Discovery work in analysis

In this post, we will deal with statements of the form $$\forall x\;\exists y \bullet P(x,y).$$ To prove these statements, we need some preliminary 'discovery work' before embarking on the proof.

Convergence of sequences
Definition: The sequence $(a_n)$ converges to $l$ if, for any $\epsilon>0$, there exists $N\in \Bbb Z^+$ such that $$n>N \Rightarrow |a_n-l|<\epsilon.$$ We can write either $a_n\to l$ as $n\to \infty$ or $\lim\limits_{n\to \infty}a_n=l$.

Example
A sequence $(a_n)$ is defined by $a_n=\dfrac{3n^2-4n}{(n+1)(n+2)}$ for $n\in \Bbb Z^+$. Show that $\lim\limits_{n\to \infty}a_n=3$.

The definition of convergence involves two quantifiers and is of the form $$\forall \epsilon\;\exists N \bullet P(\epsilon,N)$$ where $P(\epsilon,N)$ is a propositional function and the universes for $\epsilon$ and $N$ are $\Bbb R^+$ and $\Bbb Z^+$ respectively. The structure of the proof starts with an arbitrary $\epsilon$ (in its universe) and then selects a particular $N$ (in its universe) which may depend on $\epsilon$. To complete the proof, we need to show that the propositional function $P(\epsilon,N)$ is satisfied for the arbitrary $\epsilon$ and particular $N$.

We first start by the definition. We need to find a positive integer $N$ such that for all $n>N$, $$|a_n-l|=\left|\frac{3n^2-4n}{(n+1)(n+2)}-3 \right|<\epsilon.$$ But how? We can manipulate $|a_n-l|$ until it is less than any positive $\epsilon$: $$\begin{align}\left| \dfrac{3n^2-4n}{(n+1)(n+2)}-3 \right|&=\left| \dfrac{3n^2-4n-3(n+1)(n+2)}{(n+1)(n+2)} \right|\\&=\left| \dfrac{3n^2-4n-(3n^2+9n+6)}{(n+1)(n+2)} \right|\\&=\left| \dfrac{-13n-6}{(n+1)(n+2)} \right|\\&=\dfrac{13n+6}{(n+1)(n+2)}.\end{align}$$ We want to show $\dfrac{13n+6}{(n+1)(n+2)}$ is less than or equal to some simple fraction which can still be made less than $\epsilon$. Since replacing the numerator with a larger value and replacing the denominator with a smaller value both increases the size of the expression respectively, we have $$\dfrac{13n+6}{(n+1)(n+2)}\leq \dfrac{19n}{(n+1)(n+2)}\leq \dfrac{19n}{n^2}=\dfrac{19}{n}.$$ Now to ensure $\dfrac{19}{n}<\epsilon$, we can take $n>\dfrac{19}{\epsilon}$. But we are not done yet. Recall that the definition requires $N$ to be a positive integer. To fix this, we take $N$ to be the integer part or floor of $\dfrac{19}{\epsilon}$, which is defined to be the largest integer less than or equal to $\dfrac{19}{\epsilon}$ and denoted by $\lfloor \dfrac{19}{\epsilon} \rfloor$. We can now proceed to the proof.

Proof: Let $\epsilon>0$ and $N=\lfloor \dfrac{19}{\epsilon}\rfloor \in \Bbb Z^+$. Then for all integers $n>N$, we have $$\begin{align}\left| \dfrac{3n^2-4n}{(n+1)(n+2)}-3 \right|&=\left| \dfrac{3n^2-4n-3(n+1)(n+2)}{(n+1)(n+2)} \right|\\&=\left| \dfrac{3n^2-4n-(3n^2+9n+6)}{(n+1)(n+2)} \right|\\&=\left| \dfrac{-13n-6}{(n+1)(n+2)} \right|\\&=\dfrac{13n+6}{(n+1)(n+2)}\\ &\leq \dfrac{13+6n}{(n+1)(n+2)}\\&\leq \dfrac{19n}{n^2}\\&=\dfrac{19}{n}\\&< \epsilon\quad \text{since}\;n>\dfrac{19}{\epsilon}.\end{align}$$ Therefore $\lim\limits_{n\to \infty}\dfrac{3n^2-4n}{(n+1)(n+2)}=3$.

Limits of functions
Definition: Let $f:A\subset \Bbb R \to B\subset \Bbb R$. Let $a\in A$. Then $f(x)$ tends to a limit $l$ as $x$ tends to $a$ if, for any $\epsilon>0$, there exists $\delta>0$ such that $$0<|x-a|<\delta \Rightarrow |f(x)-l|<\epsilon.$$ We can write either $f(x)\to l$ as $x\to a$ or $\lim\limits_{x\to a}f(x)=l$.

Example
Show that $\lim\limits_{x\to 2}2x^2-5x=-2$.

We need to ensure that $|f(x)-l|=|2x^2-5x+2|$ is less than any specified positive $\epsilon$ by taking $x$ such that $0<|x-2|<\delta$. So we consider $|2x^2-5x+2|$: $$|2x^2-5x+2|=|(x-2)(2x-1)|=|x-2||2x-1|.$$ We can control the size of $|x-2|$ since we can choose $\delta$ such that $|x-2|<\delta$. As for $|2x-1|$, we can rewrite it in terms of $|x-2|$ using the triangle inequality: $$\begin{align}|2x-1|&=|2(x-2)+3|\\ &\leq |2(x-2)|+|3|\\ &=2|x-2|+3.\end{align}$$ Therefore $$|2x^2-5x+2|=|x-2||2x-1|\leq |x-2|(2|x-2|+3).$$ Say $|x-2|<1$. Then $2|x-2|+3<5$. If, in addition, $|x-2|<\dfrac{\epsilon}{5}$, then $|x-2|(2|x-2|+3)$ will be less than $\epsilon$. We are now ready to start the proof.

Proof: Let $\epsilon>0$ and $\delta=\min\{1,\dfrac{\epsilon}{5}\}$. Then $$0<|x-2|<\delta \Rightarrow |x-2|<1 \wedge |x-2|<\frac{\epsilon}{5}\\ \begin{align}\Rightarrow |2x^2-5x+2|&=|x-2||2x-1|\\ &\leq |x-2|(2|x-2|+3)\\ &<5|x-2|\\ &<\epsilon \end{align}$$ Therefore $\lim\limits_{x\to 2}2x^2-5x=-2$.

Remark: The choice of $\delta=\min\{1,\dfrac{\epsilon}{5}\}$ is arbitrary. Choosing $\delta=\min\{\dfrac{1}{2},\dfrac{\epsilon}{4}\}$ will also work.