Exercise 4.4.1 (a) It is already shown in Example 4.3.5. (b) Let $x_n=n+\dfrac{1}{n}$ and $y_n=n$, then we have $$ |x_n-y_n|\to 0. $$On the other hand, we have \begin{align*} |f(x_n)-f(y_n)|=&\ \left|\Big(n+\dfrac{1}{n}\Big)^3-n^3\right|\\…
Exercise 4.3.1 See Understanding Analysis Instructors’ Solution Manual Exercise 4.3.1. Exercise 4.3.2 Solution: (a) Let $f(x)\equiv 0$ (constant function). (b) Let $f(x)=x$. Please check that this function is not onetinuous.…
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)…
Exercise 2.8.1 See Understanding Analysis Instructors’ Solution Manual Exercise 2.8.1 Exercise 2.8.2 See Understanding Analysis Instructors’ Solution Manual Exercise 2.8.2 Exercise 2.8.3 See Understanding Analysis Instructors’ Solution Manual Exercise 2.8.3…
Exercise 2.7.1 See Understanding Analysis Instructors’ Solution Manual Exercise 2.7.1 Exercise 2.7.2 (a) Converges. Since $\sum\limits_{n=0}^\infty \dfrac{1}{2^n}$ converges, see Example 2.7.5 (Geometric Series), and $\dfrac{1}{2^n+n}\leqslant \dfrac{1}{2^n}$, by Theorem 2.7.4 (Comparison…
Exercise 2.6.1 Let $\varepsilon > 0$ be arbitrary. Since $(x_n)$ converges to $x$, there exits $N\in\mathbf N$ such that \begin{equation}\label{eq2.6.1.1}|x_n-x|<\frac{\varepsilon}{2}\end{equation} for all $n\geqslant N$. Whenever $n,m\geqslant N$, we have\begin{align*}|x_n-x_m|\leqslant &~|x_n-x|+|x-x_m|\\…
Exercise 2.5.1 (a) Impossible by Theorem 2.5.5. (b) Consider the sequence given by\[a_n=\frac{1+(-1)^n+\frac{1}{n}}{2}.\]Then we have\[a_{2n}=1+\frac{1}{4n},~a_{2n-1}=\frac{1}{2(2n-1)}.\]It is clear that $(a_{2n})$ converges to $1$ and $(a_{2n-1})$ converges to $0$. Moreover, it is…
Exercise 2.4.1 (a) We show by induction that $(x_n)$ is decreasing by proving $x_n > x_{n+1}$. Note that $x_1=3$ and $x_2=\dfrac{1}{4-3}=1$. Hence $x_1>x_2$. Suppose now that\[3=x_1> x_2>\cdots >x_n> x_{n+1},\]we show…
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…
Exercise 2.2.1 Example: Take the sequence (1,0,1,0,1,0,1,0,…….). To check this sequence verconges $0$, take e.g. $\varepsilon=2$. The example defines a divergent sequence. It is not hard to see the sequence…