The Squeeze Theorem

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The Squeeze theorem allows us to exchange difficult functions for easy functions.

In mathematics, sometimes we can study complex functions by exchanging them for simplier functions. The Squeeze Theorem tells us one situation where this is possible.

Squeeze Theorem Suppose that $g(x) \le f(x) \le h(x)$ for all $x$ close to $a$ but not necessarily equal to $a$ . If $\lim _{x\to a} g(x) = L = \lim _{x\to a} h(x),$ then $\lim _{x\to a} f(x) = L$ .

I’m thinking of a function $f$ . I know that for all $x$ $0 \le f(x) \le x^2.$ What is $\lim _{x\to 0} f(x)$ ?

$f(x)$ $f(0)$ $0$ impossible to say

An continuous function $f$ satisfies the property that $8x-13 \leq f(x) \leq x^2+2x-4$ . What is $\lim _{x\to 3} f(x)$ ?

$\lim _{x\to 3} \left (8x-13\right ) = \answer [given]{11}.$ $\lim _{x\to 3} \left (x^2+2x-4\right ) = \answer [given]{11}.$ Then $\lim _{x\to 3} f(x) = \answer [given]{11}.$

Consider the function $f(x) = \begin {cases} \sqrt [5]{x}\sin \left (\frac {1}{x}\right ) & \text {if $x \ne 0$,}\\ 0 & \text {if $x = 0$,} \end {cases}$

Is this function continuous at $x=0$ ?

We must show that $\lim _{x\to 0} f(x) = \answer [given]{0}$ . Note $-|\sqrt [5]{x}|\le f(x) \le |\sqrt [5]{x}|.$ Since $\lim _{x\to 0} -|\sqrt [5]{x}| = \answer [given]{0} = \lim _{x\to 0}|\sqrt [5]{x}|,$ we see by the Squeeze Theorem, Theorem, that $\lim _{x\to 0} f(x) = \answer [given]{0}$ . Hence $f(x)$ is continuous.

Here we see how the informal definition of continuity being that you can “draw it” without “lifting your pencil” differs from the formal definition.