If you find any mistakes, please make a comment! Thank you.

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 $x^2-c^2=(x-c)(x+c)$ and\[\frac{1}{x^2}-\frac{1}{c^2}=\frac{c^2-x^2}{x^2c^2}=\frac{(c-x)(c+x)}{x^2c^2}.\]We have\begin{align*}h'(c)&=\lim_{x\to c}\frac{h(x)-h(c)}{x-c}\\ &=\lim_{x\to c}\frac{4x^2+\frac{1}{x^2}-4c^2-\frac{1}{c^2}}{x-c}\\ &= \lim_{x\to c}\left(\frac{4(x^2-c^2)}{x-c}+\frac{\frac{1}{x^2}-\frac{1}{c^2}}{x-c}\right)\\ &= \lim_{x\to c}\left(\frac{4(x^2-c^2)}{x-c}+\frac{\frac{(c-x)(c+x)}{x^2c^2}}{x-c}\right)\\ &=\lim_{x\to c}\left(4(x+c)-\frac{x+c}{x^2c^2}\right)=8c-\frac{2}{c^3}.\end{align*}


Exercise 7.2

Solution: Suppose there exist differentiable functions $f$ and $g$ such that  $x=f(x)g(x)$ and $f(0)=g(0)=0$. Differentiating both sides, we obtain\[(x)’=(f(x)g(x))’\Longrightarrow 1=f'(x)g(x)+f(x)g'(x).\]Let $x=0$, we have\[1=f'(0)g(0)+f(0)g'(0)=0,\]which is impossible. Hence we are done.


Exercise 7.3

Solution: We have\begin{align*}\left(\frac{f}{g}\right)'(c)&=\lim_{x\to c}\frac{\frac{f(x)}{g(x)}-\frac{f(c)}{g(c)}}{x-c}\\ &=\lim_{x\to c}\frac{f(x)g(c)-f(c)g(x)}{(x-c)g(x)g(c)}\\&=\lim_{x\to c}\frac{(f(x)g(c)-f(c)g(c))-(f(c)g(x)-f(c)g(c))}{(x-c)g(x)g(c)}\\&=\lim_{x\to c}\frac{(f(x)g(c)-f(c)g(c))}{(x-c)g(x)g(c)}-\lim_{x\to c}\frac{(f(c)g(x)-f(c)g(c))}{(x-c)g(x)g(c)}\\&= \lim_{x\to c}\frac{f(x)-f(c)}{x-c}\frac{g(c)}{g(x)g(c)}-\lim_{x\to c}\frac{g(x)-g(c)}{x-c}\frac{f(c)}{g(x)g(c)}\\&=\frac{f'(c)}{g(c)}-\frac{g'(c)f(c)}{(g(c))^2}=\frac{f'(c)g(c)-g'(c)f(c)}{(g(c))^2}.\end{align*}


Exercise 7.4

Solution: We show that the limit $\lim_{x\to 0}\frac{f(x)-f(0)}{x-0}$ does not exist.

Clearly, we have $f(0)=0$. Let $(a_n)$ be a sequence of irrational numbers converging to $0$, then\[\lim_{n\to\infty}\frac{f(a_n)-f(0)}{a_n-0}=\lim_{n\to \infty}\frac{a_n-0}{a_n-0}=1.\]Let $(b_n)$ be a sequence of rational numbers converging to $0$, then\[\lim_{n\to\infty}\frac{f(b_n)-f(0)}{b_n-0}=\lim_{n\to \infty}\frac{\frac{1}{2}b_n-0}{b_n-0}=\frac{1}{2}.\]Since those two limits are different, we conclude that $\lim_{x\to 0}\frac{f(x)-f(0)}{x-0}$ does not exist.


Exercise 7.5

Solution: We show it by induction on $n$. If $n=1$, this is clear. Then we suppose that $(x^n)’=nx^{n-1}$. For $n+1$, we use the product rule,$$(f \cdot g)'(x)=f'(x)g(x)+f(x)g'(x).$$Let $f(x)=x^n$ and $g(x)=x$, then we have\begin{align*}&\ (x^{n+1})’=(x^n\cdot x)’\\=&\ (x^n)’\cdot x+x^n \cdot (x)’\\ =&\ nx^{n-1} \cdot x+x^n\cdot 1\\ =&\ nx^n+x^n=(n+1)x^n. \end{align*}Hence it also holds for $n+1$. We are done.


Exercise 7.6

Solution: Since $f’$ is bounded, there exists $m>0$ such that $|f'(a)|< m$ for all $a\in I$. By Theorem 7.22, for any $[x_1,x_2]\subset I $, we have\begin{align*}\frac{f(x_1)-f(x_2)}{x_1-x_2}=f'(x_0)\end{align*}for some $x_0\in (x_1,x_2)$.

For $\e >0$, take $\delta=\frac{\e}{m}$. Then for any $x_1 ,x_2\in I$ such that $|x_1-x_2| < \delta$, we have\begin{align*}|f(x_1)-f(x_2)|&=|f'(x_0)||x_1-x_2|\\ & < m \frac{\e}{m}=\e. \end{align*}Hence $f$ is uniformly continuous.


Exercise 7.7

Solution:

(a) We require that $\lim_{x\to 0} x^a=0$. It is clear this happens if and only if $a >0$.

(b) If $f$ is differentiable, then we require that $\lim_{x\to 0}\frac{f_a(x)}{x}$ exists. This means $$\lim_{x\to 0^+}\frac{f_a(x)-f_a(0)}{x-0}=\lim_{x\to 0^-}\frac{f_a(x)-f_a(0)}{x-0}.$$ Namely, $$\lim_{x\to 0^+}\frac{x^a}{x}=0 \iff \lim_{x\to 0^+} x^{a-1}=0.$$Hence it follows from part (a) that this happens if and only if $a >1$.


Exercise 7.8

Solution: Since $f’$ is continuous on $[a,b]$, it follows from Corollary 6.30 that $f’$ is bounded. Hence there exist $C >0$ such that $|f'(\alpha)| < C$ for all $\alpha\in[a,b]$.

For all $ x ,y \in [a,b]$, by Theorem 7.22, we have\[f(x)-f(y)=f'(\alpha)(x-y)\]for some $\alpha$ in between of $x$ and $y$. Hence\[|f(x)-f(y)|\leqslant C|x-y|\]since $|f'(\alpha)| < C$ for all $\alpha\in[a,b]$. Therefore $f$ is Lipschitz.


Exercise 7.9

Solution: Since $f’$ is bounded on $A$, there exist $C >0$ such that $|f'(\alpha)| < C$ for all $\alpha\in A$.

For all $ x ,y \in A$, by Theorem 7.22, we have\[f(x)-f(y)=f'(\alpha)(x-y)\]for some $\alpha$ in between of $x$ and $y$. Hence\[|f(x)-f(y)|\leqslant C|x-y|\]since $|f'(\alpha)| < C$ for all $\alpha\in A$. Therefore $f$ is Lipschitz.


Exercise 7.10

Solution:

(a) Let $f(x)=x$, then $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$. But $f'(c)\equiv 1$.

(b) Let $f(x)=\begin{cases}x, &\text{for }0\leqslant x\leqslant 1,\\ 2-x, &\text{for }1\leqslant x\leqslant 2\end{cases}$. Put $a=0$ and $b=2$. Then you can prove that $f'(c)=1 $ or $-1$, but never zero. In particular, it is not differentiable at $x=1$.

(c) Let $f(x)=\begin{cases}1, &\text{for }x=0,\\ 1/x, &\text{otherwise }\end{cases}$. Put $a=0$ and $b=1$.


Exercise 7.11

Solution: ??? No such values…


Exercise 7.12

Solution: Let $x,y$ be the length and width respectively. Suppose $p$ is the perimeter, which is a fixed parameter. Then \[2x+2y=p\Longrightarrow x=\frac{p-2x}{2}.\]The area is given by $$A(x)=xy=\frac{x(p-2x)}{2}.$$We compute the derivative of $A(x)$,\[A'(x)=\frac{p-4x}{2}.\]If $0 < x< p/4$, $A'(x)>0$ and hence $A(x)$ is increasing on $(0,p/4)$ by Corollary 7.25.

If $p/4 < x< p$, $A'(x)<0$ and hence $A(x)$ is decreasing on $(p/4,p/2)$ by Corollary 7.25.

Therefore, $A(x)$ takes the maximal value at $x=p/4$. In particular, at this time, $y=\frac{p-2x}{2}=p/4$ as well. Hence it is a square. This shows that the square has the maximal area.


Exercise 7.13

Solution: We would like to show that if $x >0$, then $x+\frac{1}{x}\geqslant 2$.

Let $f(x)=x+\frac{1}{x}$, then $f'(x)=1-\frac{1}{x^2}=\frac{x^2-1}{x^2}$.

If $ 0 < x< 1$, then $f'(x) < 0$ and hence by Corollary 7.25 that $f(x)$ is decreasing on $(0,1)$.

If $ 1 < x$, then $f'(x) > 0$ and hence by Corollary 7.25 that $f(x)$ is increasing on $(1,\infty)$.

Hence $f(x)$ takes the minimal value at $x=1$, where $f(1)=2$. Therefore $x+\frac{1}{x}\geqslant 2$ for all $x >0$.


Exercise 7.14

Solution: Define a new function\[h(x)=(f(b)-f(a))(g(x)-g(a))-(g(b)-g(a))(f(x)-f(a)).\]Then  $h$ is continuous on $[a,b]$ and differentiable on $(a,b)$. Moreover, we have $h(a)=0$ and $$h(b)=(f(b)-f(a))(g(b)-g(a))-(g(b)-g(a))(f(b)-f(a))=0.$$Using Rolle’s Theorem, there exists $x\in (a,b)$ such that $h'(x)=0$. Namely,\[(f(b)-f(a))g'(x)-(g(b)-g(a))f'(x)=0.\]


Exercise 7.15

Solution: Note that $\lim_{x\to a}\frac{f'(x)}{g'(x)}=L$, for any $\e>0$, there exist $\delta>0$ such that $$\left|\frac{f'(x)}{g'(x)}-L\right|<\e$$ for all $0<|x-a|<\delta$.

For $x$ such that $0<|x-a|<\delta$. Since $f(a)=g(a)=0$, by Exercise 7.14, we have\[(f(x)-f(a))g'(x_0)=(g(x)-g(a))f'(x_0)\]for some $x_0$ between $a$ and $x$. Hence $f(x)g'(x_0)=g(x)f'(x_0)$ for some $x_0$ such that $0< |x_0-a|<\delta$. Therefore,\[\left|\frac{f(x)}{g(x)}-L\right|=\left|\frac{f'(x_0)}{g'(x_0)}-L\right|<\e.\] Therefore $\lim_{x\to a}\frac{f(x)}{g(x)}=L$.


Exercise 7.16

Solution: See Proof of L’Hopital’s Rule.

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.

Leave a Reply

Close Menu