The Sandwich Theorem or Squeeze Play Theorem

It helps to think about the core concepts of Calculus from a young age, if you want to develop your expertise or talents further in math, pure or applied, engineering or mathematical sciences. At a tangible level, it helps you attack more or many questions of the IIT JEE Advanced Mathematics. Let us see if you like the following proof, or can absorb/digest it:

Reference: Calculus and Analytic Geometry by Thomas and Finney, 9th edition.

The Sandwich Theorem:

Suppose that $g(x) \leq f(x) \leq h(x)$ for all x in some open interval containing c, except possibly at $x=c$ itself. Suppose also that $\lim_{x \rightarrow c}g(x)= \lim_{x \rightarrow c}h(x)=L$ . Then, $\lim_{x \rightarrow c}f(x)=c$ .

Proof for Right Hand Limits:

Suppose $\lim_{x \rightarrow c^{+}}g(x)=\lim_{x \rightarrow c^{+}}h(x)=L$ . Then, for any $\in >0$ , there exists a $\delta >0$ such that for all x, the inequality $c<x<c+\delta$ implies $L-\in<g(x)<L+\in$ and $L-\in<h(x)<L+\in$ ….call this (I)

These inequalities combine with the inequality $g(x) \leq f(x) \leq h(x)$ to give

$L-\in <g(x) \leq f(x) \leq h(x)<L+\in$

$L-\in <f(x)<L+\in$

$-\in <f(x)-L<\in$ ….call this (II)

Therefore, for all x, the inequality $c<x<c+\delta$ implies $|f(x)-L|<\in$ . …call this (III)

Proof for LeftHand Limits:

Suppose $\lim_{x \rightarrow c^{-}} g(x)=\lim_{x \rightarrow c^{-}}=L$ . Then, for $\in >0$ there exists a $\delta >0$ such that for all x, the inequality $c-\delta <x<c$ implies $L-\in<g(x)<L+\in$ and $L-\in<h(x)<L+\in$ …call this (IV).

We conclude as before that for all x, $c-\delta <x<c$ implies $|f(x)-L|<\in$ .

Proof for Two sided Limits:

If $\lim_{x \rightarrow c}g(x) = \lim_{x \rightarrow c}h(x)=L$ , then $g(x)$ and $h(x)$ both approach L as $x \rightarrow c^{+}$ and as $x \rightarrow c^{-}$ so $\lim_{x \rightarrow c^{+}}f(x)=L$ and $\lim_{x \rightarrow c^{-}}f(x)=L$ . Hence, $\lim_{x \rightarrow c}f(x)=L$ . QED.