Squeeze Theorem/Functions/Proof 1

Theorem

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

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

Suppose that:

$\forall x \ne a \in I: \map g x \le \map f x \le \map h x$

$\ds \lim_{x \mathop \to a} \map g x = \lim_{x \mathop \to a} \map h x = L$

Then:

$\ds \lim_{x \mathop \to a} \ \map f x = L$

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:

$\ds \lim_{x \mathop \to a} \map f x = 0 = L$

$\Box$

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:

$\map f x = \paren {\map f x - \map g x} + \map g x \to 0 + L = L$

$\blacksquare$