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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