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Solution to Understanding Analysis Exercise 4.2


Exercise 4.2.1

See Understanding Analysis Instructors’ Solution Manual Exercise 4.2.5


Exercise 4.2.2

(a) We would like $|(5x-6)-9|<1$, that is $|5(x-3)|<1$. Hence we need\[|x-3|<\frac{1}{5}.\]Therefore, the largest possible $\delta$ is $\dfrac{1}{5}$.

(b) We would like $|\sqrt{x}-2|<1$, that is\[-1<\sqrt{x}-2<1\Longrightarrow 1<\sqrt{x}<3.\]Hence we need $1<x < 9 $, which also implies\[-3< x-4<5.\]Therefore, we must have\[|x-4|<\min\{|-3|,|5|\}=3.\]The largest possible $\delta$ is $3$.

(c) We would like $|[[x]]-3|<1$. Since $[[x]]$ is an integer, this may happen only if $[[x]]=3$, therefore $3\leqslant x<4$. We need\[3-\pi \leqslant x-\pi < 4-\pi.\]Hence\[|x-\pi|<\min \{|3-\pi|,|4-\pi|\}=\pi-3.\]The largest possible $\delta$ is $\pi-3$.

(d) We would like $|[[x]]-3|<0.01$. Since $[[x]]$ is an integer, this may happen only if $[[x]]=3$, therefore $3\leqslant x<4$. We need\[3-\pi \leqslant x-\pi < 4-\pi.\]Hence\[|x-\pi|<\min \{|3-\pi|,|4-\pi|\}=\pi-3.\]The largest possible $\delta$ is $3$.


Exercise 4.2.3

See Understanding Analysis Instructors’ Solution Manual Exercise 4.2.4


Exercise 4.2.4

(a) We would like\[\left|\frac{1}{[[x]]}-\frac{1}{10}\right|<\frac{1}{2},\]namely\[-\frac12<\frac{1}{[[x]]}-\frac{1}{10}<\frac{1}{2}.\]Therefore\[-\frac{2}{5}<\frac{1}{[[x]]}<\dfrac{3}{5}.\]Since $[[x]]$ cannot be zero, we have $x\geqslant 1$. Then we have $[[x]]>\frac{3}{5}$. Because $[[x]]$ is an integer, $[[x]]\geqslant 2$. Thus $x\geqslant 2$ and $x-10\geqslant -8$. We have $|x-10|\leqslant 8$. Therefore the largest possible $\delta$ is $8$.

(b) We would like\[\left|\frac{1}{[[x]]}-\frac{1}{10}\right|<\frac{1}{50},\]namely\[-\frac{1}{50}<\frac{1}{[[x]]}-\frac{1}{10}<\frac{1}{50}.\]Therefore\[\frac{2}{25}<\frac{1}{[[x]]}<\frac{3}{25}.\]Hence \[\frac{25}{3}<[[x]]<\frac{25}{2}.\]Because $[[x]]$ is an integer, $9\leqslant [[x]]\leqslant 12$. Thus $9\leqslant x < 13$ and $-1\leqslant x-10\leqslant 3$. We have $|x-10|\leqslant 1$. Therefore the largest possible $\delta$ is $1$.

(c) The largest $\varepsilon$ satisfying the property is $\dfrac{1}{90}$. For any $V_{\varepsilon}(10)$, there is a number $x$ in it such that $9<x<10$ . Hence $[[x]]=9$, we have\[\frac{1}{[[x]]}-\frac{1}{10}=\frac{1}{90}.\]Hence there is no suitable $\delta$ response possible.


Exercise 4.2.5

(a) Let $\varepsilon>0$. Definition 4.2.1 requires that we produce a $\delta>0$ so that $0<|x-2|<\delta$ leads to the conlcusion $|(3x+4)-10|<\varepsilon$. Note that\[|(3x+4)-10|=|3x-6|=3|x-2|.\]Thus if we choose $\delta=\varepsilon/3$, then $0<|x-2|<\delta$ implies $|(3x+4)-10|<\varepsilon$.

(b) Let $\varepsilon>0$. Definition 4.2.1 requires that we produce a $\delta>0$ so that $0<|x|<\delta$ leads to the conlcusion $|x^3|<\varepsilon$. Note that\[|x^3|=|3x-6|=|x|^3.\]Thus if we choose $\delta=\sqrt[3]{\varepsilon}$, then $0<|x|<\delta$ implies $$|x^3-0|=|x|^3<(\sqrt[3]{\varepsilon})^3=\varepsilon.$$(c) Let $\varepsilon>0$. Definition 4.2.1 requires that we produce a $\delta>0$ so that $0<|x-2|<\delta$ leads to the conlcusion $|(x^2+x-1)-5|<\varepsilon$. Note that\[|(x^2+x-1)-5|=|x-2|\cdot |x+3|.\]We can choose $\delta\leqslant 1$, then $|x+3|\leqslant 6$. Thus if we choose $\delta=\min \{1, \varepsilon/6\}$, then $0<|x|<\delta$ implies $$|(x^2+x-1)-5|=|x-2|\cdot |x+3|<\frac{\varepsilon}{6}\cdot 6=\varepsilon.$$(d) Let $\varepsilon>0$. Definition 4.2.1 requires that we produce a $\delta>0$ so that $0<|x-3|<\delta$ leads to the conlcusion $|1/x-1/3|<\varepsilon$. Note that\[\left|\frac{1}{x}-\frac{1}{3}\right|=\frac{|x-3|}{3|x|}.\]We can choose $\delta\leqslant 1$, then $2\leqslant |x|\leqslant 4$. Thus if we choose $\delta=\min \{1, 6\varepsilon\}$, then $0<|x-3|<\delta$ implies $$\left|\frac{1}{x}-\frac{1}{3}\right|=\frac{|x-3|}{3|x|}<\frac{6\varepsilon}{3\cdot 2}=\varepsilon.$$


Exercise 4.2.6

(a) True. A property is true for some set, then it is also true for a subset of this set.

(b) False. In the Definition 4.2.1, the value of $f(a)$ is not involved. In general, it can be any number.

(c) True by Corollary 4.2.4.

(d) False. Take the example $f(x)=x-a$ and $g(x)=1/(x-a)$ with domain $\mb R\setminus \{a\}$. Then $\lim_{x\to a}f(x)g(x)=1$.


Exercise 4.2.7

See Understanding Analysis Instructors’ Solution Manual Exercise 4.2.6


Exercise 4.2.8

(a) Does not exist. Note that $\lim(2-1/n)=2$ and $\lim (2+1/n)=2$, however\[\lim f(2-1/n)=-1,\quad \lim f(2+1/n)=1.\]By Corollary 4.2.5, limit $\lim_{x\to 2}f(x)$ does not exist.

(b) The limit is 1. For $0<\delta<1/4$ and any $x\in V_{\delta}(7/4)$, we have $x<2$. Hence $f(x)=-1$, for all $x\in V_{\delta}(7/4)$. Hence the limit is 1.

(c) Does not exist. Note that $\lim 1/(2n+1)=0$ and $\lim 1/(2n)=0$, however\[\lim f(1/(2n+1))=\lim (-1)^{2n+1}=-1,\]\[\lim f(1/(2n))=\lim (-1)^{2n}=1.\]By Corollary 4.2.5, limit $\lim_{x\to 0}f(x)$ does not exist.

(d) The limit is zero. For any $\varepsilon>0$, let $0<|x|<\varepsilon^3$, then\[|\sqrt[3]{x}(-1)^{[[1/x]]}|=|\sqrt[3]{x}|<\varepsilon.\]Hence the limit is zero.


Exercise 4.2.9

See Understanding Analysis Instructors’ Solution Manual Exercise 4.2.7


Exercise 4.2.10

(a) We say that $\lim_{x\to a^+}f(x)=L$ provided that, for all $\varepsilon >0$, there exists a $\delta>0$ such that whenever $0< x- a<\delta$ it follows that $|f(x)-L|<\varepsilon$.

We say that $\lim_{x\to a^-}f(x)=M$ provided that, for all $\varepsilon >0$, there exists a $\delta>0$ such that whenever $0< a-x<\delta$ it follows that $|f(x)-M|<\varepsilon$.

(b) By definition, it is clear that if $\lim_{x\to a}f(x)=L$ then both the right and left-hand limits equal $L$.

Conversely, if both the right and left-hand limits equal $L$. Since the right limit is $L$, for all $\varepsilon >0$, there exists a $\delta_1>0$ such that whenever $0< x- a<\delta_1$ it follows that $|f(x)-L|<\varepsilon$. Since the left limit is $L$, for all $\varepsilon >0$, there exists a $\delta_2>0$ such that whenever $0< a- x<\delta_2$ it follows that $|f(x)-L|<\varepsilon$.

Let $\delta=\min\{\delta_1,\delta_2\}$, then whenever $0<|x-a|< \delta$, we have $|f(x)-L|<\varepsilon$. Hence $\lim_{x\to a}f(x)=L$.


Exercise 4.2.11

See Understanding Analysis Instructors’ Solution Manual Exercise 4.2.9

Linearity

This website is supposed to help you study Linear Algebras. Please only read these solutions after thinking about the problems carefully. Do not just copy these solutions.

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