Chapter 3

Exercise 3.1

Q1

(Kwang Tae Kim)

\[ v_n=\frac{(2+\frac{1}{n})^2-2^2}{\frac{1}{n}} \\ =\frac{4+\frac{4}{n}+\frac{1}{n^2}-4}{\frac{1}{n}}\\ =4+\frac{1}{n}\] It is clear that if n is very large (that is, if the time interval is very short), vn is very close to 4. The velocity at t = 2 is 4. Thus we get the same result.

\[ v_n=\frac{2^2-(2-\frac{1}{n^2})^2}{\frac{1}{n^2}} \\ =\frac{4-(4-\frac{4}{n^2}+\frac{1}{n^4})}{\frac{1}{n^2}}\\ =4-\frac{1}{n^2}\] It is clear that if n is very large (that is, if the time interval is very short), vn is very close to 4. The velocity at t = 2 is 4. Thus we get the same result.

Exercise 3.2

Q1

\[\lim_{n\to \infty}\frac{5n^2+4}{2n^3+3} \\ = \lim_{n\to \infty}\frac{\frac{5n^2+4}{n^3}}{\frac{2n^3+3}{n^3}} \\ = \lim_{n\to \infty}\frac{\frac{5}{n}+\frac{4}{n^3}}{2+\frac{3}{n^3}} \\ = \frac{ \lim_{n\to \infty}(\frac{5}{n}+\frac{4}{n^3})}{ \lim_{n\to \infty}(2+\frac{3}{n^3})} \\ = \frac{ \lim_{n\to \infty}(\frac{5}{n})+ \lim_{n\to \infty}(\frac{4}{n^3})}{ \lim_{n\to \infty}(2)+ \lim_{n\to \infty}(\frac{3}{n^3})} \\ = \frac{0+0}{2+0}\\ = \frac{0}{2}\\ = 0\]

\[\lim_{n \to \infty}(\frac{1}{n}+(-1)^n)\\ = \lim_{n \to \infty}\frac{1}{n}+ \lim_{n \to \infty}(-1)^n \\ = 0 + \lim_{n \to \infty}(-1)^n \\ = \lim_{n \to \infty}(-1)^n \\ The \ limits \ of \ \lim_{n \to \infty}(-1)^n \ do \ not \ exist. \]

Q2

(i)\[ 50000 \times(1+\frac{0.02}{4})^4 \approx 51007.53 \]

(ii)\[ 50000 \times(1+\frac{0.02}{12})^{12} \approx 51009.22 \]

(b)\[ A_n = 50000 \times(1+\frac{0.02}{n})^n \]

n = c(1:1000)              

a_n = 50000*(1+ 0.02/n)^n

plot(n,a_n,type='l',xlab='n',ylab='50000(1+0.02/n)^n',xlim=c(0,300)) 
abline(h=50000*exp(0.02),lty=2,col=2)  #画图

\[\lim_{n \to \infty} A_n \\ = \lim_{n \to \infty}[50000 \times(1+\frac{0.02}{n})^n] \\ = 50000 \lim_{n \to \infty}(1+\frac{1}{50n})^{50n \times \frac{1}{50}}\\ let \ t=50n, then \ n \to \infty,we \ can \ get \ t=50n \to \infty \\ 50000 \lim_{n \to \infty}(1+\frac{1}{50n})^{50n \times \frac{1}{50}}\\ =50000 \lim_{t \to \infty}(1+\frac{1}{t})^{t \times \frac{1}{50}}\\ =50000 (\lim_{t \to \infty}(1+\frac{1}{t})^{t})^{\frac{1}{50}}\\ = 50000e^{\frac{1}{50}}\\ \approx 51010.07 \\ Note: \lim_{n \to \infty}(1+\frac{1}{n})^n = e\]

Thus the limit if An exist ,and the value is about 51010.07.

Q3

n = c(1:1000)              

a_n = (1+ 1/n)^n

plot(n,a_n,type='l',xlab='n',ylab='(1+1/n)^n',xlim=c(0,300))   #画图
abline(h=exp(1),lty=2,col=2)

\[\lim_{n \to \infty} a_n\\ = \lim_{n \to \infty}(1+\frac{1}{n})^n\\ = e\]

n = c(1:1000)              

a_n = (1+ 2/n)^n

plot(n,a_n,type='l',xlab='n',ylab='(1+2/n)^n',xlim=c(0,300))   #画图
abline(h=exp(2),lty=2,col=2)

\[ \lim_{n \to \infty} a_n\\ = \lim_{n \to \infty}(1+\frac{2}{n})^{n}\\ = \lim_{n \to \infty}(1+\frac{1}{\frac{n}{2}})^{\frac{n}{2}\times2}\\ let \ t=\frac{n}{2}, then \ n \to \infty,we \ can \ get \ t=\frac{n}{2} \to \infty \\ \lim_{n \to \infty}(1+\frac{1}{\frac{n}{2}})^{\frac{n}{2}\times2}\\ =\lim_{t \to \infty}(1+\frac{1}{t})^{2t}\\ =(\lim_{t \to \infty}(1+\frac{1}{t})^{t})^2\\ = e^2\]

n = c(1:1000)              

a_n = n*sin(1/n)

plot(n,a_n,type='l',xlab='n',ylab='n*sin(1/n)',xlim=c(0,15))    #画图
abline(h=1,lty=2,col=2)

\[\lim_{n \to \infty} a_n\\ = \lim_{n \to \infty}nsin(\frac{1}{n})\\ = \lim_{n \to \infty}n \times \frac{1}{n}\\ = 1 \\ Note: \lim_{n \to \infty}sin(\frac{1}{n}) \approx \lim_{n \to \infty}\frac{1}{n}\]

Exercise 3.3

Q1

1. \[\lim_{x \to -\infty}\frac{|x|}{x}\\ = \lim_{x \to -\infty}\frac{-x}{x}\\ = -1\]

As you can see in the graph below, the limit does not exist.

x = c(-1000:0)                    

a_n = x*sin(x)

plot(x,a_n,type='l',xlab='x',ylab='x*sin(x)') #画图

Q3

1. \[\lim_{t \to \infty} P(t)\\ = \lim_{t \to \infty} (35000+\frac{10000}{(t+2)^2}) \\ = \lim_{t \to \infty} 35000+ \lim_{t \to \infty} \frac{10000}{(t+2)^2}\\ =35000\]

Thus the population in the long run is 35000.

t = c(0:1000)               

P = 35000+10000/(t+2)^2

plot(t,P,type='l',xlab='t',ylab='P',xlim=c(0,30)) #画图
abline(h=35000,lty=2,col=2)

From the graph, we can know that the population in the long run is 35000.

Q4

x = c(0:1000)               

y = sqrt(x+1)-sqrt(x)

plot(x,y,type='l',xlab='x',ylab='f(x)',xlim=c(0,1000))    #画图
abline(h=0,lty=2,col=2)

\[\lim_{x \to \infty} (\sqrt{x+1}-\sqrt{x}) \\ = \lim_{x \to \infty}(\sqrt{x+1}-\sqrt{x})(\sqrt{x+1}+\sqrt{x})\frac{1}{(\sqrt{x+1}+\sqrt{x})}\\ = \lim_{x \to \infty}(x+1-x)\frac{1}{(\sqrt{x+1}+\sqrt{x})}\\ = \lim_{x \to \infty}\frac{1}{(\sqrt{x+1}+\sqrt{x})}\\ =\frac{\lim_{x \to \infty}1}{\lim_{x \to \infty}(\sqrt{x+1}+\sqrt{x})}\\ = \frac{1}{\infty}\\ = 0\]

The limit exists and it is 0.

x = c(0:1000)               

y = sqrt(x^2+x)-x

plot(x,y,type='l',xlab='x',ylab='f(x)',xlim=c(0,30))    #画图
abline(h=0.5,lty=2,col=2)

\[\lim_{x \to \infty} (\sqrt{x^2+x}-x) \\ = \lim_{x \to \infty}(\sqrt{x^2+x}-x)(\sqrt{x^2+x}+x)\frac{1}{(\sqrt{x^2+x}+x)}\\ = \lim_{x \to \infty}(x^2+x-x^2)\frac{1}{\sqrt{x^2+x}+x}\\ = \lim_{x \to \infty}x\frac{1}{\sqrt{x^2+x}+x}\\ = \lim_{x \to \infty}\frac{x}{\sqrt{x^2+x}+x}\\ = \lim_{x \to \infty}\frac{\frac{x}{x}}{\frac{\sqrt{x^2+x}+x}{x}}\\ = \lim_{x \to \infty}\frac{1}{\sqrt{1+\frac{1}{x}}+1} \\ = \frac{\lim_{x \to \infty}1}{\lim_{x \to \infty}(\sqrt{1+\frac{1}{x}}+1)}\\ = \frac{1}{1+1}\\ = 0.5\]

The limit exists and it is 0.5 .

x = c(0:1000)               

y = x^99/2^x

plot(x,y,type='l',xlab='x',ylab='f(x)',xlim=c(0,800))    #画图
abline(h=0,lty=2,col=2)

\[\lim_{x \to \infty} \frac{x^{99}}{2^x}=0 \\ \]

Exercise 3.4

Q1

x = c(0:1000) 
x=x/1000

y = 1/(x^0.5)

plot(x,y,type='l',xlab='x',ylab='1/(x^0.5)') #画图
abline(v=0,lty=2,col=2)

\[ \lim_{x\to 0+}\frac{1}{\sqrt{x}} = +\infty \]

\[ \lim_{x\to 0-}sin(\frac{1}{x}) \ doesn't \ exist. \\ As \ the \ function \ y=sinx \ is \ a \ periodic \ function,\ lim_{x\to 0-}\frac{1}{x}=-\infty, \ sin(-\infty) \ doesn't \ exist. \]

x = c(-2000:0)               
x=x/1000
y = 2-3^(1/x)

plot(x,y,type='l',xlab='x',ylab='1/(x^0.5)',ylim = c(0,2),xlim=c(-2,0)) #画图
abline(v=0,lty=2,col=2)
abline(h=2,lty=2,col=2)

\[ \lim_{x\to 0-}(2-3^{\frac{1}{x}}) \\ = \lim_{x\to 0-}2-\lim_{x\to 0-}3^{\frac{1}{x}} \\ = 2-3^{\lim_{x\to 0-}\frac{1}{x}} \\ = 2-3^{-\infty} \\ = 2-0\\ = 2\]

Exercise 3.5

Q1

\[ \lim_{x\to 7}(x^2-5x-8)^3 \\ = (7^2-5\times7-8)^3 \\ = 6^3 \\ = 216\]

\[ \lim_{x\to 1}\frac{x \sqrt{x^2+1}}{x+1} \\ = \frac{\lim_{x\to 1}x \sqrt{x^2+1}}{\lim_{x\to 1}{x+1}} \\ = \frac{\sqrt{2}}{2} \]

\[ \lim_{x\to -2}\frac{x^2-4}{x^2+x-2} \\ = \lim_{x\to -2}\frac{(x+2)(x-2)}{(x+2)(x-1)} \\ = \lim_{x\to -2}\frac{x-2}{x-1} \\ = \frac{\lim_{x\to -2}x-2}{\lim_{x\to -2}{x-1}} \\ = \frac{-2-2}{-2-1} \\ = \frac{4}{3} \]

\[ \lim_{x\to 6}\frac{x^2-6x}{x^2-5x-6} \\ = \lim_{x\to 6}\frac{x(x-6)}{(x-6)(x+1)} \\ = \lim_{x\to 6}\frac{x}{x+1} \\ = \frac{\lim_{x\to 6}x}{\lim_{x\to 6}{x+1}} \\ = \frac{6}{6+1} \\ = \frac{6}{7} \]

\[ \lim_{x\to 3}\frac{x-3}{x^2+9} \\ = \frac{\lim_{x\to 3}x-3}{\lim_{x\to 3}{x^2+9}} \\ = \frac{3-3}{3^2+9} \\ = \frac{0}{18} \\ = 0\]

\[ \lim_{x\to 2}\frac{x-5}{2-\sqrt{x-1}} \\ = \frac{\lim_{x\to 2}x-5}{\lim_{x\to 2}{2-\sqrt{x-1}}} \\ = \frac{2-5}{2-\sqrt{2-1}} \\ = \frac{-3}{2-1} \\ = -3\]

Exercise 3.6

Q1

fun1 <- function(x){
  x^2
}
fun2 <- function(x){
  x-x+1
}
fun3 <- function(x){
  1/x
}
curve(fun1, 0, 1, xlim = c(0,5), ylim = c(0,2))
curve(fun2, 1, 2, add = T)
curve(fun3, 2, 5, add = T)
abline(h=0,v=0,lty=2,col="darkgray")
points(1,1,col="red",pch=19)
points(2,1,col="red")
points(2,0.5,col="red",pch=19)
text(1,1.15,expression("(1,1)"),col="red")
text(2,1.15,expression("(2,1)"),col="red")
text(2,0.35,expression("(2,0.5)"),col="red")

(b)y=x^2、y=1、y=1/x are continuous in R. From the graph, we can know that f(x) is discontinuous at the point (2,0.5).

Q2

1. \[ \left\{\begin{aligned} 1-\sqrt{x} \neq 0 \\ x \geq 0 \\ \end{aligned}\right. \\ \left\{\begin{aligned} x \neq 1 \\ x \geq 0 \\ \end{aligned}\right. \\ dom(f)=[0,1) \cup (1,\infty)\]

2. \[ \lim_{x \to 1}f(x)=\lim_{x \to 1}\frac{x^2+x-2}{1-\sqrt{x}} \\ = \lim_{x \to 1}\frac{(x^2+x-2)(1+\sqrt{x})}{(1-\sqrt{x})(1+\sqrt{x})} \\ = \lim_{x \to 1}\frac{(x+2)(x-1)(1+\sqrt{x})}{1-x} \\ = \lim_{x \to 1}-(x+2)(1+\sqrt{x}) \\ = -(1+2)(1+\sqrt{1}) \\ = -6\]

3. \[As \ \lim_{x \to 1}f(x)=-6, \ we \ can \ define \ f(1)=-6 \ to \ make \ f \ continuous.\]

Q3

1. \[ dom(f)= (-\infty,0) \cup (0,\infty) \]
2. \[ \lim_{x \to 0}f(x)=\lim_{x \to 0}sin \frac{1}{x} \ doesn't \ exist.\]

(c)As the limit doesn’t exist, we cann’t define f(0) to make f continuous at 0.

Q4

\[ p(-2)=(-2)^5-(-2)^4-5(-2)^3+(-2)^2+8(-2)+4=-16<0\\ p(0)=4>0\\ p(3)=3^5-3^4-5\times3^3+3^2+8\times3+4=64>0\]

Thus we have

p(x)<0 for x<-1;

p(x)>0 for -1<x<2;

p(x)>0 for x>2;

The solution set is \[(-1,2) \cup (2,+\infty) \]

Q5

1. \[ p(1)=1-6-3+5+7=4>0\\ p(2)=2^5-6\times2^4-3\times2^3+5\times2^2+7=-61<0\] p(1) and p(2) have opposite sighs.As p(x) is continuous, p(x)=0 has a solution between 1 and 2.

2. \[ (1+2)/2=1.5 \\ p(1.5)=1.5^5-6\times(1.5)^4-3\times(1.5)^3+5\times(1.5)^2+7=-14.65625<0\\ p(1)=4>0\]

As p(x) is continuous, p(x)=0 has a solution between 1 and 1.5.Thus the solution is closer to 1.