# Squeeze Theorem

## Theorem

## Sequences

There are two versions of this result:

- one for sequences in the set of complex numbers $\C$, and more generally for sequences in a metric space.
- one for sequences in the set of real numbers $\R$, and more generally in linearly ordered spaces (which is stronger).

### Sequences of Real Numbers

Let $\sequence {x_n}$, $\sequence {y_n}$ and $\sequence {z_n}$ be sequences in $\R$.

Let $\sequence {y_n}$ and $\sequence {z_n}$ both be convergent to the following limit:

- $\displaystyle \lim_{n \mathop \to \infty} y_n = l, \lim_{n \mathop \to \infty} z_n = l$

Suppose that:

- $\forall n \in \N: y_n \le x_n \le z_n$

Then:

- $x_n \to l$ as $n \to \infty$

that is:

- $\displaystyle \lim_{n \mathop \to \infty} x_n = l$

Thus, if $\sequence {x_n}$ is always between two other sequences that both converge to the same limit, $\sequence {x_n} $ is said to be **sandwiched** or **squeezed** between those two sequences and itself must therefore converge to that same limit.

### Sequences of Complex Numbers

Let $\left \langle {a_n} \right \rangle$ be a sequence in $\R$ which is null, that is:

- $a_n \to 0$ as $n \to \infty$

Let $\left \langle {z_n} \right \rangle$ be a sequence in $\C$.

Suppose $\left \langle {a_n} \right \rangle$ dominates $\left \langle {z_n} \right \rangle$.

That is, suppose that:

- $\forall n \in \N: \left|{z_n}\right| \le a_n$

Then $\left \langle {z_n} \right \rangle$ is a null sequence.

### Sequences in a Linearly Ordered Space

Let $\left({S, \le, \tau}\right)$ be a linearly ordered space.

Let $\left\langle{x_n}\right\rangle$, $\left\langle{y_n}\right\rangle$, and $\left\langle{z_n}\right\rangle$ be sequences in $S$.

Let $p \in S$.

Let $\left\langle{x_n}\right\rangle$ and $\left\langle{z_n}\right\rangle$ both converge to $p$

For each $n$, let $x_n \le y_n \le z_n$.

Then $\left\langle{y_n}\right\rangle$ converges to $p$.

### Sequences in a Metric Space

Let $M = \left({S, d}\right)$ be a metric space or pseudometric space.

Let $\left\langle{x_n}\right\rangle$ be a sequence in $S$.

Let $p \in S$.

Let $\left\langle{r_n}\right\rangle$ be a sequence in $\R_{\ge 0}$.

Let $\left\langle{r_n}\right\rangle$ converge to $0$.

For each $n$, let $d \left({p, x_n}\right) \le r_n$.

Then $\left\langle{x_n}\right\rangle$ converges to $p$.

## Functions

Let $a$ be a point on an open real interval $I$.

Let $f$, $g$ and $h$ be real functions defined at all points of $I$ except for possibly at point $a$.

Suppose that:

- $\forall x \ne a \in I: \map g x \le \map f x \le \map h x$
- $\ds \lim_{x \mathop \to a} \map g x = \lim_{x \mathop \to a} \map h x = L$

Then:

- $\ds \lim_{x \mathop \to a} \ \map f x = L$

## Also known as

This result is also known, in the UK in particular, as the **sandwich theorem** or the **sandwich rule**.

In that culture, the word **sandwich** traditionally means specifically enclosing food between two slices of bread, as opposed to the looser usage of the **open sandwich**, where the there is only one such slice.

Hence, in idiomatic British English, one can refer to the (often uncomfortable) situation of being between two entities as being **sandwiched** between them.

As the idiom is not universal globally, the term **squeeze theorem** is preferred on $\mathsf{Pr} \infty \mathsf{fWiki}$, for greatest comprehension.

## Also see

## Comment

A useful tool to determine the limit of a sequence or function which is difficult to calculate or analyze.

If you can prove it is always between two sequences, both converging to the same limit, whose behavior is considerably more tractable, you can save yourself the trouble of working on that awkward case.