# Squeeze Theorem/Functions

## Theorem

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:

$\forall x \ne a \in I: g \left({x}\right) \le f \left({x}\right) \le h \left({x}\right)$
$\displaystyle \lim_{x \mathop \to a} \ g \left({x}\right) = \lim_{x \mathop \to a} \ h \left({x}\right) = L$

Then:

$\displaystyle \lim_{x \mathop \to a} \ f \left({x}\right) = L$

## Proof 1

We start by proving the special case where $\forall x: g \left({x}\right) = 0$ and $L=0$, in which case $\displaystyle \lim_{x \to a} \ h \left({x}\right) = 0$.

Let $\epsilon > 0$ be a positive real number.

Then by the definition of the limit of a function:

$\exists \delta > 0: 0 < \left|{x - a}\right| < \delta \implies \left|{h \left({x}\right)}\right| < \epsilon$

Now:

$\forall x \ne a: 0 = g \left({x}\right) \le f \left({x}\right) \le h \left({x}\right)$

so that:

$\left|{f \left({x}\right)}\right| \le \left|{h \left({x}\right)}\right|$

Thus:

$0 < |x-a| < \delta \implies \left|{f \left({x}\right)}\right| \le \left|{h \left({x}\right)}\right| < \epsilon$

By the transitive property of $\le$, this proves that:

$\displaystyle \lim_{x \to a} \ f \left({x}\right) = 0 = L$

We now move on to the general case, with $g \left({x}\right)$ and $L$ arbitrary.

For $x \ne a$, we have:

$g \left({x}\right) \le f \left({x}\right) \le h \left({x}\right)$

By subtracting $g \left({x}\right)$ from all expressions, we have:

$0 \le f \left({x}\right) - g \left({x}\right) \le h \left({x}\right) - g \left({x}\right)$

Since as $x \to a, h \left({x}\right) \to L$ and $g \left({x}\right) \to L$, we have:

$h \left({x}\right) - g \left({x}\right) \to L - L = 0$

From the special case, we now have:

$f \left({x}\right) - g \left({x}\right) \to 0$

We conclude that:

$f \left({x}\right) = \left({f \left({x}\right) - g \left({x}\right)}\right) + g \left({x}\right) \to 0 + L = L$

$\blacksquare$

## Proof 2

Let $f, g, h$ be real functions defined on an open interval $\left({a \,.\,.\, b}\right)$, except possibly at the point $c \in \left({a \,.\,.\, b}\right)$.

Let:

$\displaystyle \lim_{x \mathop \to c} \ g \left({x}\right) = L$
$\displaystyle \lim_{x \mathop \to c} \ h \left({x}\right) = L$
$g \left({x}\right) \le f \left({x}\right) \le h \left({x}\right)$ except perhaps at $x = c$.

Let $\left \langle {x_n} \right \rangle$ be a sequence of points of $\left({a \,.\,.\, b}\right)$ such that:

$\forall n \in \N_{>0}: x_n \ne c$

and:

$\displaystyle \lim_{n \mathop \to \infty} \ x_n = c$
$\displaystyle \lim_{n \mathop \to \infty} \ g \left({x_n}\right) = L$

and:

$\displaystyle \lim_{n \mathop \to \infty} \ h \left({x_n}\right) = L$

Since:

$g \left({x_n}\right) \le f \left({x_n}\right) \le h \left({x_n}\right)$

it follows from the Squeeze Theorem for Real Sequences that:

$\displaystyle \lim_{n \mathop \to \infty} \ f \left({x_n}\right) = L$

The result follows from Limit of Function by Convergent Sequences.

$\blacksquare$

## Proof 3

By the definition of the limit of a real function, we have to prove that:

$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \left({\left|{x - a}\right| < \delta \implies \left|{f \left({x}\right) - L}\right| < \epsilon}\right)$

Let $\epsilon \in \R_{>0}$ be given.

We have:

$\displaystyle \lim_{x \mathop \to a} \ g \left({x}\right) = \lim_{x \mathop \to a} \ h \left({x}\right)$

Hence by Sum Rule for Limits of Functions:

$\displaystyle \lim_{x \mathop \to a} \ h \left({x}\right) - g \left({x}\right) = 0$

By the definition of the limit of a real function:

$(1): \quad \forall \epsilon' \in \R_{>0}: \exists \delta \in \R_{>0}: \left({\left|{x - a}\right| < \delta \implies \left|{h \left({x}\right) - L}\right| < \epsilon'}\right)$
$(2): \quad \forall \epsilon' \in \R_{>0}: \exists \delta \in \R_{>0}: \left({\left|{x - a}\right| < \delta \implies \left|{g \left({x}\right) - L}\right| < \epsilon'}\right)$
$(3): \quad \forall \epsilon' \in \R_{>0}: \exists \delta \in \R_{>0}: \left({\left|{x - a}\right| < \delta \implies \left|{h \left({x}\right) - g \left({x}\right)}\right| < \epsilon'}\right)$

Take $\epsilon' = \dfrac {\epsilon} 3$ in $(1)$, $(2)$, $(3)$.

Then there exists $\delta_1, \delta_2, \delta_3$ that satisfies $(1)$, $(2)$, $(3)$ with $\epsilon' = \dfrac \epsilon 3$.

Take $\delta = \min\{\delta_1, \delta_2, \delta_3\}$.

Then:

 $\displaystyle \left\vert{x - a}\right\vert$ $<$ $\displaystyle \delta$ $\quad$ $\quad$ $\displaystyle \implies \ \$ $\displaystyle \left\vert{h \left({x}\right) - L}\right\vert$ $<$ $\displaystyle \frac {\epsilon} 3$ $\quad$ $\quad$ $\, \displaystyle \land \,$ $\displaystyle \left\vert{g \left({x}\right) - L}\right\vert$ $<$ $\displaystyle \frac {\epsilon} 3$ $\quad$ $\quad$ $\, \displaystyle \land \,$ $\displaystyle \left\vert{h \left({x}\right) - g \left({x}\right)}\right\vert$ $<$ $\displaystyle \frac {\epsilon} 3$ $\quad$ $\quad$

So, if $\left\vert{x - a}\right\vert<\delta$:

 $\displaystyle \left\vert{f \left({x}\right) - L}\right\vert$ $=$ $\displaystyle \left\vert{f \left({x}\right) - g \left({x}\right) + h \left({x}\right) - L + g \left({x}\right) - h \left({x}\right)}\right\vert$ $\quad$ $\quad$ $\displaystyle$ $\le$ $\displaystyle \left\vert{f \left({x}\right) - g \left({x}\right)}\right\vert + \left\vert{h \left({x}\right) - L}\right\vert + \left\vert{h \left({x}\right) - g \left({x}\right)}\right\vert$ $\quad$ $\quad$ $\displaystyle$ $\le$ $\displaystyle \left\vert{h \left({x}\right) - g \left({x}\right)}\right\vert + \left\vert{h \left({x}\right) - L}\right\vert + \left\vert{h \left({x}\right) - g \left({x}\right)}\right\vert$ $\quad$ $\quad$ $\displaystyle$ $\le$ $\displaystyle \frac {\epsilon} 3 + \frac {\epsilon} 3 + \frac {\epsilon} 3$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \epsilon$ $\quad$ $\quad$

$\blacksquare$

## 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.