MATH2340 Week 1 and 2

## Acknowledgement of country
 I would like to start by acknowledging that we are on the traditional lands of the Pambalong clan of the Awabakal nation.
 I acknowledge and pay my respect to the Elders past, present and emerging.
 I would like to extend this acknowledgement and pay my respects to Aboriginal and/or Torres Strait Islander peoples here today.
 I recognise that First Nations sovereignty was never ceded.
 This continent always was and always will be Aboriginal Land.
 ---
 ## Welcome to MATH2340: Linearity & Continuity
 #### Linearity
 Solving systems of linear equations,
 Data compression,
 Computer graphics,
 Machine learning / AI
 #### Continuity
 Calculus done rigourously,
 Differential equations,
 Scientific laws & engineering principles
 ---
 ## What else will I learn?
 * Abstraction
 * Mathematical argument & writing
 ---
 ## How will I learn?
 * RActively reading lecture notes
 * AttemptingCompleting Canvas quizzes
 * WatchingParticipating in lectures
 * AttendingParticipating in workshops
 * TRepeatedly trying practice problems
 ---
 ### Actively reading lecture notes
 * Read first for the important concepts
 * Then read with a pen & paper
 * take notes
 * record queries
 * complete any exercises
 * Reflect
 * why is it defined that way?
 * why do we care about that theorem?
 * Reread a couple of days later
 ---
 ### Completing Canvas quizzes
 * Unlimited attempts
 * Aim to ensure you are prepared, not to give you a grade
 * If you are unsure, go back to your notes from the reading
 ---
 ### Participating in lectures
 * Come to lectures in person
 * Be ready to *do* mathematics
 * Taking notes is less important: the lecture is recorded
 ---
 ### Participating in workshops
 * Work in groups to solve problems
 * Discussion between group and workshop facilitator
 * The whole time is spent solving problems
 ---
 ### Repeatedly trying practice problems
 * After workshop, write up your solutions (in group / alone)
 * Compare & contrast with model solutions; give yourself feedback
 * Try the problems again
 * Compare & contrast again; give yourself feedback
 ---
 ## Course organisation
 All information is on Canvas, including deadlines

Course coordinator, lecturer (first half semester) 13:00 workshop facilitator
            * Dave Smith dave.smith@newcastle.edu.au

Lecturer (second half semester) 12:00 workshop facilitator
            * Bishnu Lamichhane bishnu.lamichhane@newcastle.edu.au
            ---
            ## Assignments and assessment
            Discussed on Friday
            ---
            ## Mathematical argument
            #### Theorem
            Dave is employed by University of Newcastle.
            #### Proof
            Dave lectures MATH2340.

Dave is employed by University of Newcastle.   □
 
 ---
 ## Mathematical argument
 #### Theorem
 Dave is employed by University of Newcastle.
 #### Proof
 Dave lectures MATH2340.

Therefore,

$\implies$

Dave is employed by University of Newcastle.   □
 #### Lesson
 Without clear implications, there is no argument.
 
 ---
 ### Direction of implications
 #### Correct
 Dave lectures MATH2340.
 $\implies$
 Dave works at UoN.
 #### Incorrect
 Dave works at UoN.
 $\implies$
 Dave lectures MATH2340.
 #### Also correct
 Dave works at UoN.
 $\impliedby$
 Dave lectures MATH2340.
 #### Also correct
 $a^2+b^2=0$.
 $\iff$
 $a=0$ and $b=0$.
 ---
 ![Two attempted proofs](gfx/direction-argument.png "Title")
 ---
 ## Proof by induction
 Useful for proving a statement about all positive integers.

eg: For all $n\in\mathbb{N}$, the last digit of $5^n$ is $5$.

Comprised of:
 * base case: Statement is true for number $1$. 
 eg: $5^1=5$, whose last digit is $5$.
 * step: If statement is true number $n$, then it is true for number $n+1$. 
 eg: if $5^n=10a+5$ for some $a\in\mathbb{Z}$, then $5^{n+1} = 5(10a+5)=50a+25=10(5a+2)+5$, whose last digit is $5$.
 * Conclusion: "Hence, by induction, [statement] is true for all $n\in\mathbb{N}$."
 ---
 ## Example 2 of induction
 #### Theorem
 For all $n\in\mathbb{N}$, $5^{2n}-1$ is a multiple of $3$.
 #### Proof: Base case
 $5^{2\times1}-1 = 25-1 = 24 = 3\times8$, which is divisible by $3$.
 
 ---
 ## Example 2 of induction
 #### Theorem
 For all $n\in\mathbb{N}$, $5^{2n}-1$ is a multiple of $3$.
 #### Proof: Inductive step
 Suppose that, for some particular $n\in\mathbb{N}$, $5^{2n}-1$ is a multiple of $3$.

Then $\exists c\in\mathbb{Z}:5^{2n}-1 = 3c$.

Therefore, $5^{2(n+1)}-1 = 25 (5^{2n}) - 1 = 25 (5^{2n}-1) + 25 - 1 = 25(3c) + 24 = 3(25c) + 3(8)$, which is also divisible by $3$.
 
 ---
 ## Example 2 of induction
 #### Theorem
 For all $n\in\mathbb{N}$, $5^{2n}-1$ is a multiple of $3$.
 #### Proof: Conclusion
 Hence, by induction, $5^{2n}-1$ is a multiple of $3$ is true for every $n\in\mathbb N$.
 
 ---
 ## Anatomy of an inductive proof: eg. 2
 #### Theorem $Q(n)$$Q(n)$$Q(n)$of form $\forall n \in\mathbb N$,  $Q(n)$
 For all $n\in\mathbb{N}$, $5^{2n}-1$ is a multiple of $3$.
 #### Proof: Base case $Q(n)$$Q(n)$$Q(n)$of form $Q(1)$
 $5^{2\times1}-1$ $= 25-1 = 24 = 3\times8$, which is divisible by $3$.
 #### Proof: Inductive step $Q(n)$$Q(n)$$Q(n)$of form $Q(n)\implies Q(n+1)$
 Suppose that, for some particular $n\in\mathbb{N}$, $5^{2n}-1$ is a multiple of $3$.

Then $\exists c\in\mathbb{Z}:5^{2n}-1 = 3c$.

Therefore, $5^{2(n+1)}-1$ $ = 25 (5^{2n}) - 1 = 25 (5^{2n}-1) + 25 - 1 = 25(3c) + 24 = 3(25c) + 3(8)$, which is also divisible by $3$.
 #### Proof: Conclusion $Q(n)$$Q(n)$$Q(n)$of form Hence, by induction, $\forall n \in\mathbb N$,  $Q(n)$
 Hence, by induction, $5^{2n}-1$ is a multiple of $3$ is true for every $n\in\mathbb N$.
 
 ---
 ## Anatomy of an inductive proof: eg. 1
 #### Theorem $Q(n)$$Q(n)$$Q(n)$of form $\forall n \in\mathbb N$,  $Q(n)$
 For all $n\in\mathbb{N}$, the last digit of $5^n$ is $5$.
 #### Proof: Base case $Q(n)$$Q(n)$$Q(n)$of form $Q(1)$
 $5^1$$=5$, whose last digit is $5$.
 #### Proof: Inductive step $Q(n)$$Q(n)$$Q(n)$of form $Q(n)\implies Q(n+1)$
 If $5^n=10a+5$ for some $a\in\mathbb{Z}$, then $5^{n+1}$ $= 5(10a+5)=50a+25=10(5a+2)+5$ whose last digit is $5$.
 #### Proof: Conclusion $Q(n)$$Q(n)$$Q(n)$of form Hence, by induction, $\forall n \in\mathbb N$,  $Q(n)$
 Hence, by induction, $5^{n}$ is divisible by $5$ is true for every $n\in\mathbb N$.
 
 ---
 ## Anatomy of an inductive proof: demonstration
 I need 5 volunteers!
 #### Theorem
 of form $\forall n \in\mathbb N$,  $Q(n)$
 #### Proof: Base case
 of form $Q(1)$
 #### Proof: Inductive step
 of form $Q(n)\implies Q(n+1)$
 #### Proof: Conclusion
 Hence, by induction, $\forall n \in\mathbb N$,  $Q(n)$
 
 ---
 ## Anatomy of an inductive proof: demonstration
 I need 5 volunteers!
 #### Theorem
 of form $\forall n \in\mathbb N$,  $Q(n)$
 #### Proof: Different base case
 of form $Q(3)$
 #### Proof: Inductive step
 of form $Q(n)\implies Q(n+1)$
 #### Proof: Conclusion
 Hence, by induction, $\forall$ integers $n\geqslant3$,  $Q(n)$
 
 ---
 ## More proofs(?) by induction
 ![Two attempted proofs by induction](gfx/failed-induction-1-and-2.png "Title")
 ---
 ## Another proof(?) by induction
 ![An attempted proof by induction](gfx/failed-induction-3.png "Title")
 ---
 ## Sequences: motivation
 $(x,y)$,   $(x,y,z)$ are ordered pair and ordered triplet.

$(a,b,c,d,e,f,g)$ is an ordered heptuplet.

$(a_1,a_2,a_3,\ldots,a_n)=(a_j)_{j=1}^n$ is an ordered $n$-tuplet.

$(a_1,a_2,a_3,\ldots)$, usually written $\\{a_j\\}\_{j=1}^\infty$ is an infinite *sequence*.
            ---
            ## Sequences: Examples
            * $(7,14,21,28,35,\ldots) = \\{7n\\}\_{n=1}^\infty$
            * $\\{a_n\\}\_{n=1}^\infty$ defined by $a\_n = \frac{(-1)^n}n$
            * $\\{c_n\\}\_{n=1}^\\infty$ where $c\_{2j}=j^2$ and $c\_{2j-1}=\frac{1}{j^2}$

Then

$(a_1, a_2, a_3, a_4, a_5, a\_{100}) = $ ?

$(c_1, c_2, c_3, c_4, c_5, c\_{100}) = $ ?
 
 ---
 ## Sequences: Examples
 * $(7,14,21,28,35,\ldots) = \\{7n\\}\_{n=1}^\infty$
 * $\\{a_n\\}\_{n=1}^\infty$ defined by $a\_n = \frac{(-1)^n}n$
 * $\\{c_n\\}\_{n=1}^\\infty$ where $c\_{2j}=j^2$ and $c\_{2j-1}=\frac{1}{j^2}$

Then

$(a_1, a_2, a_3, a_4, a_5, a\_{100}) = (-1,\frac12,\frac{-1}3,\frac14,\frac{-1}5,\frac1{100})$

$(c_1, c_2, c_3, c_4, c_5, c\_{100}) = (1,1,\frac14,4,\frac19,2500)$
 
 ---
 ## Assignments and assessment
 ![From the course outline](gfx/assessments-course-outline.png "Title")

Online quiz: 11 assessments, 1 per week, best BEFORE the Thursday lecture. Unlimited attempts

Weekly workshop discussion: 11 assessments, 1 per week, completed by contributing to your workshop discussion.

Each of these assessments is worth relatively little; they are designed to encourage you to manage your learning in a productive way.
 
 ---
 ## Bounded sequences
 #### Definition
 A sequence $\\{a_n\\}\_{n=1}^\infty$ is *bounded* if there exists some $M\geq0$ such that, for all $n\in\mathbb N$, $\lvert a_n \rvert \leq M$.

We call $M$ a *bound* for $\\{a_n\\}\_{n=1}^\infty$.
            #### Examples
            If $a_n = \frac3n$, then $3$ is a bound for $\\{a_n\\}\_{n=1}^\infty$.

If $b_n = \frac{3n}{n^2+1}$, then $5$ is a bound for $\\{b_n\\}\_{n=1}^\infty$, and so is $2$.

If $c_n = -n^3$, then what is a bound for $\\{c_n\\}\_{n=1}^\infty$?

If $d_n = 1-\frac1n$, then what is a bound for $\\{d_n\\}\_{n=1}^\infty$?
 #### Smallest bound
 We do not need the smallest bound for this course; see MATH2350.
 
 ---
 ## Proof of a bound
 #### Theorem
 If $d_n = 1-\frac1n$, then $1$ is a bound for $\\{d_n\\}\_{n=1}^\infty$.
 #### Proof
 We aim to show that, for all $n\in\mathbb N$, both $-1 \leqslant 1-\frac1n$ and $1-\frac1n \leqslant 1$.
 <div style="display:flex;flex-direction:row;width:100%">
 <div style="flex:1;margin-right:2vw">

\\[\begin{aligned}
 & & n & \geqslant 1 \\\\
 & \implies & 1 & \geqslant \frac1n \\\\
 & \implies & -1 & \leqslant -\frac1n \\\\
 & \implies & 0 & \leqslant 1-\frac1n
 \end{aligned} \\]
 </div>
 <div style="flex:1;margin-left:2vw">

\\[\begin{aligned}
 & & 1 & > 0 \\\\
 & \implies & \frac1n & > 0 \\\\
 & \implies & -\frac1n & < 0 \\\\
 & \implies & 1-\frac1n & < 1
 \end{aligned} \\]
 </div>
 </div>

We have shown stronger statements than we needed.
 
 ---
 ## Deriving the proof of a bound
 We aim to show that, for all $n\in\mathbb N$, $-1 \leqslant 1-\frac1n$.
 <div style="display:flex;flex-direction:row;width:100%">
 <div style="flex:1;margin-right:2vw;border-right: 2px solid var(--r-main-color);">

The final proof is
 \\[\begin{aligned}
 & & n & \geqslant 1 \\\\
 & \implies & 1 & \geqslant \frac1n \\\\
 & \implies & -1 & \leqslant -\frac1n \\\\
 & \implies & 0 & \leqslant 1-\frac1n
 \end{aligned} \\]
 </div>
 <div style="flex:1;margin-left:2vw;border-right: 2px solid var(--r-main-color);">

But to work this out we do
 \\[\begin{aligned}
 & & -1 & \leqslant 1-\frac1n \\\\
 & \implies & \frac1n & \leqslant 2 \\\\
 & \implies & \frac12 & \leqslant n,
 \end{aligned} \\]
 which is true.
 </div>
 <div style="flex:1;margin-left:2vw">

Then make this a proof by
 \\[\begin{aligned}
 & & -1 & \leqslant 1-\frac1n \\\\
 & \impliedby & \frac1n & \leqslant 2 \\\\
 & \impliedby & \frac12 & \leqslant n,
 \end{aligned} \\]
 which is true.
 </div>
 </div>

Then we can reorder and tighten the bound if we want, but it is not required.
 
 ---
 ## Deriving the proof of a bound: your turn
 We aim to show that, for all $n\in\mathbb N$, $1-\frac1n \leqslant 1$.
 <div style="display:flex;flex-direction:row;width:100%">
 <div style="flex:1;margin-right:2vw;border-right: 2px solid var(--r-main-color);">

The final proof is
 \\[\begin{aligned}
 & & 1 & > 0 \\\\
 & \implies & \frac1n & > 0 \\\\
 & \implies & -\frac1n & < 0 \\\\
 & \implies & 1-\frac1n & < 1
 \end{aligned} \\]
 as required.
 </div>
 <div class="fragment daz-switch preload" data-fragment-index="0" style="flex:1;margin-left:2vw;border-right: 2px solid var(--r-main-color);">

But to work this out we do
 \\[\begin{aligned}
 & & 1-\frac1n & \leqslant 1 \\\\
 & \implies & &? \\\\
 & & &\phantom{\leqslant 0,}
 \end{aligned} \\]
 </div>
 <div class="fragment daz-switch" data-fragment-index="1" style="flex:1;margin-left:2vw;border-right: 2px solid var(--r-main-color);">

But to work this out we do
 \\[\begin{aligned}
 & & 1-\frac1n & \leqslant 1 \\\\
 & \implies & -\frac1n & \leqslant 0 \\\\
 & \implies & \frac1n & \geqslant 0 \\\\
 & \implies & 1 & \geqslant 0,
 \end{aligned} \\]
 which is true.
 </div>
 <div class="fragment daz-switch preload" data-fragment-index="0" style="flex:1;margin-left:2vw">

Then make this a proof by
 \\[\begin{aligned}
 & & 1-\frac1n & \leqslant 1 \\\\
 & \phantom{\implies} & & ? \\\\
 & & &\phantom{\leqslant 0,}
 \end{aligned} \\]
 </div>
 <div class="fragment daz-switch" data-fragment-index="1" style="flex:1;margin-left:2vw">

Then make this a proof by
 \\[\begin{aligned}
 & & 1-\frac1n & \leqslant 1 \\\\
 & \impliedby & -\frac1n & \leqslant 0 \\\\
 & \impliedby & \frac1n & \geqslant 0 \\\\
 & \impliedby & 1 & \geqslant 0,
 \end{aligned} \\]
 which is true.
 </div>
 </div>

Then we can reorder and tighten the bound.
 
 ---
 ## Distance between real numbers
 Suppose $x,y\in\mathbb R$.
 Then $\lvert x-y \rvert$ is the *distance* between them.

Note $\lvert x-y \rvert = \lvert y-x \rvert$, as appropriate for a distance.

So "$\lvert x-y \rvert < 5$" means "$x$ and $y$ are within $5$ of each other".

cf. vectors: $\lVert\mathbf x -\mathbf y\rVert$ is the distance between two vectors, the length of the vector starting from $\mathbf y$ and going to $\mathbf x$.
            ---
            ## Limit of a sequence
            #### Definition
            A sequence $\\{a_n\\}\_{n=1}^\infty$ *converges* to *limit* $\ell\in\mathbb R$ if

for all $\varepsilon>0$,   $\exists k\in\mathbb N$ such that $n>k \implies$ $\lvert a_n-\ell \rvert < \varepsilon$.

If so, we say "$\displaystyle\lim\_{n\to\infty} a_n = \ell$" or "$a_n\to\ell$ as $n\to\infty$".

Informally, we think of a limit as "something that the terms in the sequence get closer and closer to".
            But we need a more precise understanding for this course.

#### Examples
            If $a_n = \frac{-1}n$, then $\displaystyle\lim\_{n\to\infty} a_n = 0$.

If $b_n = \frac{n+3}{5-2n}$, then $\displaystyle\lim\_{n\to\infty} b_n = \frac{-1}2$.
 
 ---
 ## Limit of a sequence
 #### Definition
 A sequence $\\{a_n\\}\_{n=1}^\infty$ *converges* to *limit* $\ell\in\mathbb R$ if

for all $\varepsilon>0$,   $\exists k\in\mathbb N$ such that $n>k \implies$ $\lvert a_n-\ell \rvert < \varepsilon$.

#### Or, in words,
 No matter how small we pick $\varepsilon>0$,

we can always go far enough (ie. $k$ far) down the sequence to ensure that

every term after that point (ie. after $k$),

is within $\varepsilon$ of $\ell$.
 #### How to prove convergence if you know the limit
 Construct a machine (a function) that tells you $k$ for any given $\varepsilon$, and that $k$ does the job.
 
 ---
 ## Proof of sequence limit: example 1
 #### Theorem
 If $a_n = \frac{-1}n$, then $\displaystyle\lim\_{n\to\infty} a_n = 0$.
 #### Definition
 A sequence $\\{a_n\\}\_{n=1}^\infty$ *converges* to *limit* $\ell\in\mathbb R$ if

#### How to prove convergence if you know the limit
 Construct a machine (a function) that tells you $k$ for any given $\varepsilon$, and that $k$ does the job.
 #### Proof
 Fix any $\varepsilon>0$.
 Let $k=\lceil\frac1\varepsilon\rceil$.
 If $n>k$, then

#### How to prove convergence if you know the limit
 Construct a machine (a function) that tells you $k$ for any given $\varepsilon$, and that $k$ does the job.
 #### Proof
 Fix any $\varepsilon>0$.
 Let $k=\max\\{10,\lceil\frac{11}{3\varepsilon}\rceil\\}$.
 If $n>k$, then

$\lvert b_n - \ell \rvert$ $= \lvert \frac{n+3}{5-2n}-\frac{-1}2 \rvert = \lvert\frac{(2n+6)+(5-2n)}{10-4n}\rvert=\lvert\frac{11}{10-4n}\rvert = \frac{11}{4n-10} < \frac{11}{3n}$ $<$ $\frac{11}{3k} \leqslant \frac{11}{3\lceil11/3\varepsilon\rceil} \leqslant \frac{11}{3(11/3\varepsilon)} =$ $\varepsilon$.   □
 
 ---
 ## Deriving proof of sequence limit: eg 1
 Theorem:
 If $a_n = \frac{-1}n$, then $\displaystyle\lim\_{n\to\infty} a_n = 0$.
 #### How to prove convergence if you know the limit
 Construct a machine (a function) that tells you $k$ for any given $\varepsilon$, and that $k$ does the job.
 #### Rough work
 Fix any $\varepsilon>0$.

Aim to define $k\in\mathbb N$ such that $n>k \implies$ $\lvert a_n-\ell \rvert < \varepsilon$.

I can calculate $\lvert a_n - \ell \rvert$ $= \lvert \frac{-1}n - 0 \rvert = \frac1n$, and I want this to be $<\varepsilon$.

I may require $n>k$, so that $\frac1n<\frac1k$, so $\lvert a_n - \ell \rvert<$ $\frac1k$, so now I just need $\frac1k\leqslant\varepsilon$.

First guess: $k = \frac1\varepsilon$.
 This gives $\frac1k\leqslant\varepsilon$ as I wanted, but probably $\frac1\varepsilon \notin \mathbb N$. So use $k = \lceil\frac1\varepsilon\rceil$.
 
 ---
 ## Deriving proof of sequence limit: eg 1
 Theorem:
 If $a_n = \frac{-1}n$, then $\displaystyle\lim\_{n\to\infty} a_n = 0$.
 #### Rough work
 Fix any $\varepsilon>0$.

Aim to define $k\in\mathbb N$ such that $n>k \implies$ $\lvert a_n-\ell \rvert < \varepsilon$.

First guess: $k = \frac1\varepsilon$.
 This gives $\frac1k\leqslant\varepsilon$ as I wanted, but probably $\frac1\varepsilon \notin \mathbb N$. So use $k = \lceil\frac1\varepsilon\rceil$.
 #### Proof
 Fix any $\varepsilon>0$.
 Let $k=\lceil\frac1\varepsilon\rceil$.
 If $n>k$, then

$\lvert a_n - \ell \rvert$ $= \lvert \frac{-1}n - 0 \rvert = \frac1n$ $<$ $\frac 1k = \frac1{\lceil1/\varepsilon\rceil} \leqslant \frac1{1/\varepsilon} =$ $\varepsilon$.   □
 
 ---
 ## Deriving proof of sequence limit: eg 2
 Theorem:
 If $b_n = \frac{n+3}{5-2n}$, then $\displaystyle\lim\_{n\to\infty} b_n = \frac{-1}2$.
 #### How to prove convergence if you know the limit
 Construct a machine (a function) that tells you $k$ for any given $\varepsilon$, and that $k$ does the job.
 #### Rough work
 Fix any $\varepsilon>0$.
 Aim to define $k\in\mathbb N$ such that $n>k \implies$ $\lvert b_n-\ell \rvert < \varepsilon$.

Calculate $\lvert b_n - \ell \rvert$ $= \lvert \frac{n+3}{5-2n}-\frac{-1}2 \rvert = \lvert\frac{(2n+6)+(5-2n)}{10-4n}\rvert=\lvert\frac{11}{10-4n}\rvert$, and I want to simplify.

If I assume $4n>10$, then $\lvert\frac{11}{10-4n}\rvert=\frac{11}{4n-10}$.

If I assume $n>10$, then $\frac{11}{4n-10}<\frac{11}{3n}$,
 and I want this to be $<\varepsilon$.

I may require $n>t$, so that $\frac{11}{3n}<\frac{11}{3t}$, so $\lvert b_n - \ell \rvert<$ $\frac{11}{3t}$, so now I just need $\frac{11}{3t}\leqslant\varepsilon$.

First guess: $t = \frac{11}{3\varepsilon}$.
 This gives $\frac{11}{3t}\leqslant\varepsilon$ as I wanted, but probably $\frac{11}{3\varepsilon} \notin \mathbb N$. So use $t = \lceil\frac{11}{3\varepsilon}\rceil$.
 
 ---
 ## Deriving proof of sequence limit: eg 2
 #### Rough work
 Fix any $\varepsilon>0$.
 Aim to define $k\in\mathbb N$ such that $n>k \implies$ $\lvert b_n-\ell \rvert < \varepsilon$.

If I assume $4n>10$, then $\lvert\frac{11}{10-4n}\rvert=\frac{11}{4n-10}$.

So I pick $k =$ $\max\\{ \frac{10}4, 10, \lceil\frac{11}{3\varepsilon}\rceil \\} =$ $\max\\{ 10, \lceil\frac{11}{3\varepsilon}\rceil \\}$.
 
 ---
 ## Deriving proof of sequence limit: eg 2
 #### Rough work
 Fix any $\varepsilon>0$.
 Aim to define $k\in\mathbb N$ such that $n>k \implies$ $\lvert b_n-\ell \rvert < \varepsilon$.

If I assume $4n>10$, then $\lvert\frac{11}{10-4n}\rvert=\frac{11}{4n-10}$.

First guess: $t = \frac{11}{3\varepsilon}$.
 This gives $\frac{11}{3t}\leqslant\varepsilon$ as I wanted, but probably $\frac{11}{3\varepsilon} \notin \mathbb N$. So use $t = \lceil\frac{11}{3\varepsilon}\rceil$.
 #### Proof
 Fix any $\varepsilon>0$.
 Let $k=\max\\{10,\lceil\frac{11}{3\varepsilon}\rceil\\}$.
 If $n>k$, then
 $\lvert b_n - \ell \rvert$ $= \lvert \frac{n+3}{5-2n}-\frac{-1}2 \rvert = \lvert\frac{(2n+6)+(5-2n)}{10-4n}\rvert=\lvert\frac{11}{10-4n}\rvert = \frac{11}{4n-10} < \frac{11}{3n}$ $<$ $\frac{11}{3k} \leqslant \frac{11}{3\lceil11/3\varepsilon\rceil} \leqslant \frac{11}{3(11/3\varepsilon)} =$ $\varepsilon$.   □
 
 ---
 ## Deriving proof of sequence limit: your turn 1
 Theorem:
 If $c_n = 1+\frac{(-1)^n}{6 n}$, then $\displaystyle\lim\_{n\to\infty} c_n = 1$.
 #### How to prove convergence if you know the limit
 Construct a machine (a function) that tells you $k$ for any given $\varepsilon$, and that $k$ does the job.
 #### Rough work
 Fix any $\varepsilon>0$.

Aim to define $k\in\mathbb N$ such that $n>k \implies$ $\lvert c_n-\ell \rvert < \varepsilon$.

I can calculate $\lvert c_n - \ell \rvert$ $=?$

So  pick $k=?$

If you finish, try: If $d_n = \frac{3+(-1)^n}{\sqrt n}$, then $\displaystyle\lim\_{n\to\infty} d_n = 0$.
 
 ---
 ## Deriving proof of sequence limit: your turn 1
 Theorem:
 If $c_n = 1+\frac{(-1)^n}{6 n}$, then $\displaystyle\lim\_{n\to\infty} c_n = 1$.
 #### How to prove convergence if you know the limit
 Construct a machine (a function) that tells you $k$ for any given $\varepsilon$, and that $k$ does the job.
 #### Rough work
 Fix any $\varepsilon>0$.
 Aim to define $k\in\mathbb N$ such that $n>k \implies$ $\lvert c_n-\ell \rvert < \varepsilon$.

I can calculate $\lvert c_n - \ell \rvert$ $=\lvert 1+\frac{(-1)^n}{6 n}-1 \rvert = \frac{1}{6 n}$, and I want this to be $<\varepsilon$.

I may require $n>k$, so that $\frac1n<\frac1k$, so $\lvert c_n - \ell \rvert<$ $\frac1{6 k}$, so now I just need $\frac1{6 k}\leqslant\varepsilon$.

First guess: $k = \frac{1}{6\varepsilon}$.
 This gives $\frac1{6 k}\leqslant\varepsilon$ as required, but probably $\frac{1}{6\varepsilon} \notin \mathbb N$.
 So pick $k = \lceil\frac{1}{6\varepsilon}\rceil$.
 
 ---
 ## Deriving proof of sequence limit: your turn 1
 Theorem:
 If $c_n = 1+\frac{(-1)^n}{6 n}$, then $\displaystyle\lim\_{n\to\infty} c_n = 1$.
 #### Rough work
 Fix any $\varepsilon>0$.
 Aim to define $k\in\mathbb N$ such that $n>k \implies$ $\lvert c_n-\ell \rvert < \varepsilon$.

First guess: $k = \frac{1}{6\varepsilon}$.
 This gives $\frac1{6 k}\leqslant\varepsilon$ as required, but probably $\frac{1}{6\varepsilon} \notin \mathbb N$.
 So pick $k = \lceil\frac{1}{6\varepsilon}\rceil$.
 #### Summary
 If $n>\lceil\frac{1}{6\varepsilon}\rceil$, then I can show that $\lvert c_n - \ell \rvert<\varepsilon$.
 So I pick $k = \lceil\frac{1}{6\varepsilon}\rceil$.
 
 ---
 ## Deriving proof of sequence limit: your turn 1
 Theorem:
 If $c_n = 1+\frac{(-1)^n}{6 n}$, then $\displaystyle\lim\_{n\to\infty} c_n = 1$.
 #### Rough work
 Fix any $\varepsilon>0$.
 Aim to define $k\in\mathbb N$ such that $n>k \implies$ $\lvert c_n-\ell \rvert < \varepsilon$.

$\lvert c_n - \ell \rvert$ $= \lvert 1+\frac{(-1)^n}{6 n}-1 \rvert = \frac{1}{6 n}$ $<$ $\frac1{6 k} = \frac1{6\left\lceil1/6\varepsilon\right\rceil} \leqslant \frac1{6(1/6\varepsilon)} =$ $\varepsilon$.   □
 
 ---
 ## Deriving proof of sequence limit: your turn 2
 Theorem:
 If $d_n = \frac{3+(-1)^n}{\sqrt n}$, then $\displaystyle\lim\_{n\to\infty} d_n = 0$.
 #### How to prove convergence if you know the limit
 Construct a machine (a function) that tells you $k$ for any given $\varepsilon$, and that $k$ does the job.
 #### Rough work
 Fix any $\varepsilon>0$.

Aim to define $k\in\mathbb N$ such that $n>k \implies$ $\lvert d_n-\ell \rvert < \varepsilon$.

I can calculate $\lvert d_n - \ell \rvert$ $=?$

So  pick $k=?$
 
 ---
 ## Deriving proof of sequence limit: your turn 2
 Theorem:
 If $d_n = \frac{3+(-1)^n}{\sqrt n}$, then $\displaystyle\lim\_{n\to\infty} d_n = 0$.
 #### How to prove convergence if you know the limit
 Construct a machine (a function) that tells you $k$ for any given $\varepsilon$, and that $k$ does the job.
 #### Rough work
 Fix any $\varepsilon>0$.
 Aim to define $k\in\mathbb N$ such that $n>k \implies$ $\lvert d_n-\ell \rvert < \varepsilon$.

I can calculate $\lvert d_n - \ell \rvert$ $=\lvert \frac{3+(-1)^n}{\sqrt n} - 0 \rvert = \frac{\lvert 3+(-1)^n \rvert}{\sqrt n} \leqslant \frac4{\sqrt n}$, and I want this to be $<\varepsilon$.

I may require $n>k$, so that $\frac1n<\frac1k$, so $\lvert d_n - \ell \rvert<$ $\frac4{\sqrt k}$, so now I just need $\frac4{\sqrt k}\leqslant\varepsilon$.

First guess: $k = \frac{16}{\varepsilon^2}$.
 This gives $\frac4{\sqrt k}\leqslant\varepsilon$ as required, but probably $\frac{16}{\varepsilon^2} \notin \mathbb N$.
 So pick $k = \lceil\frac{16}{\varepsilon^2}\rceil$.
 
 ---
 ## Deriving proof of sequence limit: your turn 2
 Theorem:
 If $d_n = \frac{3+(-1)^n}{\sqrt n}$, then $\displaystyle\lim\_{n\to\infty} d_n = 0$.
 #### Rough work
 Fix any $\varepsilon>0$.
 Aim to define $k\in\mathbb N$ such that $n>k \implies$ $\lvert d_n-\ell \rvert < \varepsilon$.

First guess: $k = \frac{16}{\varepsilon^2}$.
 This gives $\frac4{\sqrt k}\leqslant\varepsilon$ as required, but probably $\frac{16}{\varepsilon^2} \notin \mathbb N$.
 So pick $k = \lceil\frac{16}{\varepsilon^2}\rceil$.
 #### Summary
 If $n>\lceil\frac{16}{\varepsilon^2}\rceil$, then I can show that $\lvert d_n - \ell \rvert<\varepsilon$.
 So I pick $k = \lceil\frac{16}{\varepsilon^2}\rceil$.
 
 ---
 ## Deriving proof of sequence limit: your turn 2
 Theorem:
 If $d_n = \frac{3+(-1)^n}{\sqrt n}$, then $\displaystyle\lim\_{n\to\infty} d_n = 0$.
 #### Rough work
 Fix any $\varepsilon>0$.
 Aim to define $k\in\mathbb N$ such that $n>k \implies$ $\lvert d_n-\ell \rvert < \varepsilon$.

$\lvert d_n - \ell \rvert$ $= \lvert \frac{3+(-1)^n}{\sqrt n} - 0 \rvert = \frac{\lvert 3+(-1)^n \rvert}{\sqrt n} \leqslant \frac4{\sqrt n}$ $<$ $\frac4{\sqrt k} = \frac4{\sqrt{\left\lceil\frac{16}{\varepsilon^2}\right\rceil}} \leqslant \frac4{\sqrt{\frac{16}{\varepsilon^2}}} = \frac4{4/\varepsilon} =$ $\varepsilon$.   □
 
 ---
 ## Monotonic sequences: ↑
 #### Definition
 A sequence $\\{a_n\\}\_{n=1}^\infty$ is *increasing* if $a_n < a\_{n+1}$ $\forall\\,n\geqslant 1$:
 \\[
 a_1 < a_2 < a_3 < \ldots < a_n < a\_{n+1} < \ldots.
 \\]
 $\\{b_n\\}\_{n=1}^\infty$ is *non-decreasing* if $b_n \leqslant b\_{n+1}$ $\forall\\,n\geqslant 1$:
 \\[
 b_1 \leqslant b_2 \leqslant b_3 \leqslant \ldots \leqslant b_n \leqslant b\_{n+1} \leqslant \ldots.
 \\]
 
 ---
 ## Monotonic sequences: ↓
 #### Definition
 A sequence $\\{a_n\\}\_{n=1}^\infty$ is *decreasing* if $a_n > a\_{n+1}$ $\forall\\,n\geqslant 1$:
 \\[
 a_1 > a_2 > a_3 > \ldots > a_n > a\_{n+1} > \ldots.
 \\]
 $\\{b_n\\}\_{n=1}^\infty$ is *non-increasing* if $b_n \geqslant b\_{n+1}$ $\forall\\,n\geqslant 1$:
 \\[
 b_1 \geqslant b_2 \geqslant b_3 \geqslant \ldots \geqslant b_n \geqslant b\_{n+1} \geqslant \ldots.
 \\]
 
 ---
 ## Monotonic sequences: Examples
 #### Definition
 If a sequence is (non-)increasing or (non-)decreasing, it is called *monotonic*.
 #### Examples
 For sequences $\\{a_n\\}\_{n=1}^\infty$, $\\{b_n\\}\_{n=1}^\infty$, $\\{c_n\\}\_{n=1}^\infty$, $\\{d_n\\}\_{n=1}^\infty$ defined by
 <div style="display:flex;flex-direction:row;width:100%">
 <div style="flex:1;margin-right:2vw;border-right: 2px solid var(--r-main-color);">
 $\displaystyle a_n = n^2+5$
 $\displaystyle b_n = \frac1n$
 $\displaystyle c_n = 7$
 $\displaystyle d_n = 5n-n^2$
 </div>
 <div style="flex:2;margin-left:2vw;">
 $\displaystyle a_n$ is$\displaystyle \phantom{n^2}$
 $\displaystyle b_n$ is$\displaystyle \phantom{\frac1n}$
 $\displaystyle c_n$ is$\displaystyle \phantom{7}$
 $\displaystyle d_n$ is$\displaystyle \phantom{n^2}$
 </div>
 </div>
 
 ---
 ## Monotonic sequences: Examples
 #### Definition
 If a sequence is (non-)increasing or (non-)decreasing, it is called *monotonic*.
 #### Examples
 For sequences $\\{a_n\\}\_{n=1}^\infty$, $\\{b_n\\}\_{n=1}^\infty$, $\\{c_n\\}\_{n=1}^\infty$, $\\{d_n\\}\_{n=1}^\infty$ defined by
 <div style="display:flex;flex-direction:row;width:100%">
 <div style="flex:1;margin-right:2vw;border-right: 2px solid var(--r-main-color);">
 $\displaystyle a_n = n^2+5$
 $\displaystyle b_n = \frac1n$
 $\displaystyle c_n = 7$
 $\displaystyle d_n = 5n-n^2$
 </div>
 <div style="flex:2;margin-left:2vw;">
 $\displaystyle a_n$ is increasing.$\displaystyle \phantom{n^2}$
 $\displaystyle b_n$ is decreasing.$\displaystyle \phantom{\frac1n}$
 $\displaystyle c_n$ is non-increasing and non-decreasing.$\displaystyle \phantom{7}$
 $\displaystyle d_n$ is not monotonic.$\displaystyle \phantom{n^2}$
 </div>
 </div>
 
 ---
 ## Bounded monotonic sequences
 #### Definition (recall)
 A sequence $\\{a_n\\}\_{n=1}^\infty$ is *bounded* if there exists some $M\geq0$ such that, for all $n\in\mathbb N$, $\lvert a_n \rvert \leq M$.

We call $M$ a *bound* for $\\{a_n\\}\_{n=1}^\infty$.
            #### Theorem
            If a sequence is bounded and monotonic then it is convergent.
            #### Theorem (more precise version):
            If a sequence is bounded above and non-decreasing then it is convergent.

If a sequence is bounded below and non-increasing then it is convergent.
 
 ---
 ## Bounded monotonic sequences
 #### Theorem
 If a sequence is bounded and monotonic then it is convergent.
 #### Examples
 For sequences $\\{a_n\\}\_{n=1}^\infty$, $\\{b_n\\}\_{n=1}^\infty$, $\\{c_n\\}\_{n=1}^\infty$, $\\{d_n\\}\_{n=1}^\infty$ defined by
 <div style="display:flex;flex-direction:row;width:100%">
 <div style="flex:2;margin-right:2vw;border-right: 2px solid var(--r-main-color);">
 $\displaystyle a_n = n^2+5$
 $\displaystyle b_n = \frac1n$
 $\displaystyle c_n = 7$
 $\displaystyle d_n = 5n-n^2$
 </div>
 <div style="flex:5;margin-left:2vw;">
 $\displaystyle a_n$ is$\displaystyle \phantom{n^2}$
 $\displaystyle b_n$ is$\displaystyle \phantom{\frac1n}$
 $\displaystyle c_n$ is$\displaystyle \phantom{7}$
 $\displaystyle d_n$ is$\displaystyle \phantom{n^2}$
 </div>
 </div>
 
 ---
 ## Bounded monotonic sequences
 #### Theorem
 If a sequence is bounded and monotonic then it is convergent.
 #### Examples
 For sequences $\\{a_n\\}\_{n=1}^\infty$, $\\{b_n\\}\_{n=1}^\infty$, $\\{c_n\\}\_{n=1}^\infty$, $\\{d_n\\}\_{n=1}^\infty$ defined by
 <div style="display:flex;flex-direction:row;width:100%">
 <div style="flex:2;margin-right:2vw;border-right: 2px solid var(--r-main-color);">
 $\displaystyle a_n = n^2+5$
 $\displaystyle b_n = \frac1n$
 $\displaystyle c_n = 7$
 $\displaystyle d_n = 5n-n^2$
 </div>
 <div style="flex:5;margin-left:2vw;">
 $\displaystyle a_n$ is increasing but not bounded above, so inconclusive.$\displaystyle \phantom{n^2}$
 $\displaystyle b_n$ is decreasing but not bounded below, so convergent.$\displaystyle \phantom{\frac1n}$
 $\displaystyle c_n$ is constant, so bounded & monotonic, so convergent.$\displaystyle \phantom{7}$
 $\displaystyle d_n$ is not monotonic, so inconclusive.$\displaystyle \phantom{n^2}$
 </div>
 </div>
 
 ---
 ## Squeeze theorem
 #### Theorem
 If $\forall\\,n\in\mathbb N,\\, a_n \leqslant b_n \leqslant c_n$ and $\displaystyle\lim\_{n\to\infty} a_n = \ell = \lim\_{n\to\infty} c_n$, then also $\displaystyle\lim\_{n\to\infty} b_n = \ell$.
 #### Example 1
 For sequences $\\{a_n\\}\_{n=1}^\infty$, $\\{b_n\\}\_{n=1}^\infty$, $\\{c_n\\}\_{n=1}^\infty$ defined by
 $\displaystyle a_n = 0$, $\displaystyle b_n = \frac1{n^3}$, $\displaystyle c_n = \frac1n$,

$\displaystyle a_n\to0$ and $\displaystyle c_n\to0$, so $\displaystyle b_n\to0$ also.

![Demonstration of the squeeze theorem example 1](gfx/squeeze-thm-1.png "Title")

---
 ## Squeeze theorem
 #### Theorem
 If $\forall\\,n\in\mathbb N,\\, a_n \leqslant b_n \leqslant c_n$ and $\displaystyle\lim\_{n\to\infty} a_n = \ell = \lim\_{n\to\infty} c_n$, then also $\displaystyle\lim\_{n\to\infty} b_n = \ell$.
 #### Example 2
 For sequences $\\{a_n\\}\_{n=1}^\infty$, $\\{b_n\\}\_{n=1}^\infty$, $\\{c_n\\}\_{n=1}^\infty$ defined by
 $\displaystyle a_n = \frac{-1}n$, $\displaystyle b_n = \frac{(-1)^n}n$, $\displaystyle c_n = \frac1n$,

again $\displaystyle a_n\to0$ and $\displaystyle c_n\to0$, so $\displaystyle b_n\to0$ also.

![Demonstration of the squeeze theorem example 2](gfx/squeeze-thm-2.png "Title")

---
 ## Squeeze theorem
 #### Theorem
 If $\forall\\,n\in\mathbb N,\\, a_n \leqslant b_n \leqslant c_n$ and $\displaystyle\lim\_{n\to\infty} a_n = \ell = \lim\_{n\to\infty} c_n$, then also $\displaystyle\lim\_{n\to\infty} b_n = \ell$.
 #### Your turn
 For sequences $\\{a_n\\}\_{n=1}^\infty$, $\\{b_n\\}\_{n=1}^\infty$, $\\{c_n\\}\_{n=1}^\infty$ defined by
 $\displaystyle a_n = ?$, $\displaystyle b_n = \frac{n^3+20}{n^7+1}$, $\displaystyle c_n = ?$,

we need $\displaystyle a_n\to0$ and $\displaystyle c_n\to0$, so that we can argue $\displaystyle b_n\to0$ also.
 
 ---
 ## Squeeze theorem
 #### Theorem
 If $\forall\\,n\in\mathbb N,\\, a_n \leqslant b_n \leqslant c_n$ and $\displaystyle\lim\_{n\to\infty} a_n = \ell = \lim\_{n\to\infty} c_n$, then also $\displaystyle\lim\_{n\to\infty} b_n = \ell$.
 #### Your turn
 For sequences $\\{a_n\\}\_{n=1}^\infty$, $\\{b_n\\}\_{n=1}^\infty$, $\\{c_n\\}\_{n=1}^\infty$ defined by
 $\displaystyle a_n = ?$, $\displaystyle b_n = \frac{n^3+20}{n^7+1}$, $\displaystyle c_n = ?$,

we need $\displaystyle a_n\to0$ and $\displaystyle c_n\to0$, so that we can argue $\displaystyle b_n\to0$ also.

We can use $a_n=0\to0$ and $c_n = \frac{21}n$ because $\frac{21 n^3}{n^7} = \frac{21}{n^4} < \frac{21}n\to0$.
 
 ---
 ## 📷 Algebra of limits
 #### Theorem
 If $\displaystyle\lim\_{n\to\infty}a_n$, $\displaystyle\lim\_{n\to\infty}b_n$ exist and $c\in\mathbb R$, then
 1. $\displaystyle\lim\_{n\to\infty}(a_n \pm b_n) = \left(\displaystyle\lim\_{n\to\infty}a_n\right) \pm \left(\displaystyle\lim\_{n\to\infty}b_n\right)$. 
 1. $\displaystyle\lim\_{n\to\infty}(a_n b_n) = \left(\displaystyle\lim\_{n\to\infty}a_n\right) \left(\displaystyle\lim\_{n\to\infty}b_n\right)$. 
 1. $\displaystyle\lim\_{n\to\infty}\left(\frac{a_n}{b_n}\right) = \frac{\displaystyle\lim\_{n\to\infty}a_n}{\displaystyle\lim\_{n\to\infty}b_n}$, provided $\displaystyle\lim\_{n\to\infty}b_n\neq0$. 
 1. $\displaystyle\lim\_{n\to\infty}(ca_n) = c\left(\displaystyle\lim\_{n\to\infty}a_n\right)$.

#### Proof
 See notes for 1. See workshop for 4. The hardest is 3; try it if you like.
 
 ---
 ## Algebra of limits: proof of product rule
 Theorem:
 If $\displaystyle\lim\_{n\to\infty}a_n$ and $\displaystyle\lim\_{n\to\infty}b_n$ exist, then
 $\displaystyle\lim\_{n\to\infty}(a_n b_n) = \left(\displaystyle\lim\_{n\to\infty}a_n\right) \left(\displaystyle\lim\_{n\to\infty}b_n\right)$.

#### Proof
 We know $\\{a_n\\}\_{n\in\mathbb N}$ is bounded, by $M$, say.
 We know $a_n \to \ell_a$ and $b_n\to\ell_b$ as $n\to\infty$, so
 \\[\begin{aligned}
 \&\forall \varepsilon_a>0 \\; \exists \\, k_a\in\mathbb N : n>k_a \implies \lvert a_n-\ell_a\rvert < \varepsilon_a, \\\\
 \&\forall \varepsilon_b>0 \\; \exists \\, k_b\in\mathbb N : n>k_b \implies \lvert b_n-\ell_b\rvert < \varepsilon_b.
 \end{aligned}\\]
 Fix $\varepsilon>0$.
 We calculate
 \\[\begin{aligned}
 \lvert a_nb_n - \ell_a\ell_b \rvert &= \lvert a_nb_n - a_n\ell_b + a_n\ell_b - \ell_a\ell_b \rvert \\\\
 &\leqslant \lvert a_n \rvert \lvert b_n-\ell_b \rvert + \lvert \ell_b \rvert \lvert a_n - \ell_a \rvert
 < \tfrac\varepsilon2 + \tfrac\varepsilon2,
 \end{aligned}\\]

provided $\lvert a_n-\ell_a \rvert < \frac\varepsilon {2\lvert \ell_b \rvert}$ and $\lvert b_n-\ell_b \rvert < \frac\varepsilon {2\lvert a_n \rvert}$.
 So let $\varepsilon_a = \frac\varepsilon {2\lvert \ell_b \rvert}$ and $\varepsilon_b = \frac\varepsilon {2M}$, and set $k=\max\\{k_a,k_b\\}$.
 Then, $\forall n>k$, $\lvert a_nb_n - \ell_a\ell_b \rvert<\frac\varepsilon2+\frac\varepsilon2=\varepsilon$. $\\;\square$
 
 ---
 ## Algebra of limits: proof of product rule
 Theorem:
 If $\displaystyle\lim\_{n\to\infty}a_n$ and $\displaystyle\lim\_{n\to\infty}b_n$ exist, then
 $\displaystyle\lim\_{n\to\infty}(a_n b_n) = \left(\displaystyle\lim\_{n\to\infty}a_n\right) \left(\displaystyle\lim\_{n\to\infty}b_n\right)$.

provided $\lvert a_n-\ell_a \rvert < \frac\varepsilon {2(1+\lvert \ell_b \rvert)}$ and $\lvert b_n-\ell_b \rvert < \frac\varepsilon {2(1+\lvert a_n \rvert)}$.
 So let $\varepsilon_a = \frac\varepsilon {2(1+\lvert \ell_b \rvert)}$ and $\varepsilon_b = \frac\varepsilon {2(1+M)}$, and set $k=\max\\{k_a,k_b\\}$.
 Then, $\forall n>k$, $\lvert a_nb_n - \ell_a\ell_b \rvert<\frac\varepsilon2+\frac\varepsilon2=\varepsilon$. $\\;\square$
 
 ---
 ## Algebra of limits: examples
 #### Problem
 Let $z_n = \frac3n + \frac{2+n-n^2}{17-2n+4n^2} + 5$.
 Calculate $\displaystyle \lim\_{n\to\infty} z_n$.
 
 ---
 ## Algebra of limits: examples
 #### Problem
 Let $z_n = \frac3n + \frac{2+n-n^2}{17-2n+4n^2} + 5$.
 Calculate $\displaystyle \lim\_{n\to\infty} z_n$.
 #### Attempted solution
 \\[
 \displaystyle \lim\_{n\to\infty} z_n
 = \lim\_{n\to\infty} \left(\frac3n\right) + \lim\_{n\to\infty} \left(\frac{2+n-n^2}{17-2n+4n^2}\right) + \lim\_{n\to\infty} \left(5\right)
 \\]

\\[
 = 3 \lim\_{n\to\infty} \left(\frac1n\right) + \frac{\lim\_{n\to\infty} \left(2+n-n^2\right)}{\lim\_{n\to\infty} \left(17-2n+4n^2\right)} + 5
 \\]
 
 ---
 ## Algebra of limits: examples
 #### Problem
 Let $z_n = \frac3n + \frac{2+n-n^2}{17-2n+4n^2} + 5$.
 Calculate $\displaystyle \lim\_{n\to\infty} z_n$.
 #### Attempted solution
 \\[
 \displaystyle \lim\_{n\to\infty} z_n
 = \lim\_{n\to\infty} \left(\frac3n\right) + \lim\_{n\to\infty} \left(\frac{2+n-n^2}{17-2n+4n^2}\right) + \lim\_{n\to\infty} \left(5\right)
 \\]

\\[
                = 3 \lim\_{n\to\infty} \left(\frac1n\right) + \frac{\lim\_{n\to\infty} \left(2+n-n^2\right)}{\lim\_{n\to\infty} \left(17-2n+4n^2\right)} + 5
            \\]

Error! Neither of the limits in the ratio exist. We misapplied rule 3.
 
 ---
 ## Algebra of limits: examples
 #### Problem
 Let $z_n = \frac3n + \frac{2+n-n^2}{17-2n+4n^2} + 5$.
 Calculate $\displaystyle \lim\_{n\to\infty} z_n$.
 #### Attempted solution
 \\[
 \displaystyle \lim\_{n\to\infty} z_n
 = \lim\_{n\to\infty} \left(\frac3n\right) + \lim\_{n\to\infty} \left(\frac{2+n-n^2}{17-2n+4n^2}\right) + \lim\_{n\to\infty} \left(5\right)
 \\]
 
 ---
 ## Algebra of limits: examples
 #### Problem
 Let $z_n = \frac3n + \frac{2+n-n^2}{17-2n+4n^2} + 5$.
 Calculate $\displaystyle \lim\_{n\to\infty} z_n$.
 #### Attempted solution
 \\[
 \displaystyle \lim\_{n\to\infty} z_n
 = \lim\_{n\to\infty} \left(\frac3n\right) + \lim\_{n\to\infty} \left(\frac{2+n-n^2}{17-2n+4n^2}\right) + \lim\_{n\to\infty} \left(5\right)
 \\]

\\[
 = 3 \lim\_{n\to\infty} \left(\frac1n\right) + \lim\_{n\to\infty} \left(\frac{\frac2{n^2}+\frac n{n^2}-\frac{n^2}{n^2}}{\frac{17}{n^2}-\frac{2n}{n^2}+\frac{4n^2}{n^2}}\right) + 5
 \\]
 
 ---
 ## Algebra of limits: examples
 #### Problem
 Let $z_n = \frac3n + \frac{2+n-n^2}{17-2n+4n^2} + 5$.
 Calculate $\displaystyle \lim\_{n\to\infty} z_n$.
 #### Attempted solution
 \\[
 \displaystyle \lim\_{n\to\infty} z_n
 = \lim\_{n\to\infty} \left(\frac3n\right) + \lim\_{n\to\infty} \left(\frac{2+n-n^2}{17-2n+4n^2}\right) + \lim\_{n\to\infty} \left(5\right)
 \\]

\\[
 = 3 \times 0 + \lim\_{n\to\infty} \left(\frac{\frac2{n^2}+\frac 1{n}-1}{\frac{17}{n^2}-\frac{2}{n}+4}\right) + 5
 \\]
 
 ---
 ## Algebra of limits: examples
 #### Attempted solution
 \\[
 \displaystyle \lim\_{n\to\infty} z_n
 = \lim\_{n\to\infty} \left(\frac3n\right) + \lim\_{n\to\infty} \left(\frac{2+n-n^2}{17-2n+4n^2}\right) + \lim\_{n\to\infty} \left(5\right)
 \\]

\\[
                = 3 \times 0 + \lim\_{n\to\infty} \left(\frac{\frac2{n^2}+\frac 1{n}-1}{\frac{17}{n^2}-\frac{2}{n}+4}\right) + 5
            \\]

\\[
 = 0 + \frac{\displaystyle\lim\_{n\to\infty} \left(\frac2{n^2}+\frac 1{n}-1\right)}{\displaystyle\lim\_{n\to\infty} \left(\frac{17}{n^2}-\frac{2}{n}+4\right)} + 5
 \\]
 
 ---
 ## Algebra of limits: examples
 #### Attempted solution
 \\[
 \displaystyle \lim\_{n\to\infty} z_n
 = \lim\_{n\to\infty} \left(\frac3n\right) + \lim\_{n\to\infty} \left(\frac{2+n-n^2}{17-2n+4n^2}\right) + \lim\_{n\to\infty} \left(5\right)
 \\]

\\[
                = \ldots \hspace{16em}
            \\]

\\[
 = 0 + \frac{\displaystyle\lim\_{n\to\infty} \left(\frac2{n^2}+\frac 1{n}-1\right)}{\displaystyle\lim\_{n\to\infty} \left(\frac{17}{n^2}-\frac{2}{n}+4\right)} + 5
 \\]
 
 ---
 ## Algebra of limits: examples
 #### Solution
 \\[
 \displaystyle \lim\_{n\to\infty} z_n
 = \lim\_{n\to\infty} \left(\frac3n\right) + \lim\_{n\to\infty} \left(\frac{2+n-n^2}{17-2n+4n^2}\right) + \lim\_{n\to\infty} \left(5\right)
 \\]

\\[
                = \ldots \hspace{16em}
            \\]

\\[
                = 0 + \frac{\displaystyle\lim\_{n\to\infty} \left(\frac2{n^2}+\frac 1{n}-1\right)}{\displaystyle\lim\_{n\to\infty} \left(\frac{17}{n^2}-\frac{2}{n}+4\right)} + 5
            \\]

\\[
 = 5 + \frac{\displaystyle\lim\_{n\to\infty} \left(\frac2{n^2}\right)+\lim\_{n\to\infty} \left(\frac 1{n}\right)-\lim\_{n\to\infty} \left(1\right)}{\displaystyle\lim\_{n\to\infty} \left(\frac{17}{n^2}\right)-\lim\_{n\to\infty} \left(\frac{2}{n}\right)+\lim\_{n\to\infty} \left(4\right)}
 = 5 + \frac{0+0-1}{0-0+4} = \frac{19}{4}.
 \\]
 
 ---
 ## Algebra of limits: tactic for application of quotient rule
 #### Theorem
 If $\displaystyle\lim\_{n\to\infty}a_n$ and $\displaystyle\lim\_{n\to\infty}b_n$ exist and $\displaystyle\lim\_{n\to\infty}b_n\neq0$, then
 $\displaystyle\lim\_{n\to\infty}\left(\frac{a_n}{b_n}\right) = \frac{\displaystyle\lim\_{n\to\infty}a_n}{\displaystyle\lim\_{n\to\infty}b_n}$.
 #### Observation
 To use this theorem, the lower limit must both exist and be nonzero.
 #### Tactic
 Divide the numerator and denominator by the fastest growing thing in the denominator.
 #### Example & Exercise: finish the calculation
 \\[
 \lim\_{n\to\infty} \frac{3n^3 - 4n+ \frac{n+1}{n+4}}{12 - 2n^3 + 4n^2}
 = \lim\_{n\to\infty} \frac{3 - \frac{4}{n^2}+ \frac{n+1}{n^3(n+4)}}{\frac{12}{n^3} - 2 + \frac4n}
 = \frac{\lim\_{n\to\infty} \left( 3 - \frac{4}{n^2}+ \frac{n+1}{n^3(n+4)}\right)}{\lim\_{n\to\infty} \left(\frac{12}{n^3} - 2 + \frac4n\right)}
 \\]
 
 ---
 ## Algebra of limits: tactic for application of quotient rule
 #### Example & Exercise: finish the calculation
 \\[
 \lim\_{n\to\infty} \frac{3n^3 - 4n+ \frac{n+1}{n+4}}{12 - 2n^3 + 4n^2}
 = \lim\_{n\to\infty} \frac{3 - \frac{4}{n^2}+ \frac{n+1}{n^3(n+4)}}{\frac{12}{n^3} - 2 + \frac4n}
 = \frac{\lim\_{n\to\infty} \left( 3 - \frac{4}{n^2}+ \frac{n+1}{n^3(n+4)}\right)}{\lim\_{n\to\infty} \left(\frac{12}{n^3} - 2 + \frac4n\right)}
 \\]

The denominator is $\displaystyle \lim\_{n\to\infty} \left(\frac{12}{n^3}\right) - \lim\_{n\to\infty} \left(2\right) + \lim\_{n\to\infty} \left(\frac4n\right) = 0 - 2 + 0 = -2 \neq 0$, so we were justified in using the quotient rule.

Numerator is $\displaystyle \lim\_{n\to\infty} \left(3\right) \\! - \\! \lim\_{n\to\infty} \left(\frac{4}{n^2}\right) \\! + \\! \lim\_{n\to\infty} \left(\frac{n+1}{n^3(n+4)}\right) = 3+\lim\_{n\to\infty} \left(\frac{\frac1{n^3}+\frac1{n^4}}{1+\frac4n}\right)$.

$= 3+\displaystyle\frac{\lim\_{n\to\infty} \left(\frac1{n^3}\right)+\lim\_{n\to\infty} \left(\frac1{n^4}\right)}{\lim\_{n\to\infty} \left(1\right)+\lim\_{n\to\infty} \left(\frac4n\right)} = 3+\frac{0+0}{1+0} = 3$.
 Again, we use that the limit of the denominator is nonzero to justify the second application of the quotient rule.
 
 ---
 ## Algebra of limits: more examples
 Let
 \\[\begin{aligned}
 a_n \&= \frac{700n^3 - \frac{2\sin(n)}{n} + \cos(n)n^2}{3n^2(-1)^n + 25n^3}, \\\\
 b_n \&= \frac{13n^2 - \frac\pi{n^2+7}}{n^3\sin(\pi n) + 7n^2 - 4}.
 \end{aligned}\\]
 Calculate $\displaystyle\lim\_{n\to\infty} a_n$ and $\displaystyle\lim\_{n\to\infty} b_n$.

Calculate also $\displaystyle\lim\_{n\to\infty} \left(5a_nb_n\right)$, $\displaystyle\lim\_{n\to\infty} \left(a_n^2(b_n+4)\right)$ and $\displaystyle\lim\_{n\to\infty} \left(\frac{a_n-1}{7b_n}\right)$.