42 Functional Limits Practice

#todo/school : Read §4.1-2; 📅 2024-11-06 ✅ 2024-11-06
#todo/school : Do problems 4.2.2, 5, 6, 7, 8 📅 2024-11-06 ✅ 2024-11-07

4.2.2

Question

For each state limit, find the largest possible $δ$ -neighborhood that is a proper response to the given $ε$ challenge.
a. $lim_{x \to 3} (5 x - 6) = 9$ where $ε = 1$
b. $lim_{x \to 4} \sqrt{x} = 2$ where $ε = 1$
c. $lim_{x \to π} [[x ∥ x]] = 3$ where $ε = 1$ ( $[[x ∥ x]]$ returns the greatest integer $\leq x$ )
d. $lim_{x \to π} [[x ∥ x]] = 3$ where $ε = .01$ .

Remember, $δ$ is for the input, and $ε$ is for the output of the function.

Proof

a. In general for any $ε > 0$ we would want to find some $δ > 0$ where given $x \in V_{δ} (3) ⟺ | x - 3 | < δ$ then we can find:

\begin{aligned} | 5 x - 6 - 9 | & = | 5 x - 15 | \\ = 5 | x - 3 | \\ < 5 δ \\ = ε \end{aligned}

Thus we'd have to choose $δ = \frac{1}{5} ε$ , thus in this case choose $δ = \frac{1}{5}$ .

b. In general where now $| x - 4 | < δ$ :

\begin{aligned} | \sqrt{x} - 2 | & = | \frac{(\sqrt{x} - 2) (\sqrt{x} + 2)}{\sqrt{x} + 2} | \\ = \frac{| x - 4 |}{\sqrt{x} + 2} \\ \leq \frac{| x - 4 |}{2} \\ < \frac{δ}{2} \\ = ε \end{aligned}

Thus we'd want to choose $δ = 2 ε$ here, specifically $δ = 2$ .

c and d. We expect that $δ = π - 3$ should work here for both cases (since then we can easily evaluate the $[[x ∥ x]]$ part, as shown below). Notice that:

\begin{array}{r} | x - π | < π - 3 \\ ⟺ 3 - π < x - π < π - 3 \\ ⟺ 3 < x < 2 π - 3 \\ ⟺ [[3 ∥ 3]] < x < [[2 π - 3 ∥ 2 π - 3]] \\ \Rightarrow 3 < x \leq 3 \\ \Rightarrow x = 3 \end{array}

Thus:

\begin{aligned} | [[x ∥ x]] - 3 | & = | 3 - 3 | \\ = 0 \\ < ε \end{aligned}

always will work for any $ε > 0$ .
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4.2.5

Question

Use the Functional Limit definition to supply a proper proof for the following limit statements:
a. $lim_{x \to 2} (3 x + 4) = 10$
b. $lim_{x \to 0} x^{3} = 0$
c. $lim_{x \to 2} (x^{2} + x - 1) = 5$
d. $lim_{x \to 3} \frac{1}{x} = \frac{1}{3}$

For all of these we will want to have some scratch work.

a. We want for any $ε > 0$ to find $δ > 0$ where supposing $| x - 2 | < δ$ then:

\begin{aligned} | (3 x + 4) - 10 | & = | 3 x - 6 | \\ = 3 | x - 2 | \\ < 3 δ \\ = ε \end{aligned}

Thus we want $δ = \frac{1}{3} ε$ .

b. For $| x - 0 | = | x | < δ$ we want:

\begin{aligned} | x^{3} - 0 | & = | x^{3} | \\ = | x | | x | | x | \\ < δ | x | | x | \\ \dots \\ < δ δ δ \\ = δ^{3} \\ = ε \end{aligned}

Thus we want $δ = \sqrt[3]{ε}$ here.

c. For $| x - 2 | < δ$ we want:

\begin{aligned} | x^{2} + x - 1 - 5 | & = | x^{2} + x - 6 | \\ = | (x + 3) (x - 2) | \\ = | x + 3 | | x - 2 | \\ < | x + 3 | δ \end{aligned}

While it seems that we are stuck, try to notice that:

| x - 2 | < δ \Rightarrow | x + 3 | = | (x - 2) + 5 | \leq | x - 2 | + 5 < δ + 5

Thus $| x + 3 | < δ + 5$ so we can keep going:

\begin{aligned} < (δ + 5) δ \\ = δ^{2} + 5 δ \end{aligned}

Now the problem is that we can't really solve for $δ$ here, so we are going to have to find some terms larger that simplifies nicely. Notice that we can try to complete the square here:

δ^{2} + 5 δ < δ^{2} + 6 δ + 9 = (δ + 3)^{2} = ε \Rightarrow δ = \sqrt{ε} - 3

should work.

d. For $| x - 3 | < δ$ we want:

\begin{aligned} | \frac{1}{x} - \frac{1}{3} | & = | \frac{3 - x}{3 x} | \\ = \frac{| x - 3 |}{3 | x |} \end{aligned}

Similar to (c), notice that:

| x - 3 | < δ \Rightarrow | x | = | x - 3 + 3 | \leq | x - 3 | + 3 < δ + 3

Thus $| x | < δ + 3 \Rightarrow \frac{1}{| x |} > \frac{1}{δ + 3}$ . Then:

\begin{aligned} < \frac{δ}{3 | x |} \\ < ? \end{aligned}

We seem to be stuck. What would be nice is to multiply the $| x |$ out prior while we are increasing terms, but we can't. What could help is to assume that $| x |$ is ultimately greater than some positive value, like $| x | > 7$ so that that implies that $\frac{1}{| x |} < \frac{1}{7}$ so then we could continue our work. If this is the case:

| x - 3 | < δ < 1 \Rightarrow | x - 3 | < 1 ⟺ - 1 + 3 < x < 1 + 3 ⟺ 2 < x < 4 \Rightarrow | x | < 7

Gives the restriction that $δ < 1$ .

But then to evaluate we'd want:

\begin{aligned} < \frac{δ}{3 | x |} \\ < \frac{δ}{21} \\ < ε \end{aligned}

Thus we'd need $δ < 21 ε$ as well. Since $δ < 21 ε$ and $δ < 1$ then use their minimum.

Proof

a. Let $ε > 0$ . Then choose $δ = \frac{1}{3} ε$ . Then for any $x \in V_{δ} (2)$ (namely $| x - 2 | < δ$ ), then:

\begin{aligned} | (3 x + 4) - 10 | & = | 3 x - 6 | \\ = 3 | x - 2 | \\ < 3 δ \\ = 3 \cdot \frac{1}{3} ε \\ = ε \end{aligned}

b. Let $ε > 0$ . Then choose $δ = \sqrt[3]{ε}$ . Then for any $x \in V_{δ} (0)$ ( $| x | < δ$ ) then:

\begin{aligned} | x^{3} - 0 | & = | x^{3} | \\ = | x | | x | | x | \\ < δ | x | | x | \\ < δ δ | x | \\ < δ δ δ \\ = δ^{3} \\ = (\sqrt[3]{ε})^{3} \\ = ε \end{aligned}

c. Let $ε > 0$ . Then choose $δ = \sqrt{ε} - 3$ . Then for any $x \in V_{δ} (2)$ ( $| x - 2 | < δ$ ) notice first that:

| x - 2 | < δ \Rightarrow | x + 3 | = | (x - 2) + 5 | \leq | x - 2 | + 5 < δ + 5

Thus $| x - 2 | < δ$ and $| x + 3 | < δ + 5$ . Thus:

\begin{aligned} | x^{2} + x - 1 - 5 | & = | x^{2} + x - 6 | \\ = | (x + 3) (x - 2) | \\ = | x + 3 | | x - 2 | \\ < (δ + 5) δ \\ = δ^{2} + 5 δ \\ < δ^{2} + 6 δ + 9 \\ = (δ + 3)^{2} \\ = (\sqrt{ε} - 3 + 3)^{2} \\ = (\sqrt{ε})^{2} \\ = ε \end{aligned}

d. Let $ε > 0$ . Then choose $δ = min {1, 21 ε}$ . Then for any $x \in V_{δ} (3)$ ( $| x - 3 | < δ$ ) first notice that $| x - 3 | < δ < 1 \Rightarrow x \in (2, 4) \Rightarrow | x | < 7$ . Thus:

\begin{aligned} | \frac{1}{x} - \frac{1}{3} | & = | \frac{3 - x}{3 x} | \\ = \frac{| x - 3 |}{3 | x |} \\ < \frac{| x - 3 |}{3 \cdot 7} \\ < \frac{δ}{21} \\ = \frac{21 ε}{21} \\ = ε \end{aligned}

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4.2.6

Question

Decide if the following claims are true or false, and give short justifications for each conclusion.
a. If a particular $δ$ has been constructed as a suitable response to a particular $ε$ challenge, then any smaller positive $δ$ will also suffice.
b. If $lim_{x \to a} f (x) = L$ and $a$ happens to be in the domain of $f$ , then $L = f (a)$ .
c. If $lim_{x \to a} f (x) = L$ then $lim_{x \to a} 3 [f (x) - 2]^{2} = 3 (L - 2)^{2}$
d. If $lim_{x \to a} f (x) = 0$ then $lim_{x \to a} f (x) g (x) = 0$ for any function $g$ (with domain equal to the domain of $f$ ).

Proof

a. True. If such a $δ$ has been constructed such that we get a limit, then using a $δ^{'} < δ$ works. Consider $lim_{x \to c} f (x) = L$ has been shown with $ε > 0$ to work for $δ > 0$ . We want to show that it works for $δ^{'}$ . Notice that since we suppose that $0 < | x - c | < δ^{'}$ then automatically $| x - c | < δ$ so that implies that $| f (x) - L | < ε$ already!

b. False. Consider any removable discontinuity function. For example consider:

f (x) = {\begin{cases} 1 & x = 0 \\ x & x \neq 0 \end{cases}

Then at $a = 0$ clearly $lim_{x \to a} f (x) = 0$ but $f (a) = 1$ which disagrees.

c. True. Just apply the Algebraic Limit Theorem for Functional Limits:

\begin{aligned} lim_{x \to a} 3 [f (x) - 2]^{2} & = 3 lim_{x \to a} [f (x) - 2]^{2} \\ = 3 lim_{x \to a} [f (x) - 2] lim_{x \to a} [f (x) - 2] \\ = 3 (lim_{x \to a} f (x) - 2) (lim_{x \to a} f (x) - 2) \\ = 3 (L - 2) (L - 2) \\ = 3 (L - 2)^{2} \end{aligned}

But you work backwards to justify the limits existing for the intermediate steps. For example, we know that:

lim_{x \to a} f (x) - 2

has to exist since if it didn't, then since $lim_{x \to a} - 2$ exists and so does $lim_{x \to a} f (x) = L$ , then that's a contradiction. As a result $lim_{x \to a} [f (x) - 2]^{2}$ exists and thus the first limit we had as well. Using the ALT is justified here!

d. False. Consider the function $g (x) = x$ and $f (x) = \frac{1}{x}$ both defined on $A = R^{+}$ . Clearly $lim_{x \to 0} f (x) = 0$ but then:

lim_{x \to 0} f (x) g (x) = lim_{x \to 0} \frac{1}{x} \cdot x = lim_{x \to 0} 1 = 1 \neq 0

The main issue here is that we require the limits to exist if we want to split them up like this. See how that was justified at the end of (c) in this question.

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4.2.7

Question

Let $g : A \to R$ and assume that $f$ is a bounded function on $A$ ; in the sense that $\exists M > 0$ where $| f (x) | \leq M$ for all $x \in A$ .

Show that if $lim_{x \to c} g (x) = 0$ then $lim_{x \to c} g (x) f (x) = 0$ as well.

Scratch:

Trying to use the Algebraic Limit Theorem for Functional Limits doesn't work here since we don't know the value of $lim_{x \to c} f (x)$ (the limit itself might now exist, for example consider the function $\sin (\frac{1}{x})$ over the domain $R - {0}$ . The limit at $0$ doesn't exist, even though the function is indeed bounded). Thus we have to use the definition.

If we let $ε > 0$ and want to choose some $δ > 0$ then supposing $0 < | x - c | < δ$ then we want:

\begin{aligned} | f (x) g (x) - 0 | & = | f (x) g (x) | \\ = | f (x) | | g (x) | \\ \leq M | g (x) | \end{aligned}

From $lim_{x \to c} g (x) = 0$ then we can use our $ε$ to get that $\exists δ > 0$ where for our supposed $0 < | x - c | < δ$ then:

| g (x) - 0 | = | g (x) | < ε

Thus:

\begin{aligned} < M ε \end{aligned}

Isn't ideal. We want to end up with $ε$ . It'd be better if the $ε$ for the limit is instead $\frac{ε}{M} > 0$ for this. If we did then then everything follows:

\begin{aligned} | f (x) g (x) - 0 | & \leq M | g (x) | \\ < M \cdot \frac{ε}{M} \\ = ε \end{aligned}

Proof

Let $ε > 0$ . Notice that since $\frac{ε}{M} > 0$ then because $lim_{x \to c} g (x) = 0$ then we know that we can choose some $δ > 0$ where if we suppose $0 < | x - c | < δ$ then we get:

| g (x) - 0 | = | g (x) | < \frac{ε}{M}

Thus then:

\begin{aligned} | f (x) g (x) - 0 | & = | f (x) g (x) | \\ = | f (x) | | g (x) | \\ \leq M | g (x) | \\ < M \frac{ε}{M} \\ = ε \end{aligned}

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4.2.8

Question

Compute each limit or state that it does not exist. Use the tools developed in this section to justify each conclusion.
a. $lim_{x \to 2} \frac{| x - 2 |}{x - 2}$
b. $lim_{x \to \frac{7}{4}} \frac{| x - 2 |}{x - 2}$
c. $lim_{x \to 0} (- 1)^{[[\frac{1}{x}]]}$
d. $lim_{x \to 0} \sqrt[3]{x} (- 1)^{[[\frac{1}{x}]]}$

Proof

a. This is similar to a step function over 2, suggesting the limit doesn't exist. If we consider the sequences $(x_{n}) = 2 + \frac{1}{n}$ (the right side limit) where $x_{n} \to 2$ and then:

f (x_{n}) = \frac{| 2 + \frac{1}{n} - 2 |}{2 + \frac{1}{n} - 2} = \frac{n}{| n |} = \frac{n}{n} = 1, f (x_{n}) \to 1

While if we consider the sequence $(y_{n}) = 2 - \frac{1}{n}$ (the left side limit) where $y_{n} \to 2$ and then:

f (y_{n}) = \frac{| 2 - \frac{1}{n} - 2 |}{2 - \frac{1}{n} - 2} = - \frac{| \frac{- 1}{n} |}{\frac{1}{n}} = \frac{- n}{n} = - 1, f (y_{n}) \to - 1

Since the sequences converge to different values, then the limit doesn't exist via Existence of Functional Limits.

b. We'll show that it's -1. Let $ε > 0$ then choose $δ < \frac{1}{4}$ (that's the input neighborhood radius with all constant values) such that if $0 < | x - \frac{7}{4} | < δ$ then:

| x - \frac{7}{4} | < δ < \frac{1}{4} \Rightarrow x \in (\frac{3}{2}, 2) \Rightarrow f (x) = - 1 \forall x \in (\frac{3}{2}, 2)

Thus:

\begin{aligned} | f (x) + 1 | & = | - 1 + 1 | \\ = 0 \\ < ε \end{aligned}

c. This is similar to (a), where coming from the left would give $(- 1)^{- 1} = - 1$ while coming in from the right would give $(- 1)^{0} = 1$ which are different value. Let's show this. Consider the sequence $(x_{n}) = \frac{- 1}{2 n}$ . Notice $x_{n} \to 0$ as desired and:

f (x_{n}) = (- 1)^{[[\frac{1}{\frac{- 1}{2 n}}]]} = (- 1)^{[[- 2 n ∥ - 2 n]]} = (- 1)^{2 n} = 1^{n} = 1

Thus $f (x_{n}) \to 1$ . However if $(y_{n}) = \frac{1}{2 n + 1}$ then notice $y_{n} \to 0$ but:

f (y_{n}) = (- 1)^{[[\frac{1}{\frac{1}{2 n + 1}}]]} = (- 1)^{[[2 n + 1 ∥ 2 n + 1]]} = (- 1)^{2 n + 1} = - 1

Thus $f (y_{n}) \to - 1$ . These two sequences converges to different values when applied to via $f$ , so then by the Existence of Functional Limits then the functional limit doesn't exist here.

d. We know that $lim_{x \to 0} \sqrt[3]{x} = 0$ by the following argument. Let $ε > 0$ . Choose $δ = ε^{3}$ such that for any $0 < | x - 0 | = | x | < δ$ then first notice:

| x | < δ = ε^{3} \Rightarrow \sqrt[3]{| x |} < ε

Thus:

\begin{aligned} | \sqrt[3]{x} - 0 | & = | \sqrt[3]{x} | \\ = \sqrt[3]{| x |} \\ < ε \end{aligned}

The middle step showing $| \sqrt[3]{x} | = \sqrt[3]{| x |}$ is pretty easy:

If $x = 0$ then the result is trivial.
If $x < 0$ then $\sqrt[3]{| x |} = \sqrt[3]{- x} = - \underset{> 0}{\underset{⏟}{\sqrt[3]{x}}} = | \sqrt[3]{x} |$
If $x > 0$ then $\sqrt[3]{| x |} = \sqrt[3]{x} = | \sqrt[3]{x} |$

Now to the original question, notice that $(- 1)^{[[\frac{1}{x}]]}$ is bounded by $M = 1$ here. That means that by 42 Functional Limits Practice#4.2.7 then $lim_{x \to 0} \sqrt[3]{x} (- 1)^{[[\frac{1}{x}]]}$ exists and is just $0$ .

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