Limit of Function in Interval

Theorem
Let $$f$$ be a real function which is defined on the open interval $$\left({a \, . \, . \, b}\right)$$.

Let $$\xi \in \left({a \, . \, . \, b}\right)$$

Suppose that, $$\forall x \in \left({a \, . \, . \, b}\right)$$, either:
 * $$\xi \le f \left({x}\right) \le x$$, or:
 * $$x \le f \left({x}\right) \le \xi$$.

Then $$f \left({x}\right) \to \xi$$ as $$x \to \xi$$.

Proof
Note that $$\left|{f \left({x}\right) - \xi}\right| \le \left|{\xi - x}\right|$$.

From Limit of Absolute Value‎ we have that $$\left|{x - \xi}\right| \to 0$$ as $$x \to \xi$$.

The result follows from the Squeeze Theorem.