Sequence Limit

Definition: $latex \text{Sequence } (a_n) $
has limit a

$latex \boxed{\forall \varepsilon >0, \exists N, \forall n \geq N \text { such that } |(a_n) -a| < \varepsilon}$

$latex \Updownarrow $

$latex \displaystyle \boxed{ \lim_{n\to\infty} (a_n) = a }$

What if we reverse the order of the definition like this:

∃ N such that ∀ε > 0, ∀n ≥ N,
$latex |(a_n) -a| < \varepsilon$

This means:

$latex \boxed {\forall n \geq N, (a_n) = a }$

Example:

$latex \displaystyle (a_n) = \frac{3n^{2} + 2n +1}{n^{2}-n-3}$

$latex \displaystyle\text{Prove: } (a_n) \text { convergent? If so, what is the limit ?}$

Proof:
$latex \displaystyle (a_n) = 3 + \frac{5n +10}{n^{2}-n-3}$

$latex n \to \infty, (a_n) \to 3$

Let's prove it.

$latex \text {Let } \varepsilon >0$
$latex \text{Choose N such that } \forall n \geq N, $
$latex \displaystyle |(a_n) -3| = \Bigr|\frac{5n +10}{n^{2}-n-3}\Bigr| < \varepsilon$

$latex \text{Simplify: } \displaystyle \Bigr|\frac{5n +10}{n^{2}-n-3}\Bigr|$
$latex \text{Let } n > 10 $
$latex \displaystyle \Bigr|\frac{5n +10}{n^{2}-n-3}\Bigr| < \frac{6n}{\frac{1}{2}n^{2}}= \frac{12}{n} < \varepsilon$

$Latex \text{Choose } N = \max (10, \frac{12}{\varepsilon})$
$Latex \displaystyle\forall n \geq N,
|(a_n) -3 | < \frac{12}{n} < \varepsilon$

Therefore,
$latex \displaystyle \boxed{ \lim_{n\to\infty} (a_n) = 3 }$ [QED]

[Source]: Excellent Introduction in Modern Math:
"A Concise Introduction to Pure Mathematics (3rd Edition)"
by Martin Liebeck
CRC Press @ 2011

Multi-variable Limit

Analysis is the study of Functions, using the main tool "Limit".

Limit problems appear in:
1. Continuity
2. Derivative
3. Integral
4. Sequence

Multi-variable Functions have different approach of Limit compared to Single-value functions.
Eg. L'Hôpital Rule is not applicable to Multi-variable Functions.

Case 1: Find the Limit (L) of
$latex \displaystyle f(x,y)= \frac{xy}{x^{2}+y^{2}} \text{ at point (0,0)} $

Solution:
Consider the point P(x,y) on f(x,y)
$latex \displaystyle \lim_{x\to 0} f(x,y)=f(0,y)= 0 \text{...(1)}$
$latex \displaystyle\lim_{y\to 0} f(x,y)=f(x,0)=0 \text{...(2)}$

but when P moves along y=x straight line approaching (0,0),
ie. x->0, y=x->0,
$latex \displaystyle\lim_{x\to 0} \frac{xy}{x^{2}+y^{2}}= \frac{x^2}{2.x^2} = \frac{1}{2} \text{...(3)}$

From (1),(2),(3) there are 3 limits {0, 0, 1/2}, hence the Limit L does not exist.

Case 2:
Find limit L of
$latex \displaystyle f(x,y)= \frac{x^{2}+y^{2}}{x^{4}+y^{4}}
\text { at point} (\infty, \infty)$

Solution:
Let y= kx
$latex \displaystyle f(x,y)= \frac{x^{2}+k^{2}y^{2}}{x^{4}+k^{4}y^{4}} = \frac{1+k^{2}}{(1+k^{4})x^2}$
When x -> $latex \infty$,
f(x,y) independent of k => possible limit of 0

Prove: $latex \displaystyle\lim_{(x,y)\to \infty} f(x,y)= L = 0$

$latex \displaystyle \left| \frac{x^{2}+y^{2}}{x^{4}+y^{4}} - 0 \right| = \frac{x^{2}+y^{2}}{x^{4}+y^{4}}
\leq \frac{x^{2}+y^{2}}{2x^{2}y^{2}} = \frac{1}{2x^{2}} + \frac{1}{2y^{2}}$

Note:
$latex (x^{2} - y^{2})^2 \geq 0$
$latex x^{4}-2x^{2}y^{2}+y^{4}\geq 0 $
$latex x^{4}+y^{4} \geq 2x^{2}y^{2}$

$latex \displaystyle\forall \epsilon > 0 \text{, take } \delta \geq \frac{1}{\sqrt{\epsilon}} \text{ such that}$
$latex |x|>\delta, |y|>\delta$
$latex \implies \displaystyle \left| \frac{x^{2}+y^{2}}{x^{4}+y^{4}} - 0 \right| \leq \frac{1}{2x^{2}} + \frac{1}{2y^{2}}
< \frac{1}{2{\delta}^{2}}+\frac{1}{2{\delta}^{2}}$

$latex \implies \displaystyle \left| \frac{x^{2}+y^{2}}{x^{4}+y^{4}} - 0 \right| \leq \frac{1}{2}\left({\frac{1}{\sqrt{\epsilon}}}\right)^{-2}+\frac{1}{2}\left({\frac{1}{\sqrt{\epsilon}}}\right)^{-2} = \frac{\epsilon}{2}+ \frac{\epsilon} {2}$
$latex \implies \displaystyle \left| \frac{x^{2}+y^{2}}{x^{4}+y^{4}} - 0 \right| \leq \epsilon$
$latex \implies \displaystyle\lim_{(x,y)\to \infty} f(x,y)= L =0 $
[QED]

Source: Prof Zhang ShiZao 章士藻(1940-) "Collected Works of Mathematics Education" 数学教育文集
for Ecole Normale Supérieure in China 高等师范学院

εδ Confusion in Limit & Continuity

1. Basic:
|y|= 0 or > 0 for all y

2. Limit: $latex \displaystyle\lim_{x\to a}f(x) = L$ ; x≠a
|x-a|≠0 and always >0
hence
$latex \displaystyle\lim_{x\to a}f(x) = L$
$latex \iff $
For all ε >0, there exists δ >0 such that
$latex \boxed{0<|x-a|<\delta}$
$latex \implies |f(x)-L|< \epsilon$

3. Continuity: f(x) continuous at x=a
Case x=a: |x-a|=0
=> |f(a)-f(a)|= 0 <ε (automatically)
So by default we can remove (x=a) case.

Also from 1) it is understood: |x-a|>0
Hence suffice to write only:
$latex |x-a|<\delta$

f(x) is continuous at point x = a
$latex \iff $
For all ε >0, there exists δ >0 such that
$latex \boxed{|x-a|<\delta}$
$latex \implies |f(x)-f(a)|< \epsilon$

Limit: ε-δ Analysis

For x->0, find limit L of

f(x)= (x³+5x)/x

1) guess L:

f(x)= x(x²+5)/x= x²+5

=> L= 5  when x->0

2) epsilon-delta Proof: find δ in function of ε such that:

|f(x)-5| < ε

|(x²+5)-5| <ε

|x|< √ε

Choose δ=√ε

For all ε, there is δ=√ε such that |x-0|< δ =>|f(x)-5|< ε

If ε=0.5, δ=√0.5=0.25

Rigorous Calculus: ε-δ Analysis

Rigorous Analysis epsilon-delta (ε-δ)
Cauchy gave epsilon-delta the rigor to Analysis, Weierstrass 'arithmatized' it to become the standard language of modern analysis.

1) Limit was first defined by Cauchy in "Analyse Algébrique" (1821)

2) Cauchy repeatedly used 'Limit' in the book Chapter 3 "Résumé des Leçons sur le Calcul infinitésimal" (1823) for 'derivative' of f as the limit of

$latex \frac{f(x+i)-f(x)}{i}$  when i -> 0

3) He introduced ε-δ in Chapter 7 to prove 'Mean Value Theorem': Denote by (ε , δ) 2 small numbers, such that 0< i ≤ δ , and for all x between (x+i) and x,

f '(x)- ε < $latex \frac{f(x+i)-f(x)}{i}$ < f'(x)+ ε

4) These ε-δ Cauchy's proof method became the standard definition of Limit of Function in Analysis.

5) They are notorious for causing widespread discomfort among future math students. In fact, when it was first introduced by Cauchy in the Ecole Polytechnique Lecture, the French Napoleon top students booed at him and Cauchy received warning from the school.

Note 1: From the textbook 'Calculus' (1980, USA):
"If can't understand the 'ε-δ' definition, just memorize it like a poem - at least better than saying it wrongly."
E.g. “For all ε>0, there exists δ>0, …”

Note 2: George Polya: "The students are not trained in 'ε-δ', teaching them Calculus is like dropping these rules from the sky..."

Newtonian Calculus not rigorous !

Why Newton’s Calculus Not Rigorous?

$latex f(x ) = \frac {x(x^2+ 5)} { x}$ ...[1]

cancel x (≠0)from upper and below => $latex f(x )=x^2 +5 $

$latex \mathop {\lim }\limits_{x \to 0} f(x) =x^2 +5= L=5 $ ...[2]

In [1]: we assume x ≠ 0, so cancel upper & lower x
But In [2]: assume x=0 to get L=5
[1] (x ≠ 0) contradicts with [2] (x = 0)

This is the weakness of Newtonian Calculus, made rigorous later by Cauchy's ε-δ 'Analysis'.