Pinching Theorem

Q: Does the Squeeze Theorem work for limits at infinity?

Yes. The theorem applies to $\lim_{x \to \infty} f(x)$ and $\lim_{x \to -\infty} f(x)$ as well. As long as $g(x) \leq f(x) \leq h(x)$ for all sufficiently large (or sufficiently negative) $x$, and the outer limits agree, the conclusion follows. It also works for sequences: if $a_n \leq b_n \leq c_n$ and $a_n, c_n \to L$, then $b_n \to L$.

Pinching Theorem
Sandwich Theorem
Squeeze Theorem

A theorem which allows the computation of the limit of an expression by trapping the expression between two other expressions which have limits that are easier to compute.

If g(x) ≤ f(x) ≤ h(x) for all x in a deleted neighborhood of a, and lim g(x) = lim h(x) = L, then lim f(x) = L, including L=±∞.

Example proving lim(x→∞) sin(x)/x = 0 using Pinching Theorem: -1/x ≤ sin(x)/x ≤ 1/x, with lim(-1/x)=0 and lim(1/x)=0.

See also

Deleted neighborhood

Key Formula

\text{If } g(x) \leq f(x) \leq h(x) \text{ for all } x \text{ in a deleted neighborhood of } c, \text{ and } \lim_{x \to c} g(x) = \lim_{x \to c} h(x) = L, \text{ then } \lim_{x \to c} f(x) = L.

Where:

$f(x)$ = The function whose limit you want to find
$g(x)$ = The lower bounding function (squeeze from below)
$h(x)$ = The upper bounding function (squeeze from above)
$c$ = The point at which the limit is being evaluated
$L$ = The common limit of g(x) and h(x), which becomes the limit of f(x)

Worked Example

Problem: Find the limit:

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

Step 1: Identify the difficulty. As

x \to 0

\sin(1/x)

oscillates wildly between

-1

and

1

, so you cannot simply substitute. However, the

x^2

factor shrinks toward

0

Step 2: Establish bounds using the fact that

-1 \leq \sin(1/x) \leq 1

for all

x \neq 0

. Multiply through by

x^2

(which is non-negative).

-x^2 \leq x^2 \sin\!\left(\frac{1}{x}\right) \leq x^2

Step 3: Find the limits of the bounding functions as

x \to 0

\lim_{x \to 0} (-x^2) = 0 \quad \text{and} \quad \lim_{x \to 0} x^2 = 0

Step 4: Apply the Pinching Theorem. Since

-x^2 \leq x^2 \sin(1/x) \leq x^2

and both outer limits equal

0

, the squeezed function must also have limit

0

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

Answer: The limit is

0

Another Example

This example differs because the bounding functions converge to a non-zero limit ( $1$ instead of $0$ ), and it demonstrates the classic geometric proof behind one of the most important limits in calculus.

Problem: Use the Pinching Theorem to evaluate

\lim_{x \to 0} \dfrac{\sin x}{x}

by establishing appropriate bounds.

Step 1: Consider the unit circle for

0 < x < \pi/2

. A geometric argument (comparing areas of triangles and a circular sector) yields the inequality:

\cos x \leq \frac{\sin x}{x} \leq 1

Step 2: Note that

\sin x / x

is an even function, so the same inequality holds for

-\pi/2 < x < 0

as well. The bounds apply in a full deleted neighborhood of

0

Step 3: Compute the limits of the bounding functions as

x \to 0

\lim_{x \to 0} \cos x = 1 \quad \text{and} \quad \lim_{x \to 0} 1 = 1

Step 4: Apply the Pinching Theorem. Both bounds converge to

1

, so the squeezed function does too.

\lim_{x \to 0} \frac{\sin x}{x} = 1

Answer: The limit is

1

Frequently Asked Questions

Why is it called the Squeeze Theorem, Sandwich Theorem, and Pinching Theorem?

All three names describe the same idea: you 'squeeze,' 'sandwich,' or 'pinch' a function between two others. The terminology varies by textbook and region. In North America 'Squeeze Theorem' is most common, while 'Pinching Theorem' appears frequently in older and European-style texts. They are identical in statement and proof.

When should you use the Pinching Theorem instead of direct substitution or L'Hôpital's Rule?

Use the Pinching Theorem when the function oscillates or behaves erratically near the limit point, making direct substitution and algebraic simplification impossible. A classic signal is a bounded oscillating factor (like

\sin(1/x)

\cos(1/x)

) multiplied by a factor that goes to

0

. L'Hôpital's Rule requires a

0/0

\infty/\infty

indeterminate form and differentiable functions, so it does not always apply.

Does the Squeeze Theorem work for limits at infinity?

Yes. The theorem applies to

\lim_{x \to \infty} f(x)

and

\lim_{x \to -\infty} f(x)

as well. As long as

g(x) \leq f(x) \leq h(x)

for all sufficiently large (or sufficiently negative)

x

, and the outer limits agree, the conclusion follows. It also works for sequences: if

a_n \leq b_n \leq c_n

and

a_n, c_n \to L

, then

b_n \to L

Pinching (Squeeze) Theorem vs. L'Hôpital's Rule

	Pinching (Squeeze) Theorem	L'Hôpital's Rule
Core idea	Trap a function between two simpler functions with the same limit	Replace a 0/0 or ∞/∞ limit with the limit of the ratio of derivatives
Requirements	Two bounding functions with equal, known limits	An indeterminate form (0/0 or ∞/∞) and differentiable numerator/denominator
Best used when	The function oscillates or a bounded factor is multiplied by a vanishing factor	The function is a ratio that yields an indeterminate form on substitution
Works for sequences?	Yes, directly	Not directly; must first convert to a continuous function

Why It Matters

The Pinching Theorem is essential in introductory calculus because it provides the rigorous foundation for limits that cannot be computed by substitution or algebra alone. The cornerstone limit

\lim_{x \to 0} (\sin x)/x = 1

, which underlies all trigonometric derivative formulas, is proved using this theorem. You will also encounter it frequently when analyzing sequences in series convergence tests and in multivariable calculus when bounding expressions in terms of distance from a point.

Common Mistakes

Mistake: Using bounds that do not share the same limit. For instance, concluding a limit exists because

0 \leq f(x) \leq x

x \to 5

, where the lower bound gives

0

and the upper bound gives

5

Correction: The theorem only applies when both bounding functions converge to the same value

L

. If the outer limits differ, the theorem tells you nothing — you only know the limit of

f

(if it exists) lies between those two values.

Mistake: Forgetting that the inequality

g(x) \leq f(x) \leq h(x)

must hold in a deleted neighborhood of the limit point, not just at isolated points.

Correction: Verify that the inequality holds for all

x

sufficiently close to

c

(except possibly at

c

itself). A common approach is to use known bounds like

-1 \leq \sin(\theta) \leq 1

and then multiply by a non-negative factor to preserve the inequality direction.

Related Terms

Limit — The fundamental concept the theorem helps evaluate
Theorem — General term for a proven mathematical statement
Expression — The mathematical quantity being bounded and evaluated
Deleted Neighborhood — Region around c where the bounding inequalities must hold
Compute — The act of evaluating the limit using this theorem
Continuity — Limits proved by squeezing often establish continuity
L'Hôpital's Rule — Alternative technique for indeterminate-form limits