Squeeze Theorem

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

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

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

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.

Corollary
Let $$y_n \to 0$$ as $$n \to \infty$$.

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

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

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

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

Let $$\epsilon > 0$$.

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

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.

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 $$n \to \infty$$.

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

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

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

Q.E.D.

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.