Squeeze Theorem/Functions/Proof 2

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

Let $f, g, h$ be real functions defined on an open interval $\openint a b$, except possibly at the point $c \in \openint a b$.

Let:

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

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

$\map g x \le \map f x \le \map h x$ except perhaps at $x = c$.

Let $\sequence {x_n}$ be a sequence of points of $\openint a b$ such that:

$\forall n \in \N_{>0}: x_n \ne c$

and:

$\ds \lim_{n \mathop \to \infty} \ x_n = c$

By Limit of Function by Convergent Sequences:

$\ds \lim_{n \mathop \to \infty} \map g {x_n} = L$

and:

$\ds \lim_{n \mathop \to \infty} \map h {x_n} = L$

Since:

$\map g {x_n} \le \map f {x_n} \le \map h {x_n}$

it follows from the Squeeze Theorem for Real Sequences that:

$\ds \lim_{n \mathop \to \infty} \map f {x_n} = L$

The result follows from Limit of Function by Convergent Sequences.

$\blacksquare$

Sources

• 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): Appendix: $\S 18.6$: Limits of functions