WARNING
Homework
Math 555

Many of you have been making (easily fixable since you have the right ideas) errors in writing up your proofs that involve limits. Basically, you have been writing lim n&rarr&infin zn, and assuming that the limit exists, before you actually show the limit exists. It is time to get it down right. Below are some ideas that (hopefully) will help prevent such troubles. The idea behind the fix is to use that the lim-sup and lim-inf of any sequence exist in the extended real numbers R ∪ {+∞, -∞} . So work with the lim-inf & lim-sup (keeping in mind that might be +/- ∞, but atleast they exist) instead of working with lim n&rarr&infin zn (which you don't even know if it exists). So read the below. You do not have to quote any of the Lemma/Theorems in your proofs. Just understand the ideas and use the ideas as needed in your proofs.

A Basic Lemma
Let { zn } be a sequence from R and M ε R. Then the following are equivalent.


Proof of Basic Lemma
Follows easily from the following facts. All done ♦

Squeeze Theorem (utility grade)
Let { zn } be a sequence from R and M ε R.
If there exists squences { an } and { bn } from R and N0 ε N such that

( ∀ n≥ N0)     [ an ≤ zn ≤ bn ]
and
M = lim n&rarr&infin an    and     lim n&rarr&infin bn = M
then the limit, as n&rarr&infin, of { zn } exists and
lim n&rarr&infin zn = M    .

Proof of Squeeze Theorem (utility grade)
Theorem 2.2.4 from textbook. ♦

Squeeze Theorem (working version)
Let { zn } be a sequence from R and M ε R.
If, for each ε >0 , there exists squences { anε } and { bnε } from R and Nε ε N such that

( ∀ n≥ Nε)     [ anε ≤ zn ≤ bnε ]
and
M - ε ≤ liminf n&rarr&infin an ε    and     limsup n&rarr&infin bnε ≤ M + ε
then the limit, as n&rarr&infin, of { zn } exists and
lim n&rarr&infin zn = M    .

Proof of Squeeze Theorem (working version)
Let the givens (and the If) be given. Then, for each ε >0 ,
liminf n&rarr&infin an ε ≤ liminf n&rarr&infin zn ≤ limsup n&rarr&infin zn ≤ limsup n&rarr&infin bn ε
and so
M- ε ≤ liminf n&rarr&infin zn ≤ limsup n&rarr&infin zn ≤ M + ε   .
Since ε >0 was arbitrary,
liminf n&rarr&infin zn = M = limsup n&rarr&infin zn   .
All done ♦


Findable from URL: http://www.math.sc.edu/~girardi/