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Solutions to Real Analysis: A Long-Form Mathematics Textbook Chapter 7

Exercise 7.1 Solution: (a) Note that$\sqrt{x+3}-\sqrt{c+3}=\frac{(x+3)-(c+3)}{\sqrt{x+3}+\sqrt{c+3}}=\frac{x-c}{\sqrt{x+3}+\sqrt{c+3}}.$We have\begin{align*}f'(c)&=\lim_{x\to c}\frac{f(x)-f(c)}{x-c}\\ &=\lim_{x\to c}\frac{\sqrt{x+3}-\sqrt{c+3}}{x-c}\\ &=\lim_{x\to c}\frac{\sqrt{x+3}-\sqrt{c+3}}{x-c}\\&=\lim_{x\to c}\frac{1}{x-c}\frac{x-c}{\sqrt{x+3}+\sqrt{c+3}} \\ &=\lim_{x\to c}\frac{1}{\sqrt{x+3}+\sqrt{c+3}}\\&=\frac{1}{2\sqrt{c+3}}. \end{align*}(b) Note that$\frac{1}{x}-\frac{1}{c}=\frac{c-x}{xc}.$We have\begin{align*}g'(c)&=\lim_{x\to c}\frac{g(x)-g(c)}{x-c}\\ &=\lim_{x\to c}\frac{\frac{1}{x}-\frac{1}{c}}{x-c}\\ &= \lim_{x\to c}\frac{\frac{c-x}{xc}}{x-c}\\ &=\lim_{x\to c}\frac{-1}{xc}=-\frac{1}{c^2}.\end{align*}(c) Note that…

Solutions to Real Analysis: A Long-Form Mathematics Textbook Chapter 6

Exercise 6.5 Solution: If $a=0$, for any $\e >0$ and any $x_0\in\ R$, and any $\delta >0$, we have$|f(x)-f(x_0)|=|b-b|=0 < \e$ for all $|x-x_0|<\delta$. Hence $f$ is continuous at $x_0$.…

Solutions to Real Analysis: A Long-Form Mathematics Textbook Chapter 5

Exercise 5.1 Solution: (a) Closed. As it has no limit points, $\mathbb Z$ is closed. Clearly, it is not open. $\mathbb Z$ is not compact since $\mathbb Z$ is not…

Solutions to Real Analysis: A Long-Form Mathematics Textbook Chapter 4

Exercise 4.1 Solution: (1) Converges conditionally. It is convergent by Alternating Series Test. But $\sum_{k=1}^\infty\dfrac{1}{\sqrt{k}}$ is divergent by Proposition 4.16. (2) Converges absolutely. By Geometric Series Test and the fact…

Solutions to Real Analysis: A Long-Form Mathematics Textbook Chapter 3

Exercise 3.1 Solution: Let $\e=0.001$. Because $(a_n)$ converges to $0.001$, there exist $N$ such that for all $n> N$ we have$|a_n-0.001| < 0.001,$$\Longrightarrow -0.001 < a_n- 0.001 < 0.001,$\[\Longrightarrow 0<…

Solutions to Real Analysis: A Long-Form Mathematics Textbook Chapter 1

Exercise 1.8 Solution: (a) Because $a < b$, we have $b-a\in P$. Similarly, we have $d-c\in P$. We have\begin{align*}(b+d)-(a+c)&=b+d-a-c\\ \text{ by Axiom 6 (a)} \quad&=(b-a)+(d-c)\in P.\end{align*}Hence $a+c < b+d$. (b)…

Solutions to Real Analysis: A Long-Form Mathematics Textbook

Please feel free to submit your solutions  by comments. It supposrts LaTeX  vis Mathjax! Here is my motivation to do this! Chapter 1 The Reals #1, #2, #3, #4, #5,…