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## Solution to Understanding Analysis Exercise 4.4

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|\\…

## Solution to Understanding Analysis Exercise 4.3

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.…

## 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)…

## Solution to Understanding Analysis Exercise 2.8

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…

## Solution to Understanding Analysis Exercise 2.7

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…

## Solution to Understanding Analysis Exercise 2.6

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|\\…

## Solution to Understanding Analysis Exercise 2.5

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…

## Solution to Understanding Analysis Exercise 2.4

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…

## 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…

## Solution to Understanding Analysis Exercise 2.2

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…