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## Solution to Understanding Analysis Exercise 2.3

##### Exercise 2.3.1

(a) Let $\varepsilon > 0$ be arbitrary. Since $(x_n)\to 0$, there exists $N\in\mathbf N$ such that for all $n>N$ we have $$x_n=|x_n|<\varepsilon^2.$$Hence $\sqrt{x_n}<\varepsilon$. Therefore, for all $n>N$, we also have$|\sqrt{x_n}-0|=\sqrt{x_n}<\varepsilon.$Thus we have $(\sqrt{x_n})\to 0$.

(b) If $x=0$ by part (a) we are done. If $x\ne 0$, since $x_n\geqslant 0$ for all $n$, by Order Limit Theorem, we have $x\geqslant 0$. Since $x\ne 0$, we have $x>0$.

Let $\varepsilon > 0$ be arbitrary. Since $(x_n)\to 0$, there exists $N\in\mathbf N$ such that for all $n>N$ we have $$|x_n-x|< \varepsilon\sqrt{x}.$$ Therefore, for all $n>N$, we have\begin{align*}|\sqrt{x_n}-\sqrt{x}|=&~\frac{|x_n-x|}{\sqrt{x_n}+\sqrt{x}} \\ \leqslant &~\frac{|x_n-x|}{\sqrt{x}}\\ < &~\frac{\varepsilon\sqrt{x}}{\sqrt{x}}=\varepsilon.\end{align*}

##### Exercise 2.3.2

(a) Let $\varepsilon>0$ be arbitrary. Since $(x_n)\to 2$, there exists $N\in\mathbf N$ such that for all $n>N$ we have $$|x_n-2|<\dfrac{3}{2}\varepsilon.$$Therefore, for all $n>N$, we have \begin{align*}\left|\frac{2x_n-1}{3}-1\right|=&~\left|\frac{2(x_n-2)}{3}\right|\\ =&~\frac{2}{3}|x_n-2|\\ < & ~ \frac{2}{3}\cdot \dfrac{3}{2} \varepsilon= \varepsilon.\end{align*}Therefore $\left(\dfrac{2x_n-1}{3}\right)\to 1$.

(b) Let $1>\varepsilon>0$ be arbitrary. Since $(x_n)\to 2$, there exists $N\in\mathbf N$ such that for all $n>N$ we have $$|x_n-2|<\varepsilon.$$Since $\varepsilon < 1$, we get $$x_n > 2-\varepsilon > 1.$$Therefore, for all $n>N$, we have \begin{align*}\left|\frac{1}{x_n}-\frac{1}{2}\right|=&~\left|\frac{-(x_n-2)}{2x_n}\right|\\ =&~\frac{1}{2}|x_n-2|\\ < & ~ \frac{1}{2}\cdot \varepsilon < \varepsilon.\end{align*}Here we used that $x_n > 1$. Therefore $\left(\dfrac{1}{x_n}\right)\to \dfrac{1}{2}$.

##### Exercise 2.3.3

Let $\varepsilon>0$ be arbitrary.
Since $(x_n)\to l$, there exists $N\in\mathbf N$ such that for all $n>N$ we have
$|x_n-l|<\varepsilon.$Hence for all $n>N$ we have$$\label{eq squeeze theorem 1}-\varepsilon < x_n-l < \varepsilon \iff l-\varepsilon < x_n < l+\varepsilon.$$ Similarly, since $(z_n)\to l$, there exists $N\in\mathbf N$ such that for all $n>N$ we have
$|z_n-l|<\varepsilon.$Hence for all $n>N$ we have$$\label{eq squeeze theorem 2}-\varepsilon < z_n-l < \varepsilon \iff l-\varepsilon < z_n < l+\varepsilon.$$
Note that $x_n\leqslant y_n\leqslant z_n$, therefore by \eqref{eq squeeze theorem 1} we have $$\label{eq squeeze theorem 3}l-\varepsilon < x_n\leqslant y_n$$for all $n>N$.
By \eqref{eq squeeze theorem 2} we have $$\label{eq squeeze theorem 4} y_n \leqslant z_n < l+\varepsilon$$ for all $n>N$.
Combining \eqref{eq squeeze theorem 3} and \eqref{eq squeeze theorem 4}, we have$l-\varepsilon < y_n< l+\varepsilon \iff |y_n-l| < \varepsilon$all $n>N$. Therefore $(y_n)\to l$ as well.

##### Exercise 2.3.4

Note that $(a_n)\to 0$.

(a) By Algebraic Limit Theorem, we have $$\lim (1+2a_n)=1+2\lim a_n=1,$$ $$\lim (1+3a_n-4a_n^2)=1+3\lim a_n-4\lim a_n\lim a_n=1.$$ Therefore using Algebraic Limit Theorem again, \begin{align*}\lim\left(\frac{1+2a_n}{1+3a_n-4a_n^2}\right) = &~\frac{\lim (1+2a_n)}{\lim (1+3a_n-4a_n^2)}\\ =&~\frac{1}{1}=1.\end{align*}
(b) By Algebraic Limit Theorem, we have\begin{align*}\lim\frac{(a_n+2)^2-4}{a_n}=&~\lim\frac{(a_n^2+4a_n+4)-4}{a_n}\\ =&~ \lim\frac{a_n^2+4a_n}{a_n}\\ =&~ \lim\frac{a_n(a_n+4)}{a_n}\\ =&~\lim (a_n+4)=4.\end{align*}
(c) By Algebraic Limit Theorem and multiplying both numerator and denominator by $a_n$, we have\begin{align*}\lim\frac{2/a_n+3}{1/a_n+5}=&~\lim\frac{2+3a_n}{1+5a_n}\\ =&~ \frac{\lim (2+3a_n)}{\lim (1+5a_n)}\\ =&~ \frac{2}{1}=2.\end{align*}

##### Exercise 2.3.5

We first show “if” part. Set $\lim x_n=\lim y_n=a$. Let $\varepsilon>0$ be arbitrary. Since $\lim x_n=a$, there exists $N_1\in\mathbf N$ such that for all $n>N$ we have $$\label{eq 2.3.5.1}|x_n-a|<\varepsilon.$$Similarly, since $\lim y_n=a$, there exists $N_2\in\mathbf N$ such that for all $n>N$ we have $$\label{eq 2.3.5.2}|y_n-a|<\varepsilon.$$Let $N=2\max\{N_1,N_2\}+2$, we show that$|z_n-a| < \varepsilon$for all $n>N$. If $n=2k-1$ is odd, then $k> N_1$. By \eqref{eq 2.3.5.1} we have $|z_n-a|=|x_k-a|<\varepsilon.$If $n=2k$ is even, then $k>N_2$. By \eqref{eq 2.3.5.2} we have $|z_n-a|=|y_k-a|<\varepsilon.$Therefore $(z_n)$ is convergent to the same limit of $(x_n)$.

We then “show only if” part. Suppose $\lim z_n=a$. Let $\varepsilon>0$ be arbitrary. There exists $N\in\mathbf{N}$ such that $$\label{eq 2.3.5.3}|z_n-a|<\varepsilon$$ for all $n>N$. Note that for all $n> N$ we also have $2n-1> N$ and $2n> N$. Hence by \eqref{eq 2.3.5.3} we have$|x_n-a|=|z_{2n-1}-a|< \varepsilon$and $|y_n-a|=|z_{2n-1}-a|< \varepsilon.$Therefore $\lim x_n=\lim y_n=a$.

##### Exercise 2.3.6

We are going to using the following useful formula,$\sqrt{a}-\sqrt{b}=\frac{a-b}{\sqrt{a}+\sqrt{b}}.$
By this formula, we can rewrite $b_n$ as follows,\begin{align*}b_n=&~n-\sqrt{n^2+2n}\\=&~\frac{n^2-(n^2+2n)}{n+\sqrt{n^2+2n}}\\ =&~\frac{-2n}{n+\sqrt{n^2+2n}}\\ =&\frac{-2}{1+\sqrt{n^2+2n}/n}\\ =&~\frac{-2}{1+\sqrt{1+\frac{2}{n}}}.\end{align*}Then since $\lim \dfrac{2}{n}=2\lim\dfrac{1}{n}=2\cdot 0=0$, and by Exercise 2.3.1, we have$$\lim\sqrt{1+\frac{2}{n}}=1.$$Hence by the Algebraic Limit Theorem, we have\begin{align*}\lim b_n=&~\lim\frac{-2}{1+\sqrt{1+\frac{2}{n}}}\\ =&~\frac{-2}{1+\lim\sqrt{1+\frac{2}{n}}}\\=&~\frac{-2}{1+1}=-1.\end{align*}

##### Exercise 2.3.7

(a) Here is an example. Let $(x_n)$ be the sequence (1,0,1,0,1,0,……). Let $(y_n)$ be the sequence (-1,0,-1,0,-1,0,……). Then both $(x_n)$ and $(y_n)$ are divergent. However, the sequence $(x_n+y_n)$ is given by $(0,0,0,0,0,\cdots)$. Therefore $(x_n+y_n)$ is convergent.

(b) This is impossible. Suppose $(x_n+y_n)$ and $x_n$ are convergent. Note that $$y_n=(x_n+y_n)-x_n,$$ therefore by Algebraic Limit Theorem, we have $(y_n)$ is convergent.

(c) Let $b_n=\dfrac{(-1)^n}{n}$, then it is clear that $(b_n)\to 0$. However, we have$\frac{1}{b_n}=(-1)^n.$Since $(b_n)$ is not bounded, therefore by Theorem 2.3.2, $b_n$ is divergent.

(d) This is impossible. Suppose $a_n-b_n$ is bounded, namely there exists $M_1\in\mathbf N$ such that for all $n\in \mathbf N$, $$|a_n-b_n|\leqslant M_1.$$ Since $(b_n)$ is convergent, by Theorem 2.3.2, $(b_n)$ is bounded, namely there exists $M_2\in\mathbf N$ such that for all $n\in \mathbf N$, $$|b_n|\leqslant M_2.$$Hence for all $n\in \mathbf N$, we have$|a_n|\leqslant |a_n-b_n|+|b_n|\leqslant M_1+M_2.$Thus $(a_n)$ is bounded.

(e) Take $(a_n)$ to be the sequence (0,0,0,0,……). Take $(b_n)$ to be the sequence (1,2,3,4,5,……). Clearly, $(a_n)$ and $(a_nb_n)$ are convergent but $(b_n)$ is not since $(b_n)$ is not even bounded.

##### Exercise 2.3.8

(a) By the Algebraic Limit Theorem, we can show by induction that if $(x_n)\to x$, then $(x_n^k)\to x^k$ for any $k\in\mathbf N$. Let $p(x)$ be a polynomial,$p(x)=a_mx^m+a_{m-1}x^{m-1}+\cdots+a_1x+a_0,$ where $a_m$ are real numbers. Then by the Algebraic Limit Theorem\begin{align*}\lim p(x_n)=&~\lim(a_mx_n^m+a_{m-1}x_n^{m-1}+\cdots+a_1x_n+a_0)\\ =&~ \lim a_mx_n^m+\lim a_{m-1}x_n^{m-1}+\cdots+\lim a_1x_n+\lim a_0\\=&~ a_m\lim x_n^m+a_{m-1}\lim x_n^{m-1}+\cdots+a_1\lim x_n+ a_0\\=&~a_mx^m+a_{m-1}x^{m-1}+\cdots+a_1x+a_0=p(x).\end{align*}(b) Consider the function $f:\mathbf R\to \mathbf R$ such that $f(0)=1$ and $f(x)=0,\quad x\in (-\infty,0)\cup (0,\infty).$Then take $x=0$, it is clear that if $(x_n)\to 0$, then $$(f(x_n))\to 0\ne f(0)=1.$$

##### Exercise 2.3.9

(a) Let $\varepsilon>0$ be arbitrary. Since $(a_n)$ is bounded, there exists some $M>0$ such that $$\label{eq 2.3.9.1}|a_n| < M$$ for all $n\in\mathbf N$.
Since $\lim b_n=0$, there exists $N\in\mathbf N$ such that for all $n> N$ we have$$\label{eq 2.3.9.2}|b_n|< \dfrac{\varepsilon}{M}.$$Therefore it follows from \eqref{eq 2.3.9.1} and \eqref{eq 2.3.9.2} that for all $n>N$ \begin{align*}|a_nb_n-0|=&~|a_nb_n|\leqslant M|b_n|\\ <& ~M\frac{\varepsilon}{M}=\varepsilon.\end{align*}We cannot use Algebraic Limit Theorem as the limit of $(a_n)$ may not exist.

(b) Note that $$a_n=\frac{a_nb_n}{b_n}, \quad a_n\cdot b_n=a_nb_n$$ hence it follows from parts (iii) and (iv) of Algebraic Limit Theorem that $(a_n)$ is convergent if and only if $(a_nb_n)$ is convergent. Hence $\lim b_n\ne 0$ is very important.

(c) Since $\lim a_n=0$ and $\lim b_n$ exists, therefore $(b_n)$ is bounded. Then by part (a) we have that $\lim a_nb_n=0=a\cdot b$ as $a=0$. Note that in order to use part (a), the sequences $(a_n)$ and $(b_n)$ here correspond to the sequence $(b_n)$ and $(a_n)$ in part (a), respectively.

##### Exercise 2.3.10

(a) Counterexample. Let $(a_n)$ and $(b_n)$ be the same sequence $(1,2,3,4,5,\cdots)$. Then $(a_n-b_n)$ is the sequence $(0,0,0,\cdots)$. It is clear that $\lim(a_n-b_n)=0$. However the limits of $(a_n)$ and $(b_n)$ do not exist since they are not even bounded.

(b) We use the following inequality $||x|-|y||\leqslant |x-y|$ to prove it.
Let $\varepsilon>0$ be arbitrary. Since $(b_n)\to b$, there exists $N\in\mathbf N$ such that for all $n>N$ we have $|b_n-b| < \varepsilon.$ Then by our inequality $||x|-|y||\leqslant |x-y|$, for all $n>N$ we have$||b_n|-|b||\leqslant |b_n-b|<\varepsilon.$Hence $(|b_n|)\to |b|$.

(c) Yes. Note that$b_n=(b_n-a_n)+a_n,$hence by Algebraic Limit Theorem we have$\lim b_n=\lim (b_n-a_n)+\lim a_n=0+a=a.$
(d) Yes. Note that $|b_n-b|\leqslant a_n$ implies $a_n\geqslant 0$ and $-a_n\leqslant b_n-b\leqslant a_n.$Since $\lim a_n=0$, by Algebraic Limit Theorem we have $\lim (-a_n)=0$. Now by the Sequeeze Theorem, we have $$\lim (b_n-b)=0.$$ Hence $$\lim b_n=\lim (b_n-b)+\lim b=0+b=b.$$

##### Exercise 2.3.11

(a) Let $\varepsilon>0$ be arbitrary. Since $(x_n)$ is convergent, by Theorem 2.3.2, $(x_n)$ is bounded. Hence there exists $M>0$ such that $|x_n-x|<M$ for all $n\in\mathbf N$. Again since $(x_n)\to x$, there exists $N_0\in\mathbf N$ such that for all $n>N_0$ we have $$|x_n-x|<\frac{\varepsilon}{2}.$$ Now we take an integer $N>\dfrac{2N_0M}{\varepsilon}$, we are going to show that for all $n> N$, we have $|y_n-x|<\varepsilon$. Indeed, for $n> N$, we have$|x_1-x|+\cdots+|x_{N_0}-x|<N_0M.$ $|x_{N_0+1}-x|+\cdots+|x_{n}-x|<(n-N_0)\frac{\varepsilon}{2}.$ Therefore $$\label{eq 2.3.11.1}|x_1-x|+\cdots+|x_n-x|< N_0M+(n-N_0)\frac{\varepsilon}{2}$$ Since $n>N>\dfrac{2N_0M}{\varepsilon}$, we have $$\label{eq 2.3.11.2}N_0M<n\frac{\varepsilon}{2}.$$ Therefore\begin{align*}|y_n-x|=&\frac{|x_1+\cdots+x_n-nx|}{n}\\ \leqslant &~\frac{|x_1-x|+\cdots+|x_n-x|}{n} \\ \text{use }\eqref{eq 2.3.11.1}\quad <&~\frac{N_0M+(n-N_0)\frac{\varepsilon}{2}}{n} \\ \text{use }\eqref{eq 2.3.11.2}\quad<&~\frac{n\frac{\varepsilon}{2}+(n-N_0)\frac{\varepsilon}{2}}{n}<\varepsilon.\end{align*}Hence $(y_n)\to x$ as well.

(b) Take $x_n=(-1)^n$, then $(x_n)$ does not converge but $(y_n)\to 0$.

##### Exercise 2.3.12

(a) True. Take any element $b\in B$. Since every $a_n$ is an upper bound of $B$, we have $a_n\geqslant b$ for all $n\in\mathbf N$. Therefore by the Order Limit Theorem, we have $$a=\lim a_n\geqslant b.$$Hence $a\geqslant b$. Since $b$ is chosen arbitrarily, we see that $a$ is an upper bound of $B$.

(b) True. Suppose $a\in (0,1)$, then take $\varepsilon=\min\{a,1-a\}$, then $a-\varepsilon>0$ and $a+\varepsilon < 1$. Consider the $\varepsilon$-neighborhood $V_{\varepsilon}(a)$, then$V_{\varepsilon}(a)\subset (0,1)$. Recall the definition 2.2.3B, if $(a_n)\to a$, then there exists $N\in \mathbf N$ such that $a_n\in V_{\varepsilon}(a)$ for all $n>N$. This implies $a_n\in (0,1)$ for all $n> N$ which contradicts our assumption. Hence our assumption that $a\in(0,1)$ is impossible. Thus $a$ is in the complement of $(0,1)$.

(c) False. We are going to use Theorem 1.4.3 to construct a counter example. Let $a=\sqrt{2}$ which is irrational. By Theorem 1.4.3, for any $n\in \mathbf N$ we can find a rational number $a_n$ such that$\sqrt{2} < a_n < \sqrt{2}+\frac{1}{n}.$Then we show that $(a_n)\to\sqrt{2}$.

Let $\varepsilon > 0$ be arbitrary. Let $N> \dfrac{1}{\varepsilon}$, then $\frac{1}{N}< \varepsilon$. For all $n > N$, we have $|a_n-\sqrt{2}|<\frac{1}{n} < \frac{1}{N}=\varepsilon.$Hence $(a_n)\to a$. By our choice all $a_n$ are rational, but the limit is $\sqrt{2}$ which is irrational.

##### Exercise 2.3.13

(a) Note that $a_{mn}=\dfrac{1}{1+\frac{n}{m}}$, hence $\lim\limits_{m\to\infty}a_{mn}=1$ and$\lim_{n\to\infty}\left(\lim_{m\to\infty}a_{mn}\right)=\lim_{n\to\infty} 1=1.$Similarly, $a_{mn}=\dfrac{\frac{m}{n}}{1+\frac{m}{n}}$, hence $\lim\limits_{n\to\infty}a_{mn}=0$ and $\lim_{m\to\infty}\left(\lim_{n\to\infty}a_{mn}\right)=\lim_{m\to\infty} 0=0.$
(b) Yes. It is cleat that $\lim\limits_{m,n\to\infty}a_{mn}=0$. Similarly to part (a), we have$\lim_{m\to\infty}\left(\lim_{n\to\infty}a_{mn}\right)=\lim_{n\to\infty}\left(\lim_{m\to\infty}a_{mn}\right)=0.$The three limits are the same.
If $a_{mn}=\dfrac{mn}{m^2+n^2}$, then$\lim_{m\to\infty}\left(\lim_{n\to\infty}a_{mn}\right)=\lim_{n\to\infty}\left(\lim_{m\to\infty}a_{mn}\right)=0.$However $\lim\limits_{m,n\to\infty}a_{mn}$ does not exist. Suppose the limit exists. Taking $m=n>N$, we have $a_{nn}=\dfrac{1}{2}$. Hence this sequences has infinitely many one-half’s. By the negation of Exercise 2.2.4(b), the limit has to be $\dfrac{1}{2}$. Taking $m=2n>2N$, then we have $a_{2n,n}=\dfrac{2}{5}$. Hence this sequences has infinitely many two-fifth’s. By the negation of Exercise 2.2.4(b), the limit has to be $\dfrac{2}{5}$. But this limit should be unique, hence we get a contradiction and we are done.