Tuesday, September 13, 2016

Unizor - Derivatives - Function Limit - Squeeze Theorem

Notes to a video lecture on http://www.unizor.com

Squeeze Theorem

In this lecture we present a very important so-called "Squeeze Theorem" for function limits, which we did prove for sequences before.

Squeeze Theorem
(also known as
"Pinching Theorem" or
"Sandwich Theorem", or
"Theorem about Two Policemen and a Drunk")

IF, when
x→r or x→+∞, or x→−∞,
it is GIVEN that
f(x)→L AND
h(x)→L AND
f(x) ≤ g(x) ≤ h(x)
(under the same condition of tendency of argument x)

The variant of this theorem is when L is not a concrete real number, but can be (non-rigorously) positive or negativeinfinity or (more rigorously) both functions f(x) and h(x) can be infinitely increasing or infinitely decreasing.


Let's prove for x→r where r - any real number. The other two conditions, when x→+∞ or x→−∞, or cases, when the limit is infinite, will not present any problem as a self-study exercise, they are, generally speaking, similar to the proof below.
We will use ε-δ definition of a limit.

Fix any positive ε. We have to find δ such that, if |x−r| ≤ δ, then |g(x)−L| ≤ ε.

For this ε we can find δ1 such that in δ1-neighborhood of x=rit is true that
|f(x)−L| ≤ ε or, equivalently,
L−ε ≤ f(x) ≤ L+ε

For the same ε we can find δ2such that in δ2-neighborhood ofx=r it is true that
|h(x)−L| ≤ ε or, equivalently,
L−ε ≤ h(x) ≤ L+ε

Now choose δ equal to a minimum among δ1 and δ2. Obviously, in δ-neighborhoodof x=r it is true that
L−ε ≤ f(x) ≤ L+ε AND
L−ε ≤ h(x) ≤ L+ε

Therefore, in thisδ-neighborhood of x=r it is true that
L−ε ≤ f(x) ≤ g(x) ≤ h(x) ≤ L+ε
Hence, |g(x) − L| ≤ ε
End of Proof.

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