Squeeze Theorem

Theorem

Sequences

There are two versions of this result:

• one for sequences in the set of complex numbers $\C$;
• one for sequences in the set of real numbers $\R$ (which is stronger).

Otherwise known (particularly in the UK) as the sandwich theorem.

Sequences of Real Numbers

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

Let $\left \langle {y_n} \right \rangle$ and $\left \langle {z_n} \right \rangle$ be convergent to the following limit:

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

Suppose that $\forall n \in \N: \left \langle {y_n} \right \rangle \le \left \langle {x_n} \right \rangle \le \left \langle {z_n} \right \rangle$.

Then $x_n \to l$ as $n \to \infty$, that is, $\displaystyle \lim_{n \to \infty} x_n = l$.

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

Corollary

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

$y_n \to 0$ as $n \to \infty$

Let:

$\forall n \in \N: \left|{x_n - l}\right| \le y_n$

Then $x_n \to l$ as $n \to \infty$.

Sequences of Complex Numbers

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

$y_n \to 0$ 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.

Functions

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

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

Suppose that:

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

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

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.