From ProofWiki

[edit] Theorem

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

[edit] 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).

[edit] 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:

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

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

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.

[edit] 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 $n \to \infty$ .

[edit] 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.

[edit] 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:

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

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

[edit] Proof

[edit] Proof for Real Sequences

Note from the corollary to Negative of Absolute Value, we have $\left|{x - l}\right| < \epsilon \iff l - \epsilon < x < l + \epsilon$ .

Let $ε > 0$ .

We need to prove that $\exists N: \forall n > N: \left|{x_n - l}\right| < \epsilon$ .

As $\lim_{n \to \infty} y_n = l$ we know that $\exists N_1: \forall n > N_1: \left|{y_n - l}\right| < \epsilon$ .

As $\lim_{n \to \infty} z_n = l$ we know that $\exists N_2: \forall n > N_2: \left|{z_n - l}\right| < \epsilon$ .

Let $N = \max \left\{{N_1, N_2}\right\}$ .

Then if $n > N$ , $n > N 1$ and $n > N 2$ .

So:

$\forall n > N: l - \epsilon < y_n < l + \epsilon$ ;
$\forall n > N: l - \epsilon < z_n < l + \epsilon$ .

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

So $\forall n > N: l - \epsilon < y_n \le x_n \le z_n < l + \epsilon$

and so $\forall n > N: l - \epsilon < x_n < l + \epsilon$ .

So $\forall n > N: \left|{x_n - l}\right| < \epsilon$ .

Hence the result.

$\blacksquare$

[edit] Proof of Corollary

From the corollary to Negative of Absolute Value, we have $\left|{x_n - l}\right| \le y_n \iff l - y_n \le x_n \le l + y_n.$

From the Combination Theorem for Sequences, $l - y_n \to l$ as $n \to \infty$ , and $l + y_n \to l$ as $n \to \infty$ .

So by the Squeeze Theorem for Sequences, $x_n \to l$ as $n \to \infty$ .

$\blacksquare$

[edit] Proof for Complex Sequences

In order to show that $\left \langle {z_n} \right \rangle$ is a null sequence, we want to show that:

$\forall \epsilon > 0: \exists N: \forall n > N: \left|{z_n}\right| < \epsilon$ .

But since $\left \langle {a_n} \right \rangle$ is a null sequence:

$\exists N: \forall n > N: a_n < \epsilon$ .

So, using this value of $n$ , we have: $\left \langle {z_n} \right \rangle \le a_n < \epsilon$ .

Hence the result.

$\blacksquare$

[edit] Proof for Functions

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

Let $ε > 0$ be a positive real number.

Then by the definition of the limit of a function, $\exists \delta > 0: 0 < \left|{x - a}\right| < \delta \Longrightarrow \left|{h \left({x}\right)}\right| < \epsilon$ .

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

By the transitive property of $\le$ , this proves that $\lim_{x \to a} f \left({x}\right) = 0 = L$ .

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

$\blacksquare$

[edit] Alternative Proof for Functions

Alternatively, the result Limit of Function by Convergent Sequences can directly applied to the Squeeze Theorem for Sequences:

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:

$\lim_{x \to c} g \left({x}\right) = l$ ;
$\lim_{x \to c} h \left({x}\right) = m$ ;
$g \left({x}\right) \le f \left({x}\right) \le h \left({x}\right)$ except perhaps at $x = c$ .

Let $\left \langle {x_n} \right \rangle$ be a sequence of points of $\left({a \, . \, . \, b}\right)$ such that $\forall n \in \N^*: x_n \ne c$ and $\lim_{n \to \infty} x_n = c$ .

By Limit of Function by Convergent Sequences, $\lim_{n \to \infty} g \left({x_n}\right) = l$ and $\lim_{n \to \infty} h \left({x_n}\right) = l$ .

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 Sequences that $\lim_{n \to \infty} f \left({x_n}\right) = l$ .

The result follows from Limit of Function by Convergent Sequences.

$\blacksquare$

[edit] 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.

Squeeze Theorem

From ProofWiki

Contents

[edit] Theorem

[edit] Sequences

[edit] Sequences of Real Numbers

[edit] Corollary

[edit] Sequences of Complex Numbers

[edit] Functions

[edit] Proof

[edit] Proof for Real Sequences

[edit] Proof of Corollary

[edit] Proof for Complex Sequences

[edit] Proof for Functions

[edit] Alternative Proof for Functions

[edit] Comment

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