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.

Let me know your feedback on such stuff,

Nalin Pithwa

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