Squeeze Theorem/Functions/Proof 1

Proof
We start by proving the special case where $\forall x: \map g x = 0$ and $L = 0$, in which case:
 * $\ds \lim_{x \mathop \to a} \map h x = 0$

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

Then by the definition of the limit of a function:
 * $\exists \delta > 0: 0 < \size {x - a} < \delta \implies \size {\map h x} < \epsilon$

Now:
 * $\forall x \ne a: 0 = \map g x \le \map f x \le \map h x$

so that:
 * $\size {\map f x} \le \size {\map h x}$

Thus:
 * $0 < \size {x - a} < \delta \implies \size {\map f x} \le \size {\map h x} < \epsilon$

By the transitive property of $\le$, this proves that:
 * $\ds \lim_{x \mathop \to a} \map f x = 0 = L$

We now move on to the general case, with $\map g x$ and $L$ arbitrary.

For $x \ne a$, we have:
 * $\map g x \le \map f x \le \map h x$

By subtracting $\map g x$ from all expressions, we have:
 * $0 \le \map f x - \map g x \le \map h x - \map g x$

Since as $x \to a, \map h x \to L$ and $\map g x \to L$, we have:
 * $\map h x - \map g x \to L - L = 0$

From the special case, we now have:
 * $\map f x - \map g x \to 0$

We conclude that:
 * $\map f x = \paren {\map f x - \map g x} + \map g x \to 0 + L = L$