Module 4: Towards the Definition of a Limit #

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

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

Notice that:

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

Goal: We want to create/find a definition of the limit, i.e., we want to find a predicate/formula \(\varphi_{\lim}(f, a, L)\) which will be true exactly when \(\lim_{x \to a} f(x) = L\).

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

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

\((1,5) \setminus \{3\}\) is a “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 \(x\to3\).

Some intervals from class (for \(L = 4\), which will turn out not to be the limit): Player I (see the game below) might play \((2,6)\) (\(e=2\)), \((3,5)\) (\(e=1\)) or \((3.9, 4.1)\) (\(e=0.1\)); Player II might answer with \((2,4)\setminus\{3\}\) (\(d=1\)) or \((1,5)\setminus\{3\}\) (\(d=2\)).

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 \(d \in (0, 1)\), they’ll win)

Example Game #2: (\(L=9\)) Player I gives/plays \((9-1, 9+1) = (8,10)\)
Player II Attempts : \(f’’((1,5)\setminus\{3\}) = (4,8) \setminus\{6\} \nsubseteq (8,10)\) (\(d=2\))
\(f’’((2,4)\setminus\{3\}) = (5,7)\setminus\{6\} \nsubseteq (8,10)\) (\(d=1\))

Notice that as Player II decreases \(d\), the image will stay disjoint from \((8, 10)\), so the image will never be a subset. As \(d\) 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.)

Play the game yourself: drag the sliders. Player I picks \(L\) and \(e\) (the blue horizontal band is \((L-e, L+e)\)); Player II picks \(d\) (the green vertical band is the punctured interval \((3-d, 3+d)\setminus\{3\}\)). Player II wins when the orange graph over the green band stays inside the blue band. Try \(L = 9\) and \(L = 6\).

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

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

Unpacking the subset condition. Recall \(A \subseteq B\) iff \((\forall \alpha)(\alpha \in A \Rightarrow \alpha \in B)\). Also, \(\alpha \in f’’((a-\delta, a+\delta) \setminus \{a\})\) exactly when \(\alpha = f(x)\) for some \(x \in \text{Dom}(f)\) with \(x \in (a-\delta, a+\delta) \setminus \{a\}\). So

(If \(x \notin \text{Dom}(f)\) the antecedent is false, so the implication is true and such \(x\) cause no trouble.)

For the last step we used that for real numbers \(\beta, A\) and \(b > 0\):

In our “Limit Games”, Player I plays an interval around a value \(L\), then Player II plays a punctured interval around a value \(a\). Player II wins if whenever we take \(x\) from II’s punctured interval, \(f(x)\) stays in Player I’s interval. More symbolically, Player II wins if:

(Removing \(x = a\) from consideration “punctures” II’s interval.)

We say that \(\lim_{x \to a} f(x) = L\) exactly when II can always play to win, no matter what Player I plays:

Playing the Game Algebraically #

Let \(f(x) = (x-5)^2\). We claim \(\varphi_{\lim}(f, 5, 0)\), 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.)

(Recall: \(\sqrt{A^2} = |A|\), and \(|A| < B\) is equivalent to \(-B < A < B\). E.g. \(|A| < 17\) means \(-17 < A < 17\). Likewise \(|x - a| < \delta\) is equivalent to \(a - \delta < x < a + \delta\).)

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 \(\varphi_{\lim}(f, a, L)\) 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 \(\varphi_{\lim}(f, 5, 0)\).

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!)

Player I Plays \(\epsilon = 4\): Need \(|f(x) - 0| < 4\), i.e. \(-4 < f(x) < 4\).

Again, we’ll need to require \(\delta \leq 1\) so that we only are using \((x-5)^2\) (which is what \(f(x)\) is equal to “close” to \(a = 5\)).

As we need \(\delta \leq 1\) and \(\delta \leq 2\), we take the minimum, and we let \(\delta = 1\) (or smaller).

Generalizing: Need \(|f(x) - 0| < \epsilon\), i.e. \(-\epsilon < f(x) < \epsilon\). When \(\delta \leq 1\), we will have \(f(x) = (x-5)^2\):

As \(\delta \leq 1\) and \(\delta \leq \sqrt{\epsilon}\), we let \(\delta = \min(1, \sqrt{\epsilon})\) (or smaller).

• 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 \(\varphi_{\lim}(f, a, L)\) is the predicate:

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

We write \(\lim_{x \to a} f(x) = L\) exactly when \(\varphi_{\lim}(f, a, L)\) 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

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

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

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 \(f(x) = (x-5)^2\).

Key Fact: The minimum of a collection of values is less than or equal to each of the values:

(E.g. \(\min(1, 3, 2, 17) = 1\), which is \(\le 1\), \(\le 3\), \(\le 2\) and \(\le 17\).) We will also use that if \(a < b\) and \(b \leq c\), then \(a < c\).

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

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

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

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

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

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

\(\square\)

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

For instance \(f(\sqrt{2}) = 0\), \(f(0) = 0\), \(f(4) = 16\), \(f(\pi) = 0\) and \(f(\sqrt{9}) = 9\). We claim \(\lim_{x \to 0} f(x) = 0\), i.e.

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

Scratch work, by cases on \(x\):

So \(\delta = \sqrt{\epsilon}\) works for both cases.

\(\varphi_{\lim}(f, a, L)\) 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 \(x \in \mathbb{Z}\). (E.g. \(f(5) = 25\), \(f(-2) = 4\), \(f(1) = 1\), \(f(0) = 0\), while \(f(\frac{1}{2})\) is undefined as \(\frac{1}{2} \notin \mathbb{Z}\).)

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

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

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 \(\varphi_{\lim}(f, 2, 4)\) is true.

For the same reasoning/logic, \(\varphi_{\lim}(f, 2, 17)\) 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

Listing a few of the elements of \(D\) by \(m\):

“Picture” of \(D\):

For example, zooming in near \(1\) and \(2\):

Now let \(f : D \to \mathbb{R}\) be defined by \(x \mapsto x^2\) for all \(x \in D\). 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.

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

Examples. Let \(D = (0, 1] \cup \{2\}\).

• \(-17\): \((-18, -16) \cap D = \emptyset\), which is not infinite, so \(-17\) is not an accumulation point of \(D\).
• \(0\): \((-1, 1) \cap D = (0, 1)\), \((-0.1, 0.1) \cap D = (0, 0.1)\), \((-0.01, 0.01) \cap D = (0, 0.01)\), … always infinitely many (in fact uncountably many) elements, so \(0\) is an accumulation point (and a condensation point) of \(D\).
• \(0.5\): \((0.4, 0.6) \cap D = (0.4, 0.6)\) is uncountable, so \(0.5\) is an accumulation and condensation point of \(D\).
• \(1\): \((-1, 2) \cap D = (0, 1]\), \((0.001, 1.1) \cap D = (0.001, 1]\), \((0.9999, 1.00001) \cap D = (0.9999, 1]\), … so \(1\) is an accumulation (and condensation) point.
• \(1.5\): \((0.5, 1.6) \cap D = (0.5, 1]\) is infinite, but \((1, 2) \cap D = \emptyset\) is not. As we need every open interval about \(1.5\) to meet \(D\) in infinitely many points, \(1.5\) is not an accumulation point of \(D\).
• \(2\): \((1.9, 2.1) \cap D = \{2\}\), so \(2\) is not an accumulation point of \(D\) (even though \(2 \in D\)).
• \(3\): \((2.9, 3.1) \cap D = \emptyset\), so \(3\) is not an accumulation point of \(D\).

Example. Let \(D = \{0, \frac{1}{5}, \frac{1}{4}, \frac{1}{3}, \frac{1}{2}, 1\}\). Then \((0.249, 0.251) \cap D = \{\frac{1}{4}\}\), so \(\frac{1}{4}\) is not an accumulation point. In fact a finite set has no accumulation points.

Example. Let \(B = \{0\} \cup \{\frac{1}{n+1} \mid n \in \mathbb{N}\} = \{0, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \dots\}\).

\((-0.5, 0.5) \cap B = \{0, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \frac{1}{6}, \dots\}\) is infinite, so \(0\) is an accumulation point of \(B\). However this set is countable: \(F: \mathbb{N} \to (-0.5, 0.5) \cap B\) with \(1 \mapsto 0\) and \(n \mapsto \frac{1}{n+1}\) for \(n \ge 2\) (\(2 \mapsto \frac{1}{3}, 3 \mapsto \frac{1}{4}, \dots\)) is a bijection, so the intersection is countable, not uncountable. Hence \(0\) is not a condensation point of \(B\).

Exercises. Find all accumulation points and all condensation points of:

1. \((0, 1)\)
2. \(D = (0, \tfrac{1}{2}) \cup (1, 1\tfrac{1}{3}) \cup (2, 2\tfrac{1}{4}) \cup (3, 3\tfrac{1}{5}) \cup (4, 4\tfrac{1}{6}) \cup \dots\)

With these “new” terms, let’s examine the sets \(\mathbb{Z}\) and \(D = \mathbb{Z} \cup \{m+\frac{1}{n} \mid m \in \mathbb{Z}, n \in \mathbb{N}\}\) (both \(\mathbb{Z} \subseteq \mathbb{R}\) and \(D \subseteq \mathbb{R}\)).

In \(\mathbb{Z}\):

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 \(D\): (drag \(c\) and \(d\))

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 \(D\)) Zooming in on \((1.9, 2.3)\):

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

Example: \([e, \pi]\). Every point of \([e, \pi]\) (e.g. \(2.8\), \(3\), \(3.1\)) is an accumulation point, and so are the endpoints \(e\) and \(\pi\); the set of accumulation points is \([e, \pi]\) (these are also all condensation points).

(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:

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

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 \(\varphi_{\lim} (f, a, L)\) and \(\varphi_{\lim} (f, a, \hat{L})\), then \(L = \hat{L}\).

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

Another way to say it: Let \(D \subseteq \mathbb{R}\). A value \(a \in \mathbb{R}\) is an accumulation point of \(D\) iff

i.e., there are infinitely many points from \(D\) in the interval \((a-\beta, a+\beta)\), no matter how small \(\beta > 0\) is.

E.g. Consider \(D = \mathbb{Q}\), \(a = \pi\). Then \((\pi - 1, \pi + 1) \cap \mathbb{Q} = (2.141\ldots, 4.141\ldots) \cap \mathbb{Q}\) is infinite (\(|(\pi-1,\pi+1)\cap\mathbb{Q}| \ge \aleph_0\)), and so is \((\pi - 0.1, \pi + 0.1) \cap \mathbb{Q}\), etc. So \(\pi\) is an accumulation point of \(\mathbb{Q}\) (but not a condensation point, as \(\mathbb{Q}\) is countable).

Set-up for a proof by contradiction. The theorem has the form “\((A \land B) \Rightarrow C\)”, call it \(\Psi\). We will prove \(\Psi\) by contradiction: \(\Psi\) is equivalent to \((\neg \Psi \Rightarrow (D \land \neg D))\), and \(\neg \Psi\) is \(\neg((A \land B) \Rightarrow C)\), which is equivalent to \((A \land B) \land \neg C\). So we assume \(A\), \(B\) and \(\neg C\).

Two small facts we will use: if \(a \neq b\) then \(|a - b| > 0\) (e.g. \(|2 - 7| = 5\)); and \(\varphi_{\lim}(f, a, L)\) lets us pick any \(\epsilon > 0\) and get a corresponding \(\delta\). For example, for \(\epsilon = 15\) there is a \(\delta_L(15) > 0\) such that \((\forall x \in D \setminus \{a\})(|x - a| < \delta_L(15) \Rightarrow |f(x) - L| < 15)\); for \(\epsilon = 0.02\) there is a \(\delta_L(0.02) > 0\) such that \((\forall x \in D \setminus \{a\})(|x - a| < \delta_L(0.02) \Rightarrow |f(x) - L| < 0.02)\); and so on for \(\epsilon = 0.0000001\) or \(\epsilon = \frac{1}{4}|L - \hat{L}|\).

Logical Form of the Theorem

Shorthand: \((\forall D \subseteq \mathbb{R})(\forall E \subseteq \mathbb{R})\) means \((\forall D \in \mathcal{P}(\mathbb{R})) (\forall E \in \mathcal{P}(\mathbb{R}))\). \(\operatorname{Acc}(D)\) is the set of all accumulation points of \(D\). \(E^D\) 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:

\((\forall \varepsilon > 0)(\exists \delta_L > 0)(\forall x \in D \setminus \{a\})(|x - a| < \delta_L \Rightarrow |f(x) - L| < \varepsilon)\), and

\((\forall \varepsilon > 0)(\exists \delta_{\hat{L}} > 0)(\forall x \in D \setminus \{a\})(|x - a| < \delta_{\hat{L}} \Rightarrow |f(x) - \hat{L}| < \varepsilon)\).

In particular, as \(\hat{\varepsilon} > 0\), there exist \(\delta_L > 0\) and \(\delta_{\hat{L}} > 0\) such that:

\((\forall x \in D \setminus \{a\})(|x - a| < \delta_L \Rightarrow |f(x) - L| < \hat{\varepsilon})\), and (1)
\((\forall x \in D \setminus \{a\})(|x - a| < \delta_{\hat{L}} \Rightarrow |f(x) - \hat{L}| < \hat{\varepsilon})\). (2)

Let \(\delta = \min(\delta_L, \delta_{\hat{L}})\). Then, as \(\delta_L > 0\) and \(\delta_{\hat{L}} > 0\), 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 \(\varphi_{\lim}(f, a, L)\) and \(\varphi_{\lim}(f, a, \hat{L})\), then \(L = \hat{L}\).

As \(D, E, f: D \to E, a, L\) and \(\hat{L}\) 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 \(\hat{L}\), if \(\varphi_{\lim}(f, a, L)\) and \(\varphi_{\lim}(f, a, \hat{L})\), then \(L = \hat{L}\). \(\blacksquare\)

(The last step in detail, with \(A = |L - \hat{L}|\):)

The triangle inequality \(|A + B| \leq |A| + |B|\) in examples (equality when \(A\) and \(B\) have the same sign, strict inequality otherwise):

Worked example: \(\lim_{x \to 2} x^2 = 4\) #

Let \(f: \mathbb{R} \to \mathbb{R}\), \(f(x) = x^2\). We want, for a given \(\epsilon > 0\), a \(\delta > 0\) with \(|x - 2| < \delta \Rightarrow |f(x) - 4| < \epsilon\).

Scratch work:

The problem: the bound \(\frac{\epsilon}{|x+2|}\) depends on \(x\), and \(\delta\) may only depend on \(\epsilon\). The fix is to first insist that \(\delta \le 1\), which controls \(|x + 2|\):

Then \(|x-2|\,|x+2| < 5\,|x - 2|\), so it is enough to also have \(|x - 2| < \frac{\epsilon}{5}\). We let \(\delta = \min(1, \frac{\epsilon}{5})\).

Proof: Let \(f: \mathbb{R} \to \mathbb{R}\) be the function defined by \(f(x) = x^2\) for all \(x \in \mathbb{R}\). Let \(\epsilon > 0\) be arbitrary and let \(\delta = \min(1, \frac{\epsilon}{5})\). As \(1 > 0\) and \(\frac{\epsilon}{5} > 0\), we have \(\delta > 0\). Let \(x \in \mathbb{R} \setminus \{2\}\) be arbitrary and suppose \(|x - 2| < \delta\). As \(\delta \le 1\), \(|x - 2| < 1\), so \(-1 < x - 2 < 1\) and \(3 < x + 2 < 5\). Hence \(x + 2 > 0\), so \(|x + 2| = x + 2 < 5\). As \(\delta \le \frac{\epsilon}{5}\), \(|x - 2| < \frac{\epsilon}{5}\). Now:

Hence, \(|f(x) - 4| < \epsilon\). Thus, if \(|x - 2| < \delta\), then \(|f(x) - 4| < \epsilon\). As \(x\) was arbitrary, for all \(x \in \mathbb{R} \setminus \{2\}\), if \(|x - 2| < \delta\), then \(|f(x) - 4| < \epsilon\). As \(\delta > 0\), there exists a \(\delta > 0\) such that for all \(x \in \mathbb{R} \setminus \{2\}\), if \(|x - 2| < \delta\), then \(|f(x) - 4| < \epsilon\). As \(\epsilon > 0\) was arbitrary, for all \(\epsilon > 0\), there exists a \(\delta > 0\) such that for all \(x \in \mathbb{R} \setminus \{2\}\), if \(|x - 2| < \delta\), then \(|f(x) - 4| < \epsilon\). Therefore, \(\lim_{x \to 2} f(x) = 4\). \(\square\)

Other valid choices (from class): restricting to \(\delta \le 2\) gives \(2 < x + 2 < 6\), so \(\delta = \min(2, \frac{\epsilon}{6})\) also works; restricting to \(\delta \le \frac{1}{2}\) gives \(\frac{3}{2} < x + 2 < \frac{9}{2}\), so \(\delta = \min(\frac{1}{2}, \frac{2\epsilon}{9})\) works too. There are many winning moves for Player II.