Week 3: May 26 $^$th - June 1 $^$st

Suggested Problems

Problems you may find instructive, or that I find interesting.

2.4

#21, 24, 35, 41, 42, 45, 46, 52(C) & 56

(Only evaluate limits by the methods of Chapter 2.)

Limits toward $\infty$

Notes are on the website.

Answers for the following exercises are given below.

Compute the given limits, if they exist. Use only theorems or definitions discussed in lecture.

X1

$\lim_{x\to-\infty} \ln\left(1-\frac{1}{x}\right)$

X2

$\lim_{\theta\to\infty}\frac{\sin\theta}{\theta}$

X3

$\lim_{z\to\infty} \frac{7z^2 + 5z}{z^2-3}$

X4

$\lim_{x\to\infty} \frac{3(x-1)^2}{x+1}$

X5

$\lim_{x\to\infty} \left(2^x+5^x\right)^\frac{1}{x}$

Limits diverging to $\infty$

Notes are on the website.

Answers for the following exercises are given below.

X6

Give an $M - \delta$ proof that $\lim_{x\to 1}\frac{1}{(x-1)^2} = +\infty$ .

X7

Explain in two or three sentences why $\lim_{x\to 1}\frac{1}{(x-1)^2}$ does not exist.

X8

Use theorems discussed in lecture to prove that $\lim_{\theta\to \pi}\left\lvert \cot(\theta) \right\rvert = +\infty$ .

X9

Let $\lim_{x\to 3}g(x) = +\infty$ , and let $f(x)$ be continuous with $domain(f) = (-\infty,\infty)$ . Prove that,

$$\lim_{x\to+\infty}f(x) = L \Longrightarrow \lim_{x\to 3}\left(f\circ g\right)(x) = L.$$

Is the reverse implication $\Longleftarrow$ true?

1.2

#44, 45, 46 & 47 (about boundedness)

11.1

#9, 12, 13, 15, 18, 19, 27 & 30

Answers for the following exercises are given below.

X10

Give the GLB (if it exists) and LUB (if it exists) of,

$$S = \left\{ \ln\left(\frac{1}{x^2}\right) : x\in(1,\infty) \right\}$$

X11

Give the GLB (if it exists) and LUB (if it exists) of,

$$T = \left\{ \sec\left(\theta\right) : \theta \in (\frac{\pi}{2},\pi) \right\}$$

X12

Prove that GLB of $\{\left(\frac{1}{2}\right)^n : n\in{\mathds N}\}$ is zero.

X13

Prove that $\{x : \frac{1}{x} \in (0,1)\}$ has no LUB.

2.6

#3, 7, 9, 15, 19, 21, 26, 32 & 35 (For #35, create a function of $\theta$ and use EVT.)

Assigned Problems

None. Midterm #1 is Monday June 1 $^$st .

Answers for X exercises

X1

Answer: 0

Change $\lim_{x\to-\infty} f(x)$ to $\lim_{h\to 0^-}f(\frac{1}{h})$ and use continuity.

X2

Answer: 0

Change into $\lim_{\varphi \to 0^+} \varphi\,\sin\left(\frac{1}{\varphi}\right)$ and use Squeeze Thm.

X3

Answer: 7

Divide through by $z^2$ , and change into $\lim_{h\to 0^+}\frac{7 + 5h}{1-3h^2}$ . Then use continuity of polynomials and Reciprocal Thm.

X4

Answer: DNE

By long division, $\frac{3(x-1)^2}{x+1} = 3\left(x - 3 + \frac{4}{x+1}\right)$ . Note that $\frac{x}{2} - 3 + \frac{4}{x+1} > 0$ for $x > \frac{5+\sqrt{17}}{2}$ . Therefore,

$$\begin{aligned} \frac{3(x-1)^2}{x+1} > \frac{3}{2}x && \forall x\in\left(\frac{5+\sqrt{17}}{2},\infty\right). \end{aligned}$$

Assume the limit exists, choose any $\epsilon$ (say, $\epsilon =1$ ), then look at any $x >> N(\epsilon)$ .

X5

Answer:

Factor, $\left(2^x + 5^x\right)^\frac{1}{x} = 5 \left(\left(\frac{2}{5}\right)^x + 1\right)^\frac{1}{x}$ , and change $\lim_{x\to\infty}f(x)$ to $\lim_{h\to 0^+}f(\frac{1}{h})$ ,

$$\lim_{h\to 0^+}5 \left( \left(\frac{2}{5}\right)^\frac{1}{h} + 1 \right)^h.$$

Note, $5 (1)^h < 5 \left(\left(\frac{2}{5}\right)^\frac{1}{h}+1\right)^h < 5 (1+1)^h$ . Use continuity of $e^x$ to calculate the limit of $2^h$ , then use Squeeze Thm.

X6

(Rough Work)

$$\begin{aligned} \frac{1}{(x-1)^2} > M \Longleftrightarrow \f... ... \frac{1}{\sqrt{M}} > \left\lvert x-1 \right\rvert \end{aligned}$$

(Proof)

Let $M > 0$ be given. Choose $\delta = \frac{1}{\sqrt{M}}$ . If $0 < \left\lvert x-1 \right\rvert < \delta$ then,

$$\frac{1}{(x-1)^2} > \frac{1}{\delta^2} = \frac{1}{\left(\frac{1}{\sqrt{M}}\right)^2} = M.$$

Since $M > 0$ was arbitrary, this proves $\lim_{x\to 1}\frac{1}{(x-1)^2} = \infty$ .

X7

If $\lim_{x\to 1}\frac{1}{(x-1)^2}$ existed, then it would be a (finite) number $L$ . Since $\lim_{x\to 1}\frac{1}{(x-1)^2} = \infty$ , we could let $M=L+1$ and find $\delta(M)$ so that if $0 < \left\lvert x-1 \right\rvert < \delta$ then $\left\lvert \frac{1}{(x-1)^2} - L \right\rvert > 1$ . This makes $\lim_{x\to 1}\frac{1}{(x-1)^2} = L$ impossible for any $\epsilon \leq 1$ .

X8

Since $\left\lvert \cdot \right\rvert$ and $\cos(\theta)$ are both continuous functions,

$$\lim_{\theta \to 0}\left\lvert \cos(\theta) \right\rvert = \left... ...ght\rvert = \left\lvert \cos(0) \right\rvert = \left\lvert 1 \right\rvert = 1.$$

For basically the same reasons,

$$\lim_{\theta \to 0}\left\lvert \sin(\theta) \right\rvert = \left... ...\theta \to 0}\sin(\theta) \right\rvert = \left\lvert \sin(0) \right\rvert = 0.$$

By the Reciprocal Thm,

$$\lim_{\theta \to 0}\left(\frac{\left\lvert \sin(\theta) \right\rv... ...\theta \to 0}\left\lvert \cos(\theta) \right\rvert \right)} = \frac{0}{1} = 0.$$

We proved in class that if $h(x) > 0$ and $\lim_{x\to 0} h(x) = 0$ , then $\lim_{x\to 0}\frac{1}{h(x)} = +\infty$ . Use $h(\theta) = \frac{1}{\left\lvert \cot(\theta) \right\rvert } = \frac{\left\lvert \sin(\theta) \right\rvert }{\left\lvert \cos(\theta) \right\rvert }$ .

X9

Let $\epsilon>0$ be given.

Since $\lim_{x\to\infty} f(x) = L$ , there exists $N_f(\epsilon) > 0$ so that,

$$N_f < x \Longrightarrow \left\lvert f(x) - L \right\rvert < \epsilon.$$

Plug $N_f(\epsilon)$ into the other definition as $M$ . Since $\lim_{x\to 3}g(x) = +\infty$ , there exists $\delta_g(N_f)>0$ so that,

$$0 < \left\lvert x-3 \right\rvert < \delta_g \Longrightarrow N_f < g(x).$$

Therefore, if $0<\left\lvert x-3 \right\rvert < \delta_g$ , then $N_f < g(x)$ and $\left\lvert f\left(g(x)\right) - L \right\rvert <\epsilon$ . That is,

$$0 < \left\lvert x-3 \right\rvert < \delta_g \Longrightarrow \left\lvert \left(f\circ g\right)(x) - L \right\rvert < \epsilon.$$

Reverse implication $\Longleftarrow$ is FALSE. It is possible for $\lim_{x\to 3}(f\circ g)(x)$ to exist even though $\lim_{x\to\infty}f(x)$ does not. (You can build an example using the Dirichlet function.)

X10

Answer: GLB does not exist; LUB is zero.

Logarithm is a strictly increasing function. $\frac{1}{x^2}$ is largest for $x^2$ smallest. Thus, LUB should be $\ln\left(\frac{1}{1}\right) = 0$ . For very large $x$ , $\frac{1}{x^2}$ is a very small positive number, which has a very negative logarithm. That is, GLB should be $\ln\left(0\right)$ which DNE.

X11

Answer: GLB does not exist; LUB is zero.

In quadrant II, $cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)}$ is negative. The numerator is $\approx 0$ for $\theta \approx \frac{\pi}{2}$ , and denominator is $\approx 0$ for $\theta \approx \pi$ .

X12

Label the set $H = \left\{\left(\frac{1}{2}\right)^n : n\in{\mathds N}\right\}$ .

To begin, note that $\left(\frac{1}{2}\right)^n$ is positive for any $n\in{\mathds N}$ . This proves that zero is a lower bound for set $H$ .

Assume there is a larger lower bound, $L > 0$ . Since $\frac{1}{2}\in H$ , it must be that $L \leq \frac{1}{2} < 1$ , which means that $\ln(L) < 0$ . Calculate,

$$\log_{\frac{1}{2}}L = \frac{\ln(L)}{\ln(\frac{1}{2})} = \frac{\ln(L)}{-\ln(2)} > 0.$$

This proves that $\beta = \left\lceil\frac{\ln(L)}{-\ln(2)}\right\rceil + 1$ is a natural number. That is, $\left(\frac{1}{2}\right)^\beta \in H$ . We have,

$$\left(\frac{1}{2}\right)^\beta < \left(\frac{1}{2}\right)^{\lceil\frac{\ln(L)}{-\ln(2)}\rceil} \leq L.$$

This proves that $L$ is not a lower bound for set $H$ , which contradicts our assumption.

X13

Label the set $Q = \left\{x : \frac{1}{x} \in (0,1)\right\}$ .

Assume that $L$ is an upper bound for set $Q$ . Note first that $\frac{1}{2}\in(0,1)$ , so $2\in Q$ , which means that $L \geq 2$ . Suppose $\frac{1}{x}\in(0,1)$ . In particular $x > 0$ , so we may divide by both $x$ and $L$ ,

$$\begin{aligned} x \leq L \Longleftrightarrow \frac{1}{L} \leq... ...rac{1}{x} \in \left\lbrack\frac{1}{L},\infty\right) \end{aligned}$$

The set $(0,1)$ is not contained in the set $[\frac{1}{L},\infty)$ , so there are $\frac{1}{x}\in(0,1)$ for which $x\not\leq L$ . This proves $L$ is not an upper bound for set $Q$ , which contradicts our assumption.

Set $Q$ has no upper bound, therefore it has no least upper bound.