Squeeze Theorem/Functions/Proof 1

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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:

$\forall x \ne a \in I: \map g x \le \map f x \le \map h x$
$\displaystyle \lim_{x \mathop \to a} \ \map g x = \lim_{x \mathop \to a} \ \map h x = L$


Then:

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


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:

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

$\blacksquare$