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Definition of Limit

Video Overview

Visual explanation by Daisy of the convergence of a sequence to a limit.

General Approach.

Below we outline a high-level schematic for proofs in these notes. More specific outlines follow, and then several examples are provided in subsequent pages. As with most proofs, a rough outline is to take the following approach.
  1. 1.
    State relevant definitions and the math statement we must verify (e.g. an inequality).
  2. 2.
    Find relationships (e.g. equalities) or theorems that can tie things we know (from the assumptions) to what we want.
  3. 3.
    Combine results in Step 2 to verify the statement of Step 1.

Proof from Definition of Limit.

A sequence
{sn}\mathsf{\{s_n\}}
of real numbers converges to a limit
s∈R\mathsf{s}\in\mathbb{R}
provided, for all
ϵ>0\mathsf{\epsilon>0}
, there is a natural number
N∈N\mathsf{N\in\mathbb{N}}
such that
∣sn−s∣≤ϵ\mathsf{|s_n-s|\leq \epsilon}
, for all
n≥N.\mathsf{n\geq N.}
​
Suppose we must prove
{an}\{a_n\}
converges to a limit
LL
. Then we can take the following steps.
Step 1. Write an introduction of what must be shown like the example below.
Let
ε>0\varepsilon > 0
​ be given. It suffices to show there exists an index
N∈NN\in\mathbb{N}
such that
∣an−L∣≤ε|a_n - L|\leq\varepsilon
​ for all
n≥N.n\geq N.
​
Step 2. Do scratch work to find an upper bound for
∣an−L∣|a_n-L|
in terms of something we know how to bound by
ε\varepsilon
(e.g.
1/n1/n
as in the Archimedean examples below). Then choose a large
NN
that makes the upper bound on
∣an−L∣≤ε|a_n - L| \leq \varepsilon
for
n≥Nn\geq N
.
Step 3. Tie everything together by combining the results to verify
∣an−L∣≤ε|a_n-L|\leq\varepsilon
for all
n≥N.n\geq N.
​
​
Slides_Seq_Limit_Def.pdf
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