Module 4

Consider the function \(f:\mathbb{R}\to\mathbb{R}\) defined by

\(f(x) = \begin{cases} \frac{x^2-9}{x-3} &\text{if } x \ne 3, \ 9 &\text{if } x = 3. \end{cases}\)

Notice that:

\(\text{Im}(f|(1,5)) = f’’ (1,5) = (4,8) \cup {9}\) \(↪\) interval (open) about 3: \((1,5) = (3-2, 3+2)\)

Notice that we can “remove” 9 from the image, by excluding 3 from the set we are finding the image of:

\(\text{Im}(f|(1,5)\setminus{3})) = f’’(1,5)\setminus{3}) = (4,8)\setminus{6}\)

“Punctured” open interval about 3.

Notice that \((4,8)\setminus{6} \subseteq (4,8)\) and \((4,8)\) is an interval that contains 6, the value that we would like to be the limit as \(x\to3\).

\(l=4\)

\[ \begin{aligned} \mathrm{I}: & (2,6) (e=2) \\ & (3,5) (e=1) \\ & (3.9, 4.1) (e=0.1) \\ \mathrm{I}: & (2,4) \backslash\{3\} (d=1) \\ & (1,5) \backslash\{3\} (d=2) \end{aligned} \]

\frac{x^2-9}{x-3} = \frac{(x+3)(x-3)}{(x-3)} = x+3 (x \neq 3)

\lim_{x \to 3} f(x) = 6

f’’(2, 7) = (5, 10) \setminus {6}

f’’(1, 5) \setminus {3} = (4, 8) \setminus {6}

L=9 f(x) = \begin{cases} x+3 & \text{if } x \ne 3 \ 9 & \text{if } x = 3 \end{cases}

I’m going to describe a “game” for a number L.

In this game, there are two Players: I & II, and Player I always goes first.

I plays an open interval about L, i.e., an interval of the form \((L-e, L+e)\) for some positive (real) number \(e\).

II plays a “punctured” open interval about 3, i.e., an interval of the form \((3-d, 3+d)\setminus{3}\) for some positive (real) number \(d\).

Player II wins the game if \(f’’(3-d, 3+d) \setminus{3}) \subseteq (L-e, L+e)\)

Example Game #1: (\(L = 9\)) Player I gives/plays \((9-4, 9+4) = (5, 13)\)

Player II Attempts: \(f’’((1, 5)\setminus{3}) = (4, 8)\setminus{6} \nsubseteq (5, 13) (d=2)\) \(f’’((2, 4)\setminus{3}) = (5, 7)\setminus{6} \subseteq (5, 13) (d=1)\)

Player II can play to win!

(If Player II uses any \(e \in (0, 1)\), they’ll win)

Example Game #2: (\(L=9\)) Player I gives/plays \((9-1, 9+1) = (8,10)\)\ Player II Attempts : \(f’’((1,5)\backslash{3}) = (4,0) \backslash{6} \not\subset (8,10)\) (\(d=2\))\ \(f’’((2,4)\backslash{3}) = (5,7)\backslash{6} \not\subset (8,10)\) (\(d=1\))

Notice that as Player II decreases \(e\), the image will stay disjoint from \((8, 10)\), so the image will never be a subset. As \(e\) increases, the image will still have values that are not in \((8,10)\).\ Player II can’t play to win.

Example Game #3: (\(L=6\)) Player I gives/plays \((6-1, 6+1) = (5,7)\)\ Player II can play \((2,4)\backslash{3}\) to win:\ \(f’’((2,4)\backslash{3}) = (5,7)\backslash{6} \subseteq (5,7)\)

Example Game #4: (\(L=6\)) Player I gives/plays \((6-0.01, 6+0.01) = (5.99, 6.01)\)\ Player II can play \((2.99, 3.01) \backslash {3}\) to win:\ \(f’’((2.99, 3.01)\backslash{3}) = (5.99, 6.01) \backslash{6} \subseteq (5.99, 6.01)\)

Notice that in the games where L=9, Player I could play to win (in Game #2), by making e small enough. However, when L=6, no matter how small Player I made e, Player II could make d small enough to still win. It turns out that for any other value of L (other than 6) Player I can win (This will actually give us lead us to a definition of the limit.)

\begin{center} \begin{tikzpicture} \draw[->] (0,0) – (4,0); \draw[->] (0,0) – (0,2); \draw[orange] (-0.5,0.5) – (3.5,1.5); \draw[dashed] (0,0) – (0,1.2); \draw[dashed] (0.8,0) – (0.8,1.2); \draw[dashed] (1.5,0) – (1.5,1.2); \draw[dashed] (0,1.2) – (1.5,1.2);

    \node[below] at (0.8,0) {\\(3-d\\)};
    \node[below] at (1.5,0) {\\(3+d\\)};
    \node[left] at (0,1.2) {\\(6+e\\)};
    \node[left] at (0,0.5) {\\(6-e\\)};
    \node[left] at (0,0) {\\(6\\)};
\end{tikzpicture}

\end{center}

\begin{center} \begin{tikzpicture} \draw[->] (0,0) – (4,0); \draw[->] (0,0) – (0,2); \draw[orange] (-0.5,0.5) – (3.5,1.5); \draw[dashed] (0,0) – (0,1.2); \draw[dashed] (0.8,0) – (0.8,1.2); \draw[dashed] (1.5,0) – (1.5,1.2); \draw[dashed] (0,1.2) – (1.5,1.2);

    \node[below] at (0.8,0) {\\(3-d\\)};
    \node[below] at (1.5,0) {\\(3+d\\)};
    \node[left] at (0,1.2) {\\(6+e\\)};
    \node[left] at (0,0.5) {\\(6-e\\)};
    \node[left] at (0,0) {\\(6\\)};
\end{tikzpicture}

\end{center}

\begin{center} \begin{tikzpicture} \draw[->] (0,0) – (4,0); \draw[->] (0,0) – (0,2); \draw[orange] (-0.5,0.5) – (3.5,1.5); \draw[dashed] (0,0) – (0,1.2); \draw[dashed] (0.8,0) – (0.8,1.2); \draw[dashed] (1.5,0) – (1.5,1.2); \draw[dashed] (0,1.2) – (1.5,1.2);

    \node[below] at (0.8,0) {\\(3-d\\)};
    \node[below] at (1.5,0) {\\(3+d\\)};
    \node[left] at (0,1.2) {\\(6+e\\)};
    \node[left] at (0,0.5) {\\(6-e\\)};
    \node[left] at (0,0) {\\(6\\)};
\end{tikzpicture}

\end{center}

“No matter what I chooses for \(e\), II can find a d such that \(f”((3-d, 3+d){3}) \subseteq (6-e, 6+e)"\)

Determines who “wins” the game

Let \(\varphi_{\lim} (f, 3, 6)\) be the formula/predicate:

“For every positive value of \(\epsilon\), there exists a positive value of \(\delta\) such that \(f’’( (3-\delta, 3+\delta) \setminus {3} ) \subseteq (6-\epsilon, 6+\epsilon)\). "

We can more formally write this as follows: (Using \(\epsilon\) (epsilon) for \(\epsilon\) and \(\delta\) (delta) for \(\delta\))

\((\forall \epsilon > 0) (\exists \delta > 0) (f’’( (3-\delta, 3+\delta) \setminus {3} ) \subseteq (6-\epsilon, 6+\epsilon)) (\varphi_{\lim}(f,3,6))\)

Notation: \((\forall \epsilon > 0)\) is shorthand for \((\forall \epsilon \in (0, \infty))\)

\((\exists \delta > 0)\) is shorthand for \((\exists \delta \in (0, \infty))\)

Generalization:

\((\forall \epsilon > 0) (\exists \delta > 0) (f’’(a-\delta, a+\delta) {a} ) \subseteq (L-\epsilon, L+\epsilon)) (\varphi_{\lim}(f, a, L))\)

\[ \begin{aligned} & A \subseteq B \rightarrow (\forall x) (x \in A \Rightarrow x \in B) \\ & x \in f''A \\ & \Downarrow \\ & f(x) \in A \\ & x \in f''(A) \\ \end{aligned} \] \[ \begin{aligned} & (\forall x) (x \in f''( (a-\delta, a+\delta) \setminus \{a\}) ) \Rightarrow y \in (L-E, L+E) \\ & (\forall y) (f(y) \in (a-\delta, a+\delta) \Rightarrow y \in (L-E, L+E) \\ & x \in Dom (f) \end{aligned} \] \[ \begin{aligned} & x \notin Dom(f) \rightarrow \text{false and so } \Rightarrow \text{true} \checkmark \\ & x \in Dom(f) \rightarrow \text{imp sam.} \end{aligned} \] \[ \begin{aligned} A \subseteq B &\text{iff } (\forall \alpha) (\alpha \in A \Rightarrow \alpha \in B)\\ &[(\forall \alpha \in A) (\alpha \in A \Rightarrow \alpha \in B)]\\ (\forall \alpha) (\alpha \in f"((a-\delta, a+\delta) \setminus\{a\}) \rightarrow& \alpha \in (l-\epsilon, l+\epsilon))\\ \epsilon f"\dots\\ \alpha \in f"((a-\delta, a+\delta) \setminus \{a\})\\ &\text{iff } \underbrace{f(x)}_{\text{A278}} = f(x) \text{ for some } x \in dom(f)\\ &\text{iff } \\ &\alpha \in f"x" \text{iff }\\ &(\forall x \in dom(f)) (x \in (a-\delta, a+\delta) \setminus \{a\}) \Rightarrow f(x) \in (l-\epsilon, l+\epsilon)\\ &|x-a| < \delta \Rightarrow |f(x)-l| < \epsilon \end{aligned} \]

B \in (A \ominus b, A+b) \ A-b \leq B \leq A+b \ -b \leq B-A \leq b \ (B-A) \leq b

\[ \begin{aligned} & \sqrt{3^2+5} \le x-5 \le \sqrt{3^2+5} \\ & \sqrt{3^2+5} > |x-5| \le 0 \text{free}\\ & 3 > x^2+2 \le 0 \\ & 3 > x^2 \ge 0 \\ & 3 > \sqrt{x^2} \text{since} \\ & 3 = \sqrt{9} \\ & x \in (5-3,5+3) \\ & x \in (5-3, 5+8) \backslash \{5\} \\ & (3-3,3-3) \cup \\ &f(x) = (x-5)^2 \\ & \end{aligned} \]

\(\exists x \in (a-\delta, a+\delta) \backslash {a} \implies f(x) \notin (L-\epsilon, L+\epsilon) \)

\(\downarrow\)

\(\forall \epsilon > 0 \ \ \exists \delta > 0 \ \ \ ( \delta \le \epsilon ) \implies (0<3A)\)

\[ \begin{aligned} & \text{Open} \\ f(x) &= 4x - 2 \\ \lim_{x \to 3} f &= 10 \text{Play game} \\ & (3+0^+ (f(x) \Rightarrow (10 - \epsilon, 10 + \epsilon) \end{aligned} \]

Notice that in the games when L=9, Player I could play to win (in Game #2), by making e small enough. However, when L=6, no matter how small Player I made e, Player II could make d small enough to still win. It turns out that for any other value of L (other than 6) Player I can win. (This will actually give us/lead us to a definition of the limit.)

\begin{center} \begin{tikzpicture}[scale=0.7] \draw[->] (-1,0) – (6,0); \draw[->] (0,-1) – (0,6); \draw[orange, thick] (-1,1.5) – (6,5);

\draw[dashed] (1,0) – (1,5); \draw[dashed] (0,3) – (5.5,3); \node[below] at (1,0) {3-d}; \node[below] at (2,0) {3}; \node[below] at (3,0) {3+d}; \node[left] at (0,4) {6-e}; \node[left] at (0,5) {6}; \node[left] at (0,6) {6+e}; \draw[fill=white] (3.5,3.8) circle (0.1); \fill (5,5) circle (0.1); \end{tikzpicture}

\begin{tikzpicture}[scale=0.7] \draw[->] (-1,0) – (6,0); \draw[->] (0,-1) – (0,6); \draw[orange, thick] (-1,1.5) – (6,5);

\draw[dashed] (1,0) – (1,4); \draw[dashed] (0,3) – (3,3); \node[below] at (1,0) {3-d}; \node[below] at (2,0) {3}; \node[below] at (3,0) {3+d}; \node[left] at (0,3.5) {6-e}; \node[left] at (0,4.5) {6}; \node[left] at (0,5.5) {6+e}; \draw[fill=white] (3,3.7) circle (0.1); \fill (4,5) circle (0.1); \end{tikzpicture}

\begin{tikzpicture}[scale=0.7] \draw[->] (-1,0) – (6,0); \draw[->] (0,-1) – (0,6); \draw[orange, thick] (-1,1.5) – (6,5);

\draw[dashed] (1,0) – (1,3); \draw[dashed] (0,2) – (3,2); \node[below] at (1,0) {3-d}; \node[below] at (2,0) {3}; \node[below] at (3,0) {3+d}; \node[left] at (0,2.5) {6-e}; \node[left] at (0,3.5) {6}; \node[left] at (0,4.5) {6+e}; \draw[fill=white] (3,2.8) circle (0.1); \fill (4,5) circle (0.1); \end{tikzpicture}

\end{center}

“No matter what I chooses for e, Player II can find a d such that "

\(f”((3-d, 3+d) \backslash {3}) \subseteq (6-e, 6+e)”\)

Determines who “wins” the game

I am sorry, but I am unable to produce accurate text as the content appears to be too faint or unclear.

Let \(\varphi_{\lim}(f, 3, 6)\) be the formula/predicate:

“For every positive value of \(\epsilon\), there exists a positive value of \(\delta\), such that \(f’’( {3-\delta, 3+\delta } ) \subseteq (6-\epsilon, 6+\epsilon)\).”

We can more formally write this as follows: (Using \(\epsilon\) (epsilon) for \(\epsilon\) and \(\delta\) (delta) for \(\delta\))

\[ (\forall \epsilon>0) (\exists \delta>0) (f''( \{3-\delta, 3+\delta \}) \subseteq (6-\epsilon, 6+\epsilon)) (\varphi_{\lim} (f, 3, 6)) \]

Notation: \((\forall \epsilon > 0)\) is shorthand for \((\forall \epsilon \in (0, \infty))\)

\((\exists \delta > 0)\) is shorthand for \((\exists \delta \in (0, \infty))\)

Generalization:

\[ (\forall \epsilon > 0) (\exists \delta > 0) (f''( \{a-\delta, a+\delta \}) \subseteq (L-\epsilon, L+\epsilon)) (\varphi_{\lim} (f, a, L)) \] \[ \begin{aligned} &\text { In our "Limit Games", Player I plays an interval around a value "L", then } \\ &\text { Player II plays a punctured interval around a value "a". } \\ &\text { Player II wins if whenever we take x from II's punctured interval, f(x) stays } \\ &\text { in Player I's interval. } \\ &\text { More symbolically, Player II wins if: } \\ &\left(\forall \epsilon>0\right)\left(\exists \delta>0\right)\left(\forall x \in \operatorname{Dom}(f) \backslash\{a\}\right)(x \in(a-\delta, a+\delta) \Rightarrow f(x) \in(L-\epsilon, L+\epsilon)) \\ &\uparrow \\ &\text { removes } \\ &x=a \text { from } \\ &\text { consideration } \\ &(\text { Punctures "II's interval) } \\ &\text { II's interval } \text { I's interval } \\ &\text { We say that } \lim _{x \rightarrow a} f(x)=L \text { exactly when II can always play to win no } \\ &\text { matter what player I plays: } \\ &\lim _{x \rightarrow a} f(x)=L \text { iff } \\ &\left(\forall \epsilon>0\right)\left(\exists \delta>0\right)\left(\forall x \in \operatorname{Dom}(f) \backslash\{a\}\right)(x \in(a-\delta, a+\delta) \Rightarrow f(x) \in(L-\epsilon, L+\epsilon)) \\ &\lim (f, a, L) \end{aligned} \]

Playing the Game Algebraically \ Let \(f(x) = (x-5)^2\). We claim \(\lim_{x \to 5} (f, 5, 0)\), i.e., that \(\lim_{x \to 5} f(x) = 0\). (\(f: \mathbb{R} \to \mathbb{R}\))

If this is in fact the case, Player II can always play to win.

Let’s say I plays \(\epsilon = 2\). Player II needs to find/play a value \(\delta\) such that whenever \(x \in (5-\delta, 5+\delta) \setminus {5}\) we have \(f(x) \in (0-2, 0+2)\).

We work backwards from what we want:

\(f(x) \in (-2, 2)\)

\(-2 < f(x) < 2\)

\(-2 < (x-5)^2 < 2\)

Any real squared is greater than or equal to zero. \(0 \le (x-5)^2 < 2\)

As \((x-5)^2 \ge 0\), the square root is defined. \( 0 \le |x-5| < \sqrt{2} \) If \(|A| < B\), then \(-B < A < B\). Also: \(\sqrt{A^2} = |A|\).

\(-\sqrt{2} < x - 5 < \sqrt{2}\)

\(5 - \sqrt{2} < x < 5 + \sqrt{2}\)

\(x \in (5 - \sqrt{2}, 5 + \sqrt{2})\)

This suggests that we can let \(\delta = \sqrt{2}\) (or any smaller value.)

\[ \begin{aligned} \sqrt{x^2} &= |x| \\ 0 &\le |A| < 17 \\ |A| &< 17 \\ -17 &< A < 17 \end{aligned} \]

x \in (5 - \sqrt{e}, 5 + \sqrt{e}) \ x \in (5 - \delta, 5 + \delta)

\[ \begin{aligned} \delta &> |a - x| \\ -\delta &< x - a < \delta \\ a - \delta &< x < a + \delta \end{aligned} \]

Generalizing: (Still consider \(f(x)=(x-5)^2\), \(a=5\), \(L=0\))

We know I will play some \(\epsilon > 0\), with this information, let’s see if we can determin II’s play (\(\delta > 0\)) based off of what value I chooses for \(\epsilon\).

Again, we’ll work backwards from what we want:

\[ \begin{aligned} f(x) &\in (-\epsilon, 0 + \epsilon) \\ f(x) &\in (-\epsilon, \epsilon) \\ -\epsilon &< f(x) < \epsilon \\ -\epsilon &< (x-5)^2 < \epsilon \\ 0 &\leq (x-5)^2 < \epsilon \\ 0 &\leq |x-5| < \sqrt{\epsilon} \\ -\sqrt{\epsilon} &< x-5 < \sqrt{\epsilon} \\ 5-\sqrt{\epsilon} &< x < 5 + \sqrt{\epsilon} \end{aligned} \]

Player II needs to use \(\delta = \sqrt{\epsilon}\) (or a smaller value) to win.

Comment: \(x \in (a-\delta, a+\delta)\) is equivalent to \(|x-a| < \delta\), and

\(f(x) \in (L-\epsilon, L+\epsilon)\) is equivalent to \(|f(x)-L| < \epsilon\), so we can write \(\lim(f, a, L)\) by:

\((\forall \epsilon > 0)(\exists \delta > 0) (\forall x \in \text{Dom}(f) \setminus {a})( |x-a| < \delta \implies |f(x) - L| < \epsilon)\)

Now, consider

\[ f(x) = \begin{cases} (x-5)^2 & \text{if } x \notin \mathbb{Z}, \\ 6 & \text{if } x \in \mathbb{Z}. \end{cases} (f: \mathbb{R} \rightarrow \mathbb{R}) \]

We will again claim \(\lim (f, 5, 0)\).

We again will try to find II’s move \(\delta\) for given moves \(\epsilon > 0\) that Player I makes.

\underline{Player I Plays \(\epsilon = \frac{1}{2}\)}: Need: \(|f(x) - 0| < \frac{1}{2}\) (i.e., \(f(x) \in (0 - \frac{1}{2}, 0 + \frac{1}{2})\))

\[ -\frac{1}{2} < f(x) < \frac{1}{2} \]

For the “replacement”, we know we are excluding 5, but we’ll also want to make \(\delta\) small enough not to include 4 nor 6, as we only want to consider \(f(x)\) near \(x=5\), so that we only use \((x-5)^2\).

This means we’ll need \(\delta \le 1\).

With this restriction:

\[ \begin{aligned} -\frac{1}{2} &< (x-5)^2 < \frac{1}{2} \\ 0 &\le (x-5)^2 < \frac{1}{2} \\ 0 &\le |x-5| < \sqrt{\frac{1}{2}} \longrightarrow \text{we'll need } \delta = \sqrt{\frac{1}{2}} \text{ (or less).} \end{aligned} \]

(\(\sqrt{\frac{1}{2}} < 1\), so this is fine!)

98

\[ \begin{aligned} & \text{Player I Plays } \mathcal{E} = 4: \text{ Need } |f(x)-0| < 4 \\ & -4 < f(x) < 4 \\ & \text{Again, we'll need to require } \delta \leq 1 \text{ so that we only are using } (x-5)^2 \\ & \text{(which is what } f(x) \text{ is equal to "close" to } a=5\text{).} \\ & -4 < (x-5)^2 < 4 \\ & 0 \leq (x-5)^2 < 4 \\ & 0 \leq |x-5| < 2 \implies \text{This tells us that we need } \delta \leq 2 \\ & \text{As we need } \delta \leq 1 \text{ and } \delta \leq 2, \text{ we take the minimum, and we let } \\ & \delta = 1 \text{ (or smaller)}. \\ & \text{Generalizing:} \text{ Need } |f(x) - 0| < \mathcal{E} \\ & -\mathcal{E} < f(x) < \mathcal{E} \\ & \text{When } \delta \leq 1, \text{ we will have } f(x) = (x-5)^2 \\ & -\mathcal{E} < (x-5)^2 < \mathcal{E} \\ & 0 \leq (x-5)^2 < \mathcal{E} \\ & 0 \leq |x-5| < \sqrt{\mathcal{E}} \implies \delta \leq \sqrt{\mathcal{E}} \\ & \text{As } \delta \leq 1 \text{ and } \delta \leq \sqrt{\mathcal{E}}, \text{ we let } \delta = \min(1, \sqrt{\mathcal{E}}) \text{ (or smaller)} \end{aligned} \]

\lim_{x \to a} (f, a, L): (\forall \epsilon > 0)(\exists \delta > 0)(\forall x \in \text{Dom}(f) \setminus {a}: |x-a| < \delta \implies |f(x) - L| < \epsilon)

  • Proof Outline:
  • Proof:
  • Introduce the function \(f\) (Domain, Codomain, “rule”).
  • Let \(\epsilon > 0\) be arbitrary.
  • Let \(\delta = \dots\) (found in scratch work) \(\implies\) Work backwards from \(|f(x) - L| < \epsilon\), to get \(|x-a| < \delta\). \(\delta\) can be based on/depend on \(\epsilon\).
  • Show/State \(\delta > 0\) (and is real). May need to set \(\delta\) equal to a minimum of multiple values.
  • Let \(x \in \text{Dom}(f) \setminus {a}\) be arbitrary.
  • Assume/Suppose \(|x-a| < \delta\).
  • Show \(|f(x) - L| < \epsilon\). \(\implies\) This will generally be the scratch work in reverse, or very similar with explanations added.
  • “Hence, if \(|x-a| < \delta\), then \(|f(x) - L| < \epsilon\).”
  • “As \(x\) was arbitrary, for all \(x \in \text{Dom}(f) \setminus {a}\), if \(|x-a| < \delta\), then \(|f(x) - L| < \epsilon\).”
  • “As \(\delta > 0\), there exists a \(\delta > 0\) such that for all \(x \in \text{Dom}(f) \setminus {a}\), if \(|x-a| < \delta\), then \(|f(x) - L| < \epsilon\).”
  • “As \(\epsilon > 0\) was arbitrary, for all \(\epsilon > 0\), then exists a \(\delta > 0\) such that for all \(x \in \text{Dom}(f) \setminus {a}\), if \(|x-a| < \delta\), then \(|f(x) - L| < \epsilon\).”
  • “Therefore, \(\lim_{x \to a} f(x) = L\).”
  • QED/End Proof Symbol.

For \(f: \mathbb{R} \rightarrow \mathbb{R}\), defined by \(f(x) = (x-5)^2\), we will prove \(\lim_{x \to 5} f(x) = 0\).

Proof: Let \(f: \mathbb{R} \rightarrow \mathbb{R}\) be the function defined by \(f(x) = (x-5)^2\) for all \(x \in \mathbb{R}\). Let \(\varepsilon > 0\) be arbitrary. Let \(\delta = \sqrt{\varepsilon}\). As \(\varepsilon > 0\), we have that \(\delta > 0\). Now, let \(x \in \mathbb{R} \setminus {5}\) be arbitrary. Suppose \(|x - 5| < \delta\). Now:

\[ 0 \leq |x - 5| < \sqrt{\varepsilon} \] \[ 0 \leq (x-5)^2 < \varepsilon. \]

As \((x - 5)^2 \geq 0\), \(|(x - 5)^2| = (x - 5)^2\), so we have:

\[ |(x - 5)^2| < \varepsilon \] \[ |(x-5)^2 - 0| < \varepsilon \] \[ |f(x) - 0| < \varepsilon. \]

Hence, \(|f(x) - 0| < \varepsilon\). Thus, if \(|x - 5| < \delta\), then \(|f(x) - 0| < \varepsilon\). As \(x\) was arbitrary, for all \(x \in \mathbb{R} \setminus {5}\), if \(|x - 5| < \delta\), then \(|f(x) - 0| < \varepsilon\). As \(\delta > 0\), then exists a \(\delta > 0\) such that for all \(x \in \mathbb{R} \setminus {5}\), if \(|x - 5| < \delta\), then \(|f(x) - 0| < \varepsilon\). As \(\varepsilon > 0\) was arbitrary, for all \(\varepsilon > 0\), there exists a \(\delta > 0\) such that for all \(x \in \mathbb{R} \setminus {5}\), if \(|x - 5| < \delta\), then \(|f(x) - 0| < \varepsilon\). Therefore, \(\lim_{x \to 5} f(x) = 0\).

Recall that \(Y_{\text{lim}}(f, a, L)\) is the predicate:

\((\forall \epsilon > 0)(\exists \delta > 0)(\forall x \in \text{Dom}(f)\backslash{a})(|x - a| < \delta \implies |f(x) - L| < \epsilon)\)

We write \(\lim_{x \to a} f(x) = L\) exactly when \(Y_{\text{lim}}(f, a, L)\) is true.

Warm Up: We will prove that \(\lim_{x \to 1} (2x + 7) = 9\). (\(f(x) = 2x + 7\), \(f: \mathbb{R} \to \mathbb{R}\))

“Scratch Work”: \(|f(x) - 9| < \epsilon\) (work backwards)

\(|(2x + 7) - 9| < \epsilon\)

\(|2x - 2| < \epsilon\)

\(2|x - 1| < \epsilon\)

\(|x - 1| < \epsilon/2\) (We set \(\delta = \epsilon/2\))

We can now write the proof.

Proof: Let \(f: \mathbb{R} \to \mathbb{R}\) be the function defined by \(f(x) = 2x + 7\) for all real values of \(x\).

Let \(\epsilon > 0\) be arbitrary. Now, let \(\delta = \epsilon/2\). Notice that as \(\epsilon > 0\), we have \(\delta > 0\).

Now, let \(x \in \mathbb{R} \backslash {1}\) be arbitrary, and suppose \(|x - 1| < \delta\). Now:

\(|x - 1| < \epsilon/2\)

\(2|x - 1| < \epsilon\)

\(|2x - 2| < \epsilon\)

\(|(2x + 7) - 9| < \epsilon\)

\(|f(x) - 9| < \epsilon\).

Hence, we have \(|f(x) - 9| < \epsilon\). Therefore, if \(|x-1| < \delta\), then \(|f(x) - 9| < \epsilon\). As x was arbitrary, for all \(x \in \mathbb{R} \setminus {1}\), if \(|x-1| < \delta\), then \(|f(x) - 9| < \epsilon\). As \(\delta > 0\), there exists a \(\delta > 0\) such that for all \(x \in \mathbb{R} \setminus {1}\), if \(|x-1| < \delta\), then \(|f(x) - 9| < \epsilon\). As \(\epsilon > 0\) was arbitrary, for all \(\epsilon > 0\), then exists a \(\delta > 0\) such that for all \(x \in \mathbb{R} \setminus {1}\) if \(|x-1| < \delta\), then \(|f(x) - 9| < \epsilon\). Hence, we have \(\lim_{x \to 1} f(x) = 9\).

Now, consider \(f : \mathbb{R} \to \mathbb{R}\) defined by

\[ f(x) = \begin{cases} 2x+7 & \text{if } x \neq 1, \\ 23 & \text{if } x = 1. \end{cases} \]

Here, we still have that \(\lim_{x \to 1} f(x) = 9\), and the proof is almost identical:

  • When we introduce the function at the start, we would introduce this function instead.

  • In the “calculation” portion, after we have \(|(2x+7) - 9| < \epsilon\), we would note that as \(x \neq 1\), \(f(x) = 2x+7\), so that we can write \(|f(x) - 9| < \epsilon\). (Recall \(x \in \mathbb{R} \setminus {1}\) was arbitrary, so \(x \neq 1\))

Now, consider \(F:\mathbb{R}\rightarrow \mathbb{R}\) defined by \(x \mapsto \begin{cases} (x-5)^2 & \text{if } x \notin \mathbb{Z}, \ 9 & \text{if } x \in \mathbb{Z}. \end{cases}\)

We will show/prove \(\varphi_{lim} (F, 5, 0)\), i.e., \(\lim_{x\to 5} f(x) = 0\).

Logical Form: \((\forall \epsilon > 0) (\exists \delta > 0) (\forall x \in Dom(f)\setminus{5}) (|x-5| < \delta \implies |f(x) - 0| < \epsilon)\)

Previously: If we just had \(f(x) = (x-5)^2\), we needed to set \(\delta = \sqrt{\epsilon}\). This won’t quite work for us here.

Recall the “game”: If Player I plays \(\epsilon = 4\), Player II will need to find a value of \(\delta\) that will keep the function values in \((0-4, 0+4)\), i.e., between -4 and 4.

If II plays \(\sqrt{\epsilon} = \sqrt{4} = 2\), we are keeping \(x\) between \(5-2=3\) and \(5+2 = 7\), i.e., in the interval: \((3,7) \setminus {5}\).

This will be fine for any \(x\) that is not an integer, however, notice that if \(x=4\) or \(x=6\), we have \(f(x) = 9\), which is not in \((-4, 4)\).

(So Player II would lose)

Instead, Player II needs to make \(\delta\) small enough to avoid these values.

As long as Player I keeps \(\delta \leq 1\), there will be no other integers in \((5-\delta, 5+\delta) \setminus {5}\) so the case when \(x \in \mathbb{Z}\) won’t come up, and we just will have \(f(x) = (x-5)^2\).

% The image contains a hand-drawn x-y coordinate plane.

% * The x-axis is labeled with values, including 1 and 6. % * The y-axis is labeled with values, including 2, 3, 4, and 5. % * A curved line (approximately a parabola) is drawn on the graph. % * There are annotations near the y-axis, appearing to say:

% * z = 5-2 % * h = 4 % * t = 3

I can represent the annotated equations:

\(z = 5 - 2\) \(h = 4\) \(t = 3\)

\[ \begin{aligned} \min(1, 3, 2, 17) &= 1 \\ \min(1, 3, 2, 17) &\leq 1 \\ \min(1, 3, 2, 17) &\leq 3 \\ \min(1, 3, 2, 17) &\leq 2 \\ \min(1, 3, 2, 17) &\leq 17 \\ a &< b \\ & \& \\ & b \leq c \\ a &< c \end{aligned} \] \[ \min(a, b, c) \leq a, \min(a, b, c) \leq b, \text{and} \min(a, b, c) \leq c. \]

Also, when Player II is playing a \(\delta\), any value less than a “winning” \(\delta\), will also be will also win.

In our case, we will set \(\delta = \min(\sqrt{\epsilon}, 1)\). (Note: A min. of positive numbers is positive)

\[ f(x) = \begin{cases} (x-5)^2 & \text{if } x \notin \mathbb{Z}, \\ 9 & \text{if } x \in \mathbb{Z}. \end{cases} \]

Now, let \(x \in \mathbb{R}\).

Let \(\epsilon > 0\) be arbitrary and let \(\delta = \min(\sqrt{\epsilon}, 1)\). As \(\epsilon > 0\), we have that \(\delta > 0\), as both \(\sqrt{\epsilon} > 0\) and \(1 > 0\). Now, let \(x \in \mathbb{R} \setminus {5}\) be arbitrary, and suppose that \(|x - 5| < \delta\).

As \(\delta = \min(\sqrt{\epsilon}, 1)\), we have that \(\delta \leq 1\), so that \(|x - 5| < \delta \leq 1\) and \(|x - 5| < 1\).

From this, \(-1 < x - 5 < 1\) and \(4 < x < 6\). As \(x \neq 5\) and \(4 < x < 6\), we have that \(x \notin \mathbb{Z}\). Hence, we have that \(f(x) = (x - 5)^2\).

Now, as \(\delta = \min(\sqrt{\epsilon}, 1)\), we also have that \(\delta \leq \sqrt{\epsilon}\), so \(|x - 5| < \delta \leq \sqrt{\epsilon}\) and \(|x - 5| < \sqrt{\epsilon}\).

Now:

\[ 0 \leq |x-5| < \sqrt{\epsilon} \] \[ 0 \leq (x-5)^2 < \epsilon \]

As \((x-5)^2 \geq 0\), \(|(x-5)^2| = (x-5)^2\), so we have:

\[ |(x-5)^2| < \epsilon \] \[ |(x-5)^2 - 0| < \epsilon \]

As we found \(f(x) = (x-5)^2\), we have that \(|f(x) - 0| < \epsilon\). Thus, if \(|x-5| < \delta\), then \(|f(x)-0|<\epsilon\).

As \(x\) was arbitrary, for all \(x \in \mathbb{R} \setminus {5}\), if \(|x-5| < \delta\), then \(|f(x) - 0| < \epsilon\).

As \(\epsilon > 0\), there exists a \(\delta > 0\) such that for all \(x \in \mathbb{R} \setminus {5}\), if \(|x-5| < \delta\), then \(|f(x) - 0| < \epsilon\).

As \(\epsilon\) was arbitrary, for all \(\epsilon > 0\), there exists a \(\delta > 0\) such that for all \(x \in \mathbb{R} \setminus {5}\), if \(|x-5| < \delta\), then \(|f(x) - 0| < \epsilon\).

Therefore, \(\lim_{x \to 5} f(x) = 0\).

% \begin{figure}[h!] % \centering % \begin{tikzpicture}[scale=0.5] % \draw[red] (-1,0) – (1,0); % x axis % \draw[red] (0,-1) – (0,3); % y axis % \draw[red] (0,0) parabola bend (1,1) (2,0); % \node[red] at (0.5, -3) {3/\epsilon = 8}; % \node[red] at (-0.5,1) {y}; % \end{tikzpicture} % \end{figure} % fix this\(

\delta = \frac{\epsilon}{3}

\text{Start} |f(x) - 9| < \epsilon

23 9 7 |x - 1| < ??

\delta

Example: Consider \)f: \mathbb{R} \to \mathbb{R}\( defined by

\[ f(x) = \begin{cases} 7x^2 + 6 & \text{if } x > 0, \\ 3|x+2| & \text{if } x < 0, \\ 8 & \text{if } x = 0. \end{cases} \]

We will show \)\lim_{x \to 0} (f,0,6)\(, i.e., \)\lim_{x \to 0} f(x) = 6\(.

There are four different “parts” of the domain of \)f\(:

Cases:

  • \item[(i)] \)x < -2\( \item[(ii)] \)-2 \leq x < 0\( \item[(iii)] \)x=0\( \item[(iv)] \)x > 0\(

We choose \)x \in \mathbb{R} \setminus {0}\(, so this case won’t come up.

If we have \)\delta \leq 2\(, we will have \)-2 < x < 2\(, so case (i) won’t be relevant.

(This means we will set \)\delta = \min(2, \dots)\()

For case (ii): \)-2 \leq x < 0\(. Here: \)f(x) = 3(x+2)\( as \)-2 \leq x < 0\( implies \)x+2 \geq 0\(, so \)|x+2| = x+2\(. Also, \)0 \leq x+2 < 2\(, so

\[ \begin{aligned} |f(x) - 6| &< \epsilon \\ |3(x+2) - 6| &< \epsilon \\ |3x+6-6| &< \epsilon \\ |3x| &< \epsilon \\ 3|x| &< \epsilon \\ |x-0| &< \frac{\epsilon}{3} \longrightarrow \text{we need } \delta \leq \frac{\epsilon}{3}, \text{ so } \delta = \min(2, \frac{\epsilon}{3}, \dots) \end{aligned} \]

For case (iv): Here: \)f(x) = 7x^2 + 6\(.

\)|f(x) - 6| < \epsilon\(\ \)|7x^2 + 6 - 6| < \epsilon\(\ \)|7x^2| < \epsilon\(\ \)7|x^2| < \epsilon\(\ \)|x^2| < \frac{\epsilon}{7}\(\ \)|x| < \sqrt{\frac{\epsilon}{7}}\( \

We need: \)|x - 0| < \sqrt{\frac{\epsilon}{7}} \implies \delta \leq \sqrt{\frac{\epsilon}{7}}\(, so \)\delta = \min(2, \frac{\epsilon}{3}, \sqrt{\frac{\epsilon}{7}})\(

Proof: Let \)f:\mathbb{R}\rightarrow \mathbb{R}\( be the function defined by

\[ f(x)= \begin{cases} 7x^2 + 6 & \text{if } x > 0,\\ 3|x+2| & \text{if } x < 0,\\ 8 & \text{if } x = 0. \end{cases} \]

Let \)\epsilon > 0\( be arbitrary. Define \)\delta = \min(2, \frac{\epsilon}{3}, \sqrt{\frac{\epsilon}{7}})\(. As \)\epsilon > 0\(, we have \)\frac{\epsilon}{3} > 0\( and \)\sqrt{\frac{\epsilon}{7}}>0\(. As \)2>0\( as well, we have \)\delta>0\(. Let \)x \in \mathbb{R} \setminus {0}\( be arbitrary, and suppose \)|x - 0| < \delta\(. We will consider two cases, either \)x < 0\( or \)x > 0\(.

Case 1: Suppose \)x < 0\(. As \)x < 0\(, we have that \)f(x) = 3|x + 2|\(. As \)\delta \leq 2\( and \)|x - 0| < \delta\(, we have that \)|x - 0| < 2\( so that \)|x| < 2\( and \)-2 < x < 2\(. Hence, \)0 < x + 2 < 4\(. As \)x + 2 > 0\(, we have that \)|x + 2| = x + 2\(, so that \)f(x) = 3(x + 2)\(. Also, as \)\delta \leq \frac{\epsilon}{3}\(, we have \)|x - 0| < \frac{\epsilon}{3}\(.

Now: \)|x-0| < \frac{\epsilon}{3}\(

\[ 3|x|<\epsilon \] \[ |3x|<\epsilon \] \[ |3x+6-6| < \epsilon \] \[ |3(x+2)-6|<\epsilon \]

As \)f(x) = 3(x+2)\(, we have \)|f(x)-6|<\epsilon\(.

Case 2: Now, suppose \)x>0\(. Thus, we have that \)f(x)=7x^2+6\(. As \)\delta \leq \sqrt{\frac{\epsilon}{7}}\( and \)|x-0|<\delta\(, we have that \)|x-0|<\sqrt{\frac{\epsilon}{7}}\(. Now:

\[ |x|<\sqrt{\frac{\epsilon}{7}} \] \[ |x|^2 < \frac{\epsilon}{7} \] \[ 7|x|^2 < \epsilon \] \[ 7x^2 < \epsilon \] \[ |7x^2+6-6|<\epsilon \]

As \)f(x) = 7x^2+6\(, we have that \)|f(x)-6|<\epsilon\(.

These cases are exhaustive, hence, \)|f(x)-6|<\epsilon\(. Thus, if \)|x-0|<\delta\(, then \)|f(x)-6|<\epsilon\(. As \)x\( was arbitrary, for all \)x \in \mathbb{R} \setminus {0}\(, if \)|x-0|<\delta\(, then \)|f(x)-6|<\epsilon\(.

As \)\delta > 0\(, there exists a \)\delta > 0\( such that for all \)x \in \mathbb{R} \setminus {0}\(, if \)|x-0|<\delta\(, then \)|f(x)-6|<\epsilon\(. As \)\epsilon\( was arbitrary, for all \)\epsilon > 0\(, there exists a \)\delta > 0\( such that for all \)x \in \mathbb{R} \setminus {0}\(, if \)|x-0|<\delta\(, then \)|f(x)-6|<\epsilon\(. Therefore, \)\lim_{x \to 0} f(x) = 6\(.

\)\square\(

(\exists \epsilon > 0)(\forall \delta > 0)\left( {x \in \mathbb{R} \setminus {0} \mid |x-0| < \delta \Rightarrow |f(x) - 6| < \epsilon } \right)

\[ \begin{aligned} \text{Case (ii) } & -2 \leq x < 0 \\ \text{Here } f(x) &= 3|x+2| \\ & -2 \leq x < 0 \\ & 0 \leq x+2 < 2 \\ & x+2 > 0\\ \text{So } |x+2| &= (x+2) \\ \implies f(x) &= 3(x+2) \\ |f(x) - 6| &< \epsilon \\ |3(x+2)-6| &< \epsilon \\ |3x+6-6| &< \epsilon \\ |3x| &< \epsilon \\ 3|x| &< \epsilon \\ |x| &< \frac{\epsilon}{3} \\ |x-0| &< \frac{\epsilon}{3} \end{aligned} \] \[ \begin{aligned} &\text{Let } f: \mathbb{R} \to \mathbb{R} \text{ be defined by } x \mapsto \begin{cases} x^2, & \text{ if } x \in \mathbb{Q} \\ 0, & \text{ if } x \in \mathbb{P} \end{cases} \\ \\ &f(\sqrt{2}) = 0 \\ &f(0) = 0 \\ &f(4) = 16 \\ &f(\pi) = 0 f(\sqrt{9}) = 9 \\ \\ &\lim_{x \to 0} f(x) = 0. \end{aligned} \] \[ \begin{aligned} & \text{Any \\)\delta\\( works! } \mid \delta = N\epsilon \\ & \begin{array}{c} 3>\epsilon \\ 3>|0| \\ 3>|0-0| \\ 3>|0-f(x)| \\ \text{Case } x \in P \end{array} \begin{array}{c} \delta = \text{min}(\delta_\epsilon, ---\frac{1}{3}) \\ 3 \mu>|x| \\ 3>|2x| \\ 3>|0-x| \\ 3>|0-f(x)| \\ \text{Case } x \in Q \end{array} \\ & (3>|0-f(x)| \Rightarrow \delta > |x-0| ) ( \forall x \in R \backslash \{0 \} ) (0 < \delta \Rightarrow \exists \epsilon ) (0 < \epsilon < 3\Lambda) \end{aligned} \]

Here is the text from the image.

The image seems to represent a function with vertical asymptotes. The vertical lines likely represent the asymptotes and the curved lines shows the function going towards negative infinity when x goes towards 0.

\begin{tikzpicture} \draw[->, red] (-2,0) – (2,0); % x-axis \draw[very thin, blue] (-1,-2) – (-1,2); % First asymptote \draw[very thin, blue] (0,-2) – (0,2); % Second asymptote \draw[very thin, blue] (1,-2) – (1,2); % Third asymptote

\node[below] at (-1,0) {\)1/3\(}; % Position for 1/3 \node[below] at (1,0) {\)1/2\(}; % Position for 1/2 \fill[red] (0,0) circle (2pt); % red dot on the middle

\draw[->, red] (0,0) to[out=270,in=90] (0,-2); \end{tikzpicture}

\lim_{x\to a} f(x) = L \text{iff} (\forall \epsilon > 0) (\exists \delta > 0) (\forall x \in \text{Dom}(f) \setminus {a}) (|x - a| < \delta \implies |f(x) - L| < \epsilon)

f(5) = 25, f(-2) = 4, f(1) = 1

f(0) = 0, f(\frac{1}{2}) \text{ DNE}

{2, 3} = (1.5, 2.5) \cup \mathbb{Z}

\[ \begin{aligned} m &= 0: \{0 + \tfrac{1}{1}, 0 + \tfrac{1}{2}, 0 + \tfrac{1}{3}, 0 + \tfrac{1}{4}, \dots \}\\ &1, 1 + \tfrac{1}{2}, 1 + \tfrac{1}{3}, 1 + \tfrac{1}{4}, 1 + \tfrac{1}{5}, 1 + \tfrac{1}{6} , ...\\ m &= 1: 2, 1\tfrac{1}{2}, 1\tfrac{1}{3}, 1\tfrac{1}{4}, 1\tfrac{1}{5}, ...\\ m &= 2: 3, 2\tfrac{1}{2}, 2\tfrac{1}{3}, 2\tfrac{1}{4}, 2\tfrac{1}{5}, ...\\ m &= -1: 0, -1 + \tfrac{1}{2}, -1 + \tfrac{1}{3}, -1 + \tfrac{1}{4} , ...\\ & -\tfrac{1}{2}, -\tfrac{2}{3}, -\tfrac{3}{4}, ... \end{aligned} \]

\)\varphi \text{lim}(f, a, L)\( iff \)(\forall \epsilon > 0)(\exists \delta > 0)(\forall x \in \text{Dom}(f)\setminus{a})(|x-a|<\delta \implies |f(x) - L| < \epsilon)\(\

Let’s examine the function \)f: \mathbb{Z} \rightarrow \mathbb{Z}\( defined by \)f(x) = x^2\( for all \)x \in \mathbb{Z}\(.

We’ll take a look at the formula \)\varphi \text{lim}(f, 2, 4)\(:

\)(\forall \epsilon > 0)(\exists \delta > 0)(\forall x \in \mathbb{Z}\setminus{2})(|x-2| < \delta \implies |f(x)-4| < \epsilon)\(

In the “Game”, if Player II plays any \)\delta \leq 1\(, \)|x-2| < \delta\( will just be false:

\[ \begin{aligned} |x-2| &< 1 \\ -1 < x-2 &< 1\\ 1 < x &< 3 \end{aligned} \]

As \)x \in \mathbb{Z}\setminus{2}\(, there are no values of \)x\( which satisfy \)1 < x < 3\(.

This makes the implication (trivially) true, so \)\varphi \text{lim}(f, 2, 4)\( is true.

For the same reasoning/logic, \)\varphi \text{lim}(f, 2, 17)\( will also be true.

Using our limit notation, we would write both

\[ \begin{aligned} \lim_{x\to 2} f(x) &= 4 \text{and} \lim_{x\to 2} f(x) = 17 \end{aligned} \]

We don’t want this sort of issue to come up, we want the limit to be well defined & unique.

The issue/problem in this example was that the punctured open interval \)(2-\delta, 2+\delta) \setminus {2}\( was disjoint from the domain of the function, i.e.,:

\[ \text{Dom}(f) \cap (2-\delta, 2+\delta) \setminus \{2\} = \emptyset \text{(for } \delta \leq 1) \] \[ \mathbb{Z} \cap (2-\delta, 2+\delta) \setminus \{2\} = \emptyset \text{(for } \delta \leq 1) \]

In other words, the issue is that we could find an open interval that isolates 2 from the domain of \)f\(.

Let’s consider a similar, but different function:

\[ f : D \to \mathbb{R}, \text{ defined by } x \mapsto x^2 \text{ for all } x \in D \]

Where

\[ D = \mathbb{Z} \cup \left\{ m + \frac{1}{n} \middle| m \in \mathbb{Z}, n \in \mathbb{N} \right\} \] \[ = \left\{ \dots, -3, -2\frac{1}{2}, -2\frac{1}{3}, -2\frac{1}{4}, \dots, -2, -1\frac{1}{2}, -1\frac{1}{3}, -1\frac{1}{4}, \dots \right\} \]

% “Picture” of D: (This part is best represented with an image, as it’s a number line) %fix this

\[ \begin{aligned} &D: \\ & 1.5 \\ & \frac{1}{2} \ \frac{1}{3} \ \frac{1}{4} \ \frac{1}{5} \ \frac{1}{6} \ \frac{1}{7} \ ... \\ & ( .9, 1.1) \cap D = \{\frac{11}{12}, \frac{12}{13}, \frac{13}{14}, ... \} \\ & ( .9, 1.01) \cap D = \{ 1, \frac{1}{101}, \frac{1}{102}, \frac{1}{103}, ... \} \\ & (1.4, 1.6) \cap D = \{ 1.5 \} \\ & ( .9, 2.9) \cap D = D \end{aligned} \] \[ \begin{aligned} D &= (0,1] \cup \{2\}\\ -17&: (-18,-16) \cap D = \emptyset \implies \text{not inf. so -17 is not} \\ & \text{an acc. pt.} \\ 0&: (-1,1) \cap D = (0,1) \implies \text{inf. many \#}'s\\ &(-0.1, 0.1) \cap D = (0, 0.1)\\ &(-0.01, 0.01) \cap D \text{"} \\ & \vdots\\ & \text{0 is an acc. pt. of D} \\ &\text{\& cond. pt.}\\ 6.5&: (6.4, 6.6) \cap D \text{is an acc. pt. of D}\\ 1&: (-1,2) \cap D = (0,1] \\ & (0.001, 1.11) \cap D = (0.001, 1] \\ & (0.9999, 1.00001) \cap D = (0.9999, 1] \\ & \frac{1}{1} \text{ is an acc. pt.} \end{aligned} \]

1.5: \)(0.5, 1.6) \cap D = (0.5, 1] \in inf.\( \)(1, 2) \cap D = \emptyset \leftarrow not , inf\( 1.5 is not an acc. pt of D. 2: \)(1.9, 2.1) \cap D = {2} , not , acc. , pt , of , D\( 3: \)(2.9, 3.1) \cap D = \emptyset , not , acc. , pt. , of , D\(.

  • D
  • 0, \)\frac{1}{5}, \frac{1}{4}, \frac{1}{3}, \frac{1}{2}, 1\(
  • (0.249, 0.251) \)\cap\( D = \){\frac{1}{4}}\( Not acc.pt

F: \)\mathbb{N} \to B\(

\)1 \mapsto 0\(

\)2 \mapsto \frac{1}{3}\(

\)3 \mapsto \frac{1}{4}\(

\)4 \mapsto \frac{1}{5}\(

\)5 \mapsto \frac{1}{6}\(

\)n \mapsto \frac{1}{n+1}\(

\)B\(

inf. so 0 is acc. pt.

Countable / not uncountable.

0 is not a condensation pt. of D.

For this function \)\varphi_{\lim}(f, 2, L)\( will be true if \)L=4\(, but will be false for any \)L \neq 4\(:

For \)L=4\(, whatever Player I chooses for \)\epsilon > 0\(, Player II can “win” by playing \)\delta = \min(2, \frac{\epsilon}{6})\(.

For \)L \neq 4\(, Player I can find a value of \)\epsilon > 0\( small enough that Player II cannot play to win.

\)\leftarrow\( As we make this \)\epsilon\(-window smaller, we will always have infinitely many points in this window close to 2, and any/every \)\delta\(-window will have infinitely many points in the domain.

We can’t isolate 2.

Definition (Accumulation Point) Let \)D \subseteq \mathbb{R}\(. We say that \)a \in \mathbb{R}\( is an accumulation point of \)D\( iff

\[ (\forall c \in \mathbb{R}) (\forall d \in \mathbb{R}) (c < a < d \Rightarrow (c, d) \cap D \text{ is infinite}). \]

Equivalent to:

\[ |( (c, d) \cap D ) \setminus \{a\} | \text{ is infinite} \]

or

\[ |( (c, d) \cap D ) \setminus \{a\} | \geq \aleph_0 \]

Definition (Condensation Point) Let \)D \subseteq \mathbb{R}\(. We say that \)a \in \mathbb{R}\( is a condensation point of \)D\( iff

\[ (\forall c \in \mathbb{R}) (\forall d \in \mathbb{R}) (c < a < d \Rightarrow (c, d) \cap D \text{ is uncountable}). \]

Equivalent to:

\[ |( (c, d) \cap D ) \setminus \{a\} | \text{ is uncountable} \]

or

\[ |( (c, d) \cap D ) \setminus \{a\} | > \aleph_0 \] \[ \begin{aligned} \text{2) }&D = (0, \frac{1}{2}) \cup (\frac{1}{3}, \frac{1}{4}) \cup (2, 2\frac{1}{4}) \cup (3, 3\frac{1}{5}) \cup (4, 4\frac{1}{6}) \cup \dots \\ &\cup \left\{ -\frac{1}{3} + \frac{m}{n+1} : m \in \mathbb{N}, n \in \mathbb{N} \right\}\\ &\to \text{Acc. } \mathbb{Z}^+ \cup \{1\}\\ &\to \text{Cond } \omega \\ \text{1) }&(0, 1)\\ &\text{Cond/Acc. } \to [0, 1] \end{aligned} \] \[ (2.19999, 2.20001) \cap \mathbb{Z} = \{2 \frac{1}{5}\} \]

\usepackage[margin=1in]{geometry}

With these “new” terms, let’s examine the sets:

\) \mathbb{Z}\( and \)D = \mathbb{Z} \cup {m+\frac{1}{n} | m \in \mathbb{Z}, n \in \mathbb{N}}\( (Both \)\mathbb{Z} \subseteq \mathbb{R}\( and \)D \subseteq \mathbb{R}\()

In \)\mathbb{Z}\(:

  • 0
  • 1
  • 2
  • 3
  • 4
  • 1.5
  • 2.5

Notice 1.5 & 2.5 are both real, \)1.5 < 2 < 2.5\(, but \)(1.5, 2.5) \cap \mathbb{Z} = {2}\( which is not infinite.

This tells us that 2 is not an accumulation point (nor a condensation point) of \)\mathbb{Z}\(.

For similar reasons, no value will be an accumulation (condensation) point of \)\mathbb{Z}\(.

In \)D\(: %Diagram:

  • 1
  • c
  • 2
  • d
  • 3

Here, no matter what values of \)c\( & \)d\( are chosen with \)c < 2 < d\(, there will be infinitely many points in \)(c,d) \cap D\( (specifically the points will be to the right of 2).

This tells us that 2 is an accumulation point of \)D\(.

Similarly, every integer will be an accumulation point of \)D\(.

(Still Examining D)

\begin{center}

  1. 9 \)2\frac{1}{2}\( \)2\frac{3}{6}\( \)2\frac{5}{5}\( \)2\frac{1}{4}\( \)2\frac{1}{3}\( 2.3 \end{center}

If we were to choose \)c = 1.9\( and \)d = 2.3\( (\)1.9 < 2 < 2.3\(), we can define \)f:\mathbb{N} \xrightarrow[]{onto}_{1-1} ((1.9, 2.3) \cap D) \setminus {2}\( as follows:

\[ \begin{aligned} 1 &\mapsto 2\frac{1}{4} \\ 2 &\mapsto 2\frac{1}{5} \\ 3 &\mapsto 2\frac{1}{6} \\ 4 &\mapsto 2\frac{1}{7} \\ \vdots& \\ n &\mapsto 2\frac{1}{n+3} \end{aligned} \]

As this function is a bijection, \)((1.9, 2.3) \cap D) \setminus {2}\( is denumerable / countable.

This tells us that 2 is not a condensation point of D. (In fact, every integer in D is not a condensation point).

Also notice that any non-integer value in D will not be an accumulation point.

\[ \begin{aligned} & [e, \pi] \\ & e \curvearrowright (\cdots 2.8 \rightarrow 3 \rightarrow 3.1 \rightarrow \pi) \\ & \text{Accumulation Pts:} [e, \pi] \\ & \text{(also condensation Pts).} \end{aligned} \]

f:\mathbb{Z} \longrightarrow \mathbb{Z}, x \mapsto x^2

a=2

2 is not acc. pt of a.

(Adjusted/Fixed) Definition:

Let \)D \subseteq \mathbb{R}\( and \)E \subseteq \mathbb{R}\(. Also, let \)f: D \to E\( and suppose \)a \in \mathbb{R}\( is an accumulation point of \)D\(. Then:

\)\mathop{\mathcal{Y}\text{lim}} (f, a, L) \text{ iff } (\forall \epsilon > 0)(\exists \delta > 0)(\forall x \in \text{dom}(f) \setminus {a})( |x-a| < \delta \Rightarrow |f(x) - L| < \epsilon)\(

With this additional requirement that \)a\( be an accumulation point of the domain, we avoid the issue of the antecedent being always false.

Theorem: Let \)D \subseteq \mathbb{R}\(, \)E \subseteq \mathbb{R}\(, \)f: D \to E\(, and let \)a \in \mathbb{R}\( be an accumulation point of \)D\(. Also, let \)L \in \mathbb{R}\( and \)\hat{L} \in \mathbb{R}\(.

If \)\mathop{\mathcal{Y}\text{lim}} (f, a, L)\( and \)\mathop{\mathcal{Y}\text{lim}} (f, a, \hat{L})\(, then \)L = \hat{L}\(.

We will prove this theorem. This theorem tells us that the limit is unique (if it exists).

Let \)D \subseteq \mathbb{R}\(. A value \)a \in \mathbb{R}\( is an accumulation point of \)D\(

iff \)(\forall \beta > 0) (((\alpha-\beta, \alpha+\beta) \cap D) \setminus {a} \text{ is infinite})\(

Infinitely many points from \)D\( in this interval (no matter how small \)\beta > 0\( is)

\begin{tikzpicture} \draw[<->] (0,0) – (6,0); \draw[dashed, thick, orange] (0.5, 0.1) – (2.5, 0.1); \draw[dashed, thick, orange] (2.5, 0.1) – (4.5, 0.1); \node at (1.5, 0.3) {\)a-\beta\(}; \node at (3.5, 0.3) {\)a+\beta\(}; \draw[thick] (2.5,0) – (2.5,0.2); \node at (2.5, -0.3) {\)a\(}; \end{tikzpicture}

Eg. Consider \)D = \mathbb{Q}\(, \)a = \pi\(

\begin{tikzpicture} \draw[<->] (0,0) – (6,0); \draw[thick] (1, 0) – (1, 0.2); \draw[thick] (3, 0) – (3, 0.2); \draw[thick] (5, 0) – (5, 0.2); \node at (1, 0.3) {\)\pi - 1\(}; \node at (3, -0.3) {\)\pi \approx 3.141…\(}; \node at (5, 0.3) {\)\pi + 1\(}; \end{tikzpicture} \end{align*} \

\[ \begin{aligned} & \text{(Adjusted/Fixed) Definition:} \\ & \text{Let } D \subseteq \mathbb{R} \text{ and } E \subseteq \mathbb{R} \text{. Also, let } f: D \rightarrow E \text{ and suppose } a \in \mathbb{R} \\ & \text{is an accumulation point of } D \text{. Then:} \\ & \varphi_{\lim}(f, a, L) \text{ iff } (\forall \epsilon > 0)(\exists \delta > 0)(\forall x \in dom(f) \setminus \{a\})(0 < |x - a| < \delta \Rightarrow |f(x) - L| < \epsilon) \\ & \text{With this additional requirement that } a \text{ be an accumulation point of the} \\ & \text{domain, we avoid the issue of the antecedent being always false.} \\ & \text{Theorem: Let } D \subseteq \mathbb{R}, E \subseteq \mathbb{R}, f: D \rightarrow E, \text{ and let } a \in \mathbb{R} \text{ be an accumulation} \\ & \text{point of } D \text{. Also, let } L \in \mathbb{R} \text{ and } \hat{L} \in \mathbb{R} \text{.} \\ & \text{If } \varphi_{\lim}(f, a, L) \text{ and } \varphi_{\lim}(f, a, \hat{L}), \text{ then } L = \hat{L} \text{.} \\ & \text{We will prove this theorem. This theorem tells us that the limit is} \\ & \text{unique (if it exists).} \end{aligned} \]

“(A\land B) \Rightarrow C” : \Psi

\[ Prove \\)\Psi\\( by * \\)\Psi\\( is equivalent to \\)(\neg \Psi \Rightarrow (D \land \neg D))\\( \\)\neg \Psi\\( is \\)\neg((A \land B) \Rightarrow C)\\( Assume \\)(A \land B) \land \neg C\\( \rotatebox{90}{ \begin{minipage}{0.5\textwidth} \[ \[ \begin{aligned} a &\neq b \\ 0 &< |a-b| \end{aligned} \] \] \[ |2-7| = 5 \] \end{minipage} } \[ \begin{aligned} &\lim_{(f, a, L)} \\ &(\forall \epsilon > 0)(\exists \delta_1 > 0) (\forall x \in D \setminus \{a\}) (|x-a| < \delta_1 \implies |f(x) - L| < \epsilon) \\ &\epsilon = 15: \text{ There is a } \delta_1(15) > 0 \text{ s/t } \\ &(\forall x \in D \setminus \{a\}) (|x - a| < \delta_1(15) \implies |f(x) - L| < 15) \\ &\epsilon = 0.02: \text{ There is a } \delta_1(0.02) > 0 \text{ s/t } \\ &(\forall x \in D \setminus \{a\}) (|x - a| < \delta_1(0.02) \implies |f(x) - L| < 0.02) \\ &\epsilon = 0.0000001 \\ &12 - 714 = 3 \end{aligned} \] For \\)\epsilon > 0\\(, \\)\exists \delta > 0\\( There is a corresponding \\)\delta > 0\\( s/t \\)(\forall x \in D \setminus \{a\})(\left|x-a\right|<\delta \Rightarrow \left|f(x) - L\right| < \epsilon)\\( \\)Y_{\lim} (f,a, L)\\( Logical Form of the Theorem \[ (\forall \delta \subseteq \mathbb{R}) (\forall E \subseteq \mathbb{R}) (\forall f \in E^D) (N_a \in Acc(D)) (\forall L \in \mathbb{R}) (\forall L' \in \mathbb{R}) \left[ \left( \varphi_{\lim} (f_a, L) \land \varphi_{\lim} (f_a, L') \right) \Rightarrow L = L' \right] \] Shorthand for: \[ (\forall D \in \mathcal{P}(\mathbb{R})) (\forall E \in \mathcal{P}(\mathbb{R})) \] Acc(D) is the set of all accumulation points of D. E is the set of all functions with domain D \& codomain E. Recall the equivalences: \\)\neg (A \Rightarrow B)\\( and \\)(A \land \neg B)\\( and: \\)C\\( and \\)(\neg C \Rightarrow (D \land \neg D))\\( We will do this proof by contradiction, so we assume: i.e.: \[ \neg \left[ \left( \varphi_{\lim} (f_a, L) \land \varphi_{\lim} (f_a, L') \right) \Rightarrow L = L' \right] \] \[ \left[ \left( \varphi_{\lim} (f_a, L) \land \varphi_{\lim} (f_a, L') \right) \land L \neq L' \right] \] And then, we will produce a contradiction: "(\\)\Psi \land \neg \Psi\\()" "Idea" of the Proof: \begin{tikzpicture} \draw[<->] (0,0) -- (8,0); \draw[dashed] (1,0) -- (1,2); \draw[dashed] (3,0) -- (3,2); \draw[dashed] (5,0) -- (5,2); \draw[dashed] (7,0) -- (7,2); \node at (1,-0.3) {\\)a-\delta\\(}; \node at (3,-0.3) {\\)a\\(}; \node at (5,-0.3) {\\)a+\delta\\(}; \draw[<->] (0,2) -- (8,2); \node at (0,0.3) {\\)a-\delta\\(}; \node at (3,0.3) {\\)a\\(}; \node at (5,0.3) {\\)a+\delta\\(}; \node at (2,2.3) {\\)L\\(}; \node at (6,2.3) {\\)\hat{L}\\(}; \node at (2,-0.5) {\\)L-\epsilon\\(}; \node at (6,-0.5) {\\)\hat{L}-\epsilon\\(}; \node at (2,2.5) {\\)L\\(}; \node at (6,2.5) {\\)\hat{L}\\(}; \node at (2,-0.5) {\\)L-\epsilon\\(}; \node at (6,-0.5) {\\)\hat{L}-\epsilon\\(}; \fill[yellow!50, pattern=north east lines] (1,0) rectangle (3,2); \fill[yellow!50, pattern=north east lines] (5,0) rectangle (7,2); \end{tikzpicture} For a given \\)\epsilon\\(, there will be \\)\delta\\( & \\)\hat{\delta}\\( windows that keep the function inside the windows around \\)\hat{L}\\( & \\)L\\(. If the windows around \\)\hat{L}\\( & \\)L\\( are disjoint, and there are values in the domain in this \\)\delta, \hat{\delta}\\( windows we'll have a contradiction (each corresponding point would need to be in both windows). In our proof, we will be assuming the following (for a contradiction): i) \\)\lim_{x \to a} f(x) = L: (\forall \epsilon > 0) (\exists \delta > 0) (\forall x \in dom(f) \setminus \{a\} ) (|x - a| < \delta \Rightarrow |f(x) - L| < \epsilon)\\( ii) \\)\lim_{x \to a} f(x) = \hat{L}: (\forall \epsilon > 0) (\exists \delta > 0) (\forall x \in dom(f) \setminus \{a\} ) (|x - a| < \delta \Rightarrow |f(x) - \hat{L}| < \epsilon)\\( iii) \\)L \ne \hat{L}\\( We have these "facts" to work with (we're assuming these); we don't need to show them. For (i) & (ii), this means that we can choose any \\)\epsilon > 0\\(, and we can state that there is a corresponding \\)\delta\\( (for each \\)L\\( & \\)\hat{L}\\(). In the proof we will use an \\)\epsilon > 0\\( that is small enough to make sure that the \\)\epsilon\\(-windows don't overlap. x \in (a - \delta, a + \delta) \\ a - \delta \leq x \leq a + \delta \\ - \delta < x - a < \delta \\ |x - a| \leq \delta **Proof:** Let \\)D \subseteq \mathbb{R}\\( and \\)E \subseteq \mathbb{R}\\( be arbitrary sets and let \\)f:D\rightarrow E\\( be an arbitrary function. Also, let \\)a\\( be an arbitrary accumulation point of \\)D\\(. Finally, let \\)L\\( and \\)\hat{L}\\( be arbitrary real numbers. We will show that if \\)\lim_{x \to a} f(x) = L\\( and \\)\lim_{x \to a} f(x) = \hat{L}\\(, then \\)L = \hat{L}\\(. Towards a contradiction, suppose \\)\lim_{x \to a} f(x) = L\\(, \\)\lim_{x \to a} f(x) = \hat{L}\\(, and that \\)L \neq \hat{L}\\(. As \\)L \neq \hat{L}\\(, we have that \\)|L - \hat{L}| > 0\\(. Define \\)\hat{\varepsilon}\\( to be \\)\frac{1}{4} |L - \hat{L}|\\(, i.e., \\)\hat{\varepsilon} = \frac{1}{4}|L - \hat{L}|\\(. Note that as \\)|L - \hat{L}| > 0\\(, we have that \\)\hat{\varepsilon} = \frac{1}{4}|L - \hat{L}| > 0\\(. Since \\)\lim_{x \to a} f(x) = L\\( and \\)\lim_{x \to a} f(x) = \hat{L}\\(, we have that: \\)(\forall \varepsilon > 0)(\exists \delta_1 > 0)(\forall x \in D \setminus \{a\})(|x - a| < \delta_1 \Rightarrow |f(x) - L| < \varepsilon)\\(, and \\)(\forall \hat{\varepsilon} > 0)(\exists \delta_2 > 0)(\forall x \in D \setminus \{a\})(|x - a| < \delta_2 \Rightarrow |f(x) - \hat{L}| < \hat{\varepsilon})\\(. In particular, as \\)\hat{\varepsilon} > 0\\(, there exist \\)\delta_1 > 0\\( and \\)\delta_2 > 0\\( such that: \\)(\forall x \in D \setminus \{a\})(|x - a| < \delta_1 \Rightarrow |f(x) - L| < \hat{\varepsilon})\\(, and (1) \\)(\forall x \in D \setminus \{a\})(|x - a| < \delta_2 \Rightarrow |f(x) - \hat{L}| < \hat{\varepsilon})\\(. (2) Let \\)\delta = \min(\delta_1, \delta_2)\\(. Then, as \\)\delta_1 > 0\\( and \\)\delta_2 > 0\\(, we have that \\)\delta > 0\\(. Since \\)\delta > 0\\( and since \\)a\\( is an accumulation point of \\)D\\(, we have that \\)((a - \delta, a + \delta) \cap D) \setminus \{a\}\\( is infinite, so that there is some \\)\hat{x} \in ((a - \delta, a + \delta) \cap D) \setminus \{a\}\\(. As \\)\delta = \min(\delta_L, \delta_{\hat{L}})\\(, we have \\)\delta \le \delta_L\\( and \\)\delta \le \delta_{\hat{L}}\\( so that: \[ \begin{aligned} (a - \delta, a + \delta) \subseteq (a - \delta_L, a + \delta_L), \text{ and } \\ (a - \delta, a + \delta) \subseteq (a - \delta_{\hat{L}}, a + \delta_{\hat{L}}). \end{aligned} \] From this, and as \\)\hat{x} \in ((a - \delta, a + \delta) \cap D) \setminus \{a\}\\(, we have that: \[ \begin{aligned} \hat{x} \in ((a - \delta_L, a + \delta_L) \cap D) \setminus \{a\} \text{, and } \\ \hat{x} \in ((a - \delta_{\hat{L}}, a + \delta_{\hat{L}}) \cap D) \setminus \{a\}. \end{aligned} \] Hence, we have that \\)\hat{x} \in D \setminus \{a\}\\( and that \\)|\hat{x} - a| < \delta_L\\( and \\)|\hat{x} - a| < \delta_{\hat{L}}\\(. By (1) and (2), we have that \\)|f(\hat{x}) - L| < \hat{\varepsilon}\\( and \\)|f(\hat{x}) - \hat{L}| < \hat{\varepsilon}\\(. \hfill (3) Now: \[ \begin{aligned} |L - \hat{L}| &= |L - f(\hat{x}) + f(\hat{x}) - \hat{L}| \\ &\le |L - f(\hat{x})| + |f(\hat{x}) - \hat{L}| \text{(By the Triangle Inequality)} \\ &= |f(\hat{x}) - L| + |f(\hat{x}) - \hat{L}| \\ &< \hat{\varepsilon} + \hat{\varepsilon} \\ &= 2 \hat{\varepsilon} = 2(\frac{1}{4}|L - \hat{L}|) = \frac{|L - \hat{L}|}{2}. \text{(By (3))} \text{(As } \hat{\varepsilon} = \frac{1}{4}|L - \hat{L}| \text{)} \end{aligned} \] Now, \\)|L - \hat{L}| < \frac{1}{2}|L - \hat{L}|\\(, so that \\)\frac{1}{2}|L - \hat{L}| < 0\\(, and \\)|L - \hat{L}| < 0\\(. As \\)|L - \hat{L}| > 0\\(, we have a contradiction. Hence, if \\)\varphi_{\text{lim}}(f, a, L)\\( and \\)\varphi_{\text{lim}}(f, a, L')\\(, then \\)L=L'\\(. As \\)D, E, f: D \to E, a, L\\( and \\)L'\\( were arbitrary, for all \\)D \subseteq \mathbb{R}\\(, \\)E \subseteq \mathbb{R}\\(, \\)f: D \to E\\(, and for all accumulation points \\)a\\( of \\)D\\( and any real numbers \\)L\\( and \\)L'\\(, if \\)\varphi_{\text{lim}}(f, a, L)\\( and \\)\varphi_{\text{lim}}(f, a, L')\\(, then \\)L=L'\\(. \[ \begin{aligned} |L - \hat{L}| &< \frac{1}{2}|L - \hat{L}| \\ A &< \frac{1}{2}A \\ A - \frac{1}{2}A &< 0 \\ (1-\frac{1}{2})A &< 0 \\ \frac{1}{2}A &< 0 \end{aligned} \] \[ \begin{aligned} |L-L| &\geq 0 \\ \\ |L - L| &= |(L - f(x)) + (f(x) - L)| \\ &\leq |L-f(x)| + |f(x) - L| \\ &= |f(x) - L| + |f(x) - L| \\ &< \epsilon + \epsilon \\ &= 2\epsilon\\ &= 2(\frac{\epsilon}{4}) \end{aligned} \] \hrulefill \[ \begin{aligned} |A + B| &\leq |A| + |B| \\ |2 + 7| &= |2| + |7| \\ |(-2) + (-7)| &= |-2| + |-7| \\ |2 + (-7)| &< |2| + |-7| \\ |(-2) + 7| &< |-2| + |7| \end{aligned} \] \hrulefill |7 - 4| = 3 \[ |L - L| < \frac{\epsilon}{4} \] \[ 0 < |L - L| \] \[ |L - L| < 0 \] \[ \begin{aligned} & F: A \rightarrow B, g: A \rightarrow B \\ & (f+g)(x) \\ & f: \mathbb{R} \rightarrow \mathbb{R} \\ & x \mapsto x^2 \\ & f(2) = 4 \\ & g: \mathbb{R} \rightarrow \mathbb{R} \\ & x \mapsto 2x-1 \\ & g(2) = 3 \\ & (f+g)(2) = f(2) + g(2) = 4 + 3 = 7 \\ & (f+g): A \rightarrow B \\ & x \mapsto f(x) + g(x) \end{aligned} \] \lim_{x \to 2} f(x) = 4 where \(f : \mathbb{R} \to \mathbb{R}, f(x) = x^2\) for all \(x \in \mathbb{R}\) *Goal*: \(|x - 2| < \delta\) |f(x) - 4| < 3 \\ |x^2 - 4| < 3 \\ |(x - 2)(x + 2)| < 3 \\ |x - 2| |x + 2| < 3 \\ |x - 2| < \frac{3}{|x + 2|} |x - 2| < \delta \\ \delta \leq \frac{3}{|x+2|} Let \(A > |x+2|\). Then we require \( \delta \le \frac{3}{A} \). Also \( \frac{3}{|x+2|} \ge \delta \) |x+2| < A \\ -1 < x - 2 < 1 Then \( 1 < x < 3 \) 3 < x + 2 < 5 \\ |x + 2| = x + 2 \\ |x + 2| < 5 \\ A > |x + 2| \delta = \min(1, \frac{3}{5}) \\ A |x+2| \newline \delta \le \frac{3}{A} \newline \frac{3}{|x+2|} \ge \delta \newline |x+2| < A \newline -1 < x - 2 < 1 \newline 3 < x + 2 < 5 \newline |x + 2| = x + 2 \newline |x + 2| < 5 \newline A > |x + 2| \newline \delta = \min(1, \frac{3}{5}) \newline A 0) \ (\forall x \in \text{Dom}(f \setminus \{a\}) ) \ ( |x - a| < \delta \implies |f(x) - L| < \epsilon ) \] \[ \delta = \min(1, \frac{5}{3}) \] \[ |2 - x| < \delta : \ 8 > |2-x| \] \[ 1 > |2-x| \] \[ 1 > 2 - x > -1 \] \[ \frac{5}{3} > |2 - x| \implies 8 > \frac{5}{3} \] \[ 3 < x+2 < 5 \] \[ \begin{array}{c} x+2<5 \\ \swarrow \searrow \\ |x+2| = x+2 \end{array} \] \[ |x+2| < 5 \] \[ |x+2| < \frac{5}{3} \] \[ 3 \cdot \frac{5}{3} > |2 + x| \implies 5 > |2 + x| \] \[ 3 > |2 + x| \cdot \frac{1}{3} \] \[ 3 > |2 - x| \] \[ 3 > |(2 - x)(2+x)| \] \[ 3 > |4-x^2| \] \[ 3 > |4 - (x)f| \[ \begin{aligned} & |x+2| < 4 \\ & \text{Prob.} \\ & \downarrow \\ & \delta = \min \left( 2 , \frac{2}{3} \right) \\ & |x-2| < 2 \\ & -2 < x-2 < 2 \\ & 2 < x+2 < 6 \\ & \delta = \min \left( \frac{1}{2} , \frac{6}{2} \right) \\ & |x-2| < \frac{1}{2} \\ & -\frac{1}{2} < x-2 < \frac{1}{2} \\ & \frac{3}{2} < x+2 < \frac{9}{2} \end{aligned} \] \\)|2x+1|\\(\\ \\)x \rightarrow 0\\( \\)\frac{1}{2}\\( prob < \\)\delta\\( \]

\lim_{x \to 2} (f+g)(x) = 12 \ \lim_{x \to 2} (f+g)(x) = 12 \ | (f+g)(x) - 12 | < \epsilon \ | f(x) + g(x) - 12 | < \epsilon \ | f(x) + g(x) - 4 - 8 | < \epsilon \ | (f(x) - 4) + (g(x) - 8) | < \epsilon \ | f(x) - 4 | + | g(x) - 8 | < \epsilon \ | f(x) - 4 | < \frac{2}{3} \text{ and } | g(x) - 8 | < \frac{2}{3}

\[ \]

f(x) = x^2 \ g(x) = x^3 \ | A + B | \leq |A| + |B| \text{ Triangle ineq.} \ | (f(x) - 4) + (g(x) - 8) | \leq |f(x) - 4| + |g(x) - 8|

\[ If I have: \]

| f(x) - 4 | < \frac{2}{3} \ | g(x) - 8 | < \frac{2}{3}

\[ For \\)\frac{2}{3} > 0\\(, if \\)(0 < \delta \leq \epsilon)(0 < 3A)\\( I can find \\)\delta > 0$ such that \]

| x - 2 | < \delta \text{ then } | f(x) - 4 | < \frac{2}{3} $$