x}\)
\(\lim_{x\to 0}3\cdot\dfrac{\sin 3x}{3x}\)\(=\)\(3\cdot 1=3\)

limit equals 3

Example 3 — Ratio of two sines
Evaluate \(\displaystyle\lim_{x\to 0}\dfrac{\sin 5x}{\sin 2x}\).
Solution

Build a \(\sin\theta/\theta\) piece for each angle.

\(\dfrac{\sin 5x}{\sin 2x}\)\(=\)\(\dfrac{\sin 5x}{5x}\cdot\dfrac{2x}{\sin 2x}\cdot\dfrac{5}{2}\)
\(\lim_{x\to 0}\)\(=\)\(1\cdot 1\cdot\dfrac{5}{2}=\dfrac{5}{2}\)

limit equals five halves

Example 4 — The \((1-\cos x)/x\) limit
Evaluate \(\displaystyle\lim_{x\to 0}\dfrac{1-\cos x}{x}\).
Solution

This is one of the two named special limits (angles in radians).

\(\lim_{x\to 0}\dfrac{1-\cos x}{x}\)\(=\)\(0\)

For comparison, \(\lim_{x\to 0}\dfrac{1-\cos x}{x^2}=\dfrac{1}{2}\).

limit equals 0

Common pitfalls

The bounds must have the same limit. If \(g\) and \(h\) approach different values, the squeeze theorem tells you nothing.
Work in radians. \(\dfrac{\sin x}{x}\to 1\) is only true for radian measure; in degrees the limit is a different number.
Match the angle to the denominator. For \(\dfrac{\sin 3x}{x}\), you need \(3x\) on the bottom — insert the constant, do not just cancel.

Frequently asked questions

What is the squeeze theorem?

If a function is trapped between two others, \(g\le f\le h\), and both outer functions approach the same limit \(L\), then \(f\) also approaches \(L\).

What is the limit of sin x over x as x approaches 0?

It equals \(1\), with \(x\) in radians. It is proved by squeezing \(\dfrac{\sin x}{x}\) between \(\cos x\) and \(1\).

When should I use the squeeze theorem?

When a function oscillates or has no easy formula for its limit — for example \(x^2\sin(1/x)\). Bound it between two functions with the same limit instead of computing it directly.

Do the special limits work in degrees?

No. \(\dfrac{\sin x}{x}\to 1\) and \(\dfrac{1-\cos x}{x}\to 0\) hold only in radians. Calculus always uses radian measure.

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