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 \(x\to3\).

\(l=4\)

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.)

“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:

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

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

\(\downarrow\)

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

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.)

“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 \(\delta\))

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

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

Generalization:

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.)

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:

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

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.

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

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:

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

• 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:

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

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

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:

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

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)

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:

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

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

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

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:

• (i) \(x < -2\)
• (ii) \(-2 \leq x < 0\)
• (iii) \(x=0\)
• (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

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

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

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:

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, \(\lim_{x \to 0} f(x) = 6\).

\(\square\)

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}

\(\varphi \text{lim}(f, a, L)\) 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:

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

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.,:

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:

Where

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

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

Equivalent to:

or

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

Equivalent to:

or

\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)

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:

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.

(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)

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

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

Shorthand for:

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.:

And then, we will produce a contradiction: “(\(\Psi \land \neg \Psi\))”

“Idea” of the Proof:

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.

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:

From this, and as \(\hat{x} \in ((a - \delta, a + \delta) \cap D) \setminus {a}\), we have that:

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 \(|f(\hat{x}) - \hat{L}| < \hat{\varepsilon}\). (3)

Now:

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 \(L=L’\).

By the Triangle Inequality: \(| A + B | \leq |A| + |B|\)

So: \[ | (f(x) - 4) + (g(x) - 8) | \leq |f(x) - 4| + |g(x) - 8| \]