Squeeze Theorem/Functions

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

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

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

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

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

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

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

By Limit of Function by Convergent Sequences:
 * $\displaystyle \lim_{n \to \infty} \ g \left({x_n}\right) = L$

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

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

The result follows from Limit of Function by Convergent Sequences.

Alternative Proof 2
We have to prove that: $\forall \epsilon >0 \exists \delta >0 (|x-a|<\delta \implies |f(x)-L|<\epsilon)$

Therefore, let $\epsilon >0$ be given.

We have:

(1)$\forall \epsilon >0 \exists \delta >0 (|x-a|<\delta \implies |h(x)-L|<\epsilon)$

(2)$\forall \epsilon >0 \exists \delta >0 (|x-a|<\delta \implies |g(x)-L|<\epsilon)$

And, since $\displaystyle \lim_{x \to a} \ g \left({x}\right)=\displaystyle \lim_{x \to a} \ h \left({x}\right)$, we have: $\displaystyle \lim_{x \to a} \ h \left({x}\right) - g \left({x}\right) =0$, and then:

(3)$\forall \epsilon >0 \exists \delta >0 (|x-a|<\delta \implies |h(x)-g(x)|<\epsilon$

Take $\frac{\epsilon}{3}$ in (1),(2),(3). Therefore, there exists $\delta_1, \delta_2, \delta_3$ that satisfies (1),(2),(3) with $\frac{\epsilon}{3}$. Take $\delta=min\{\delta_1, \delta_2, \delta_3\}$

This way,

$|x-a|<\delta \implies |h(x)-L|<\frac{\epsilon}{3}, |g(x)-L|<\frac{\epsilon}{3}, |h(x)-g(x)|<\frac{\epsilon}{3}$

So, if $|x-a|<\delta$ We have that:

$|f(x)-L|=|f(x)-g(x)+h(x)-L+g(x)-h(x)|\leq |f(x)-g(x)|+|h(x)-L|+|h(x)-g(x)| \leq |h(x)-g(x)|+|h(x)-L|+|h(x)-g(x)|<\frac{\epsilon}{3}+\frac{\epsilon}{3}+\frac{\epsilon}{3}=\epsilon$