Solution to Understanding Analysis 2nd Edition Exercise of Chapter 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\begin{equation}\label{eq squeeze theorem 1}-\varepsilon < x_n-l < \varepsilon \iff l-\varepsilon < x_n < l+\varepsilon.\end{equation} 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\begin{equation}\label{eq squeeze theorem 2}-\varepsilon < z_n-l < \varepsilon \iff l-\varepsilon < z_n < l+\varepsilon.\end{equation}

Note that $x_n\leqslant y_n\leqslant z_n$, therefore by \eqref{eq squeeze theorem 1} we have \begin{equation}\label{eq squeeze theorem 3}l-\varepsilon < x_n\leqslant y_n\end{equation}for all $n>N$.

By \eqref{eq squeeze theorem 2} we have \begin{equation}\label{eq squeeze theorem 4} y_n \leqslant z_n < l+\varepsilon\end{equation} 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 \begin{equation}\label{eq 2.3.5.1}|x_n-a|<\varepsilon.\end{equation}Similarly, since $\lim y_n=a$, there exists $N_2\in\mathbf N$ such that for all $n>N$ we have \begin{equation}\label{eq 2.3.5.2}|y_n-a|<\varepsilon.\end{equation}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 \begin{equation}\label{eq 2.3.5.3}|z_n-a|<\varepsilon\end{equation} 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 \begin{equation}\label{eq 2.3.9.1}|a_n| < M\end{equation} 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\begin{equation}\label{eq 2.3.9.2}|b_n|< \dfrac{\varepsilon}{M}.\end{equation}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 \begin{equation}\label{eq 2.3.11.1}|x_1-x|+\cdots+|x_n-x|< N_0M+(n-N_0)\frac{\varepsilon}{2}\end{equation} Since $n>N>\dfrac{2N_0M}{\varepsilon}$, we have \begin{equation}\label{eq 2.3.11.2}N_0M<n\frac{\varepsilon}{2}.\end{equation} 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.

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.

You may also like...

1 Response

Leave a Reply

Your email address will not be published. Required fields are marked *