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Advanced Mathematics (Seventh Edition) Tongji University exercises 1-2 personal solutions

2022-06-13 03:57:00 【Navigator_ Z】

Advanced mathematics （ The seventh edition ） Tongji University exercises 1-2

 

1. Next Column various topic in , which some Count Column closed Convergence , which some Count Column Hair scattered ？ Yes closed Convergence Count Column , through too view Observe ∣ X n ∣ Of change turn Trend potential , Write Out it People Of extremely limit ： \begin{aligned}&1. Of the following questions , Which sequences converge , Which series diverge ？ On convergent sequence , Through observation |X_n| Change trend of ,\\\\& Write their limits ：&\end{aligned} ​1. Next Column various topic in , which some Count Column closed Convergence , which some Count Column Hair scattered ？ Yes closed Convergence Count Column , through too view Observe ∣Xn​∣ Of change turn Trend potential , Write Out it People Of extremely limit ：​​

   ( 1 )   { 1 2 n } ;                  ( 2 )   { ( − 1 ) n 1 n } ;    ( 3 )   { 2 + 1 n 2 } ;           ( 4 )   { n − 1 n + 1 } ;    ( 5 )   { n ( − 1 ) n } ;              ( 6 )   { 2 n − 1 3 n } ;    ( 7 )   { n − 1 n } ;             ( 8 )   { [ ( − 1 ) n + 1 ] n + 1 n } \begin{aligned} &\ \ (1)\ \left\{\frac{1}{2^n}\right\};\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)\ \left\{(-1)^n\frac{1}{n}\right\};\\\\ &\ \ (3)\ \left\{2+\frac{1}{n^2}\right\};\ \ \ \ \ \ \ \ \ (4)\ \left\{\frac{n-1}{n+1}\right\};\\\\ &\ \ (5)\ \{n(-1)^n\};\ \ \ \ \ \ \ \ \ \ \ \ (6)\ \left\{\frac{2^n-1}{3^n}\right\};\\\\ &\ \ (7)\ \left\{n-\frac{1}{n}\right\};\ \ \ \ \ \ \ \ \ \ \ (8)\ \left\{[(-1)^n+1]\frac{n+1}{n}\right\} & \end{aligned} ​  (1) { 2n1​};                (2) { (−1)nn1​};  (3) { 2+n21​};         (4) { n+1n−1​};  (5) { n(−1)n};            (6) { 3n2n−1​};  (7) { n−n1​};           (8) { [(−1)n+1]nn+1​}​​

Explain ：

   ( 1 )   When n → ∞ when , 1 2 n Trend towards On 0 , because this Count Column yes closed Convergence Of , lim ⁡ n → ∞ 1 2 n = 0    ( 2 )   When n → ∞ when , ( − 1 ) n 1 n Trend towards On 0 , because this Count Column yes closed Convergence Of , lim ⁡ n → ∞ ( − 1 ) n 1 n = 0    ( 3 )   When n → ∞ when , 1 n 2 Trend towards On 0 , namely 2 + 1 n 2 Trend towards On 2 , because this Count Column yes closed Convergence Of , lim ⁡ n → ∞ ( 2 + 1 n 2 ) = 2    ( 4 )   n − 1 n + 1 = 1 − 1 n 1 + 1 n , When n → ∞ when , 1 − 1 n 1 + 1 n Trend towards On 1 , because this Count Column yes closed Convergence Of , lim ⁡ n → ∞ n − 1 n + 1 = 1    ( 5 )   When n → ∞ when , n ( − 1 ) n Trend towards On ∞ , because this Count Column yes Hair scattered Of .    ( 6 )   2 n − 1 3 n = ( 2 3 ) n − 1 3 n , When n → ∞ when , ( 2 3 ) n − 1 3 n Trend towards On 0 , because this Count Column yes closed Convergence Of , lim ⁡ n → ∞ 2 n − 1 3 n = 0    ( 7 )   When n → ∞ when , n − 1 n Trend towards On ∞ , because this Count Column yes Hair scattered Of .    ( 8 )   [ ( − 1 ) n + 1 ] n + 1 n = [ ( − 1 ) n + 1 ] ( 1 + 1 n ) , When n → ∞ when , [ ( − 1 ) n + 1 ] ( 1 + 1 n ) Trend towards On back complex stay Count word 0 , 2 jump dynamic ,    because this Count Column yes Hair scattered Of . \begin{aligned} &\ \ (1)\ When n \rightarrow \infty when ,\frac{1}{2^n} Tend to 0, So the sequence is convergent ,\lim_{n \rightarrow \infty}\frac{1}{2^n}=0\\\\ &\ \ (2)\ When n \rightarrow \infty when ,(-1)^n\frac{1}{n} Tend to 0, So the sequence is convergent ,\lim_{n \rightarrow \infty}(-1)^n\frac{1}{n}=0\\\\ &\ \ (3)\ When n \rightarrow \infty when ,\frac{1}{n^2} Tend to 0, namely 2+\frac{1}{n^2} Tend to 2, So the sequence is convergent ,\lim_{n \rightarrow \infty}\left(2+\frac{1}{n^2}\right)=2\\\\ &\ \ (4)\ \frac{n-1}{n+1}=\frac{1-\frac{1}{n}}{1+\frac{1}{n}}, When n \rightarrow \infty when ,\frac{1-\frac{1}{n}}{1+\frac{1}{n}} Tend to 1, So the sequence is convergent ,\lim_{n \rightarrow \infty}\frac{n-1}{n+1}=1\\\\ &\ \ (5)\ When n \rightarrow \infty when ,n(-1)^n Tend to \infty, So the sequence is divergent .\\\\ &\ \ (6)\ \frac{2^n-1}{3^n}=\left(\frac{2}{3}\right)^n-\frac{1}{3^n}, When n \rightarrow \infty when ,\left(\frac{2}{3}\right)^n-\frac{1}{3^n} Tend to 0, So the sequence is convergent ,\lim_{n \rightarrow \infty}\frac{2^n-1}{3^n}=0\\\\ &\ \ (7)\ When n \rightarrow \infty when ,n-\frac{1}{n} Tend to \infty, So the sequence is divergent .\\\\ &\ \ (8)\ [(-1)^n+1]\frac{n+1}{n}=[(-1)^n+1]\left(1+\frac{1}{n}\right), When n \rightarrow \infty when ,[(-1)^n+1]\left(1+\frac{1}{n}\right) Tends to repeat in numbers 0,2 beating ,\\\\&\ \ So the sequence is divergent .\\\\ & \end{aligned} ​  (1)  When n→∞ when ,2n1​ Trend towards On 0, because this Count Column yes closed Convergence Of ,n→∞lim​2n1​=0  (2)  When n→∞ when ,(−1)nn1​ Trend towards On 0, because this Count Column yes closed Convergence Of ,n→∞lim​(−1)nn1​=0  (3)  When n→∞ when ,n21​ Trend towards On 0, namely 2+n21​ Trend towards On 2, because this Count Column yes closed Convergence Of ,n→∞lim​(2+n21​)=2  (4) n+1n−1​=1+n1​1−n1​​, When n→∞ when ,1+n1​1−n1​​ Trend towards On 1, because this Count Column yes closed Convergence Of ,n→∞lim​n+1n−1​=1  (5)  When n→∞ when ,n(−1)n Trend towards On ∞, because this Count Column yes Hair scattered Of .  (6) 3n2n−1​=(32​)n−3n1​, When n→∞ when ,(32​)n−3n1​ Trend towards On 0, because this Count Column yes closed Convergence Of ,n→∞lim​3n2n−1​=0  (7)  When n→∞ when ,n−n1​ Trend towards On ∞, because this Count Column yes Hair scattered Of .  (8) [(−1)n+1]nn+1​=[(−1)n+1](1+n1​), When n→∞ when ,[(−1)n+1](1+n1​) Trend towards On back complex stay Count word 0,2 jump dynamic ,   because this Count Column yes Hair scattered Of .​

2.   ( 1 ) Count Column Of Yes world sex yes Count Column closed Convergence Of What Well strip Pieces of ？      ( 2 ) nothing world Count Column yes no One set Hair scattered ？      ( 3 ) Yes world Count Column yes no One set closed Convergence ？ \begin{aligned}&2. \ (1) The boundedness of a sequence of numbers is a condition for the convergence of a sequence of numbers ？\\\\&\ \ \ \ (2) Whether the unbounded sequence of numbers must diverge ？\\\\&\ \ \ \ (3) Whether a bounded sequence of numbers must converge ？&\end{aligned} ​2. (1) Count Column Of Yes world sex yes Count Column closed Convergence Of What Well strip Pieces of ？    (2) nothing world Count Column yes no One set Hair scattered ？    (3) Yes world Count Column yes no One set closed Convergence ？​​

Explain ：

   ( 1 )   Count Column Of Yes world sex yes Count Column closed Convergence Of have to want strip Pieces of .    ( 2 )   nothing world Count Column One set Hair scattered .    ( 3 )   Yes world Count Column No One set closed Convergence . \begin{aligned} &\ \ (1)\ The boundedness of sequence is a necessary condition for the convergence of sequence .\\\\ &\ \ (2)\ Unbounded sequences must diverge .\\\\ &\ \ (3)\ Bounded sequences do not necessarily converge .\\\\ & \end{aligned} ​  (1)  Count Column Of Yes world sex yes Count Column closed Convergence Of have to want strip Pieces of .  (2)  nothing world Count Column One set Hair scattered .  (3)  Yes world Count Column No One set closed Convergence .​

3.   Next Column Turn off On Count Column ∣ x n ∣ Of extremely limit yes a Of set The righteous , which some yes Yes Of , which some yes wrong Of ？ Such as fruit yes Yes Of , try say bright The reason is from ; Such as fruit yes wrong Of ,      try to Out One individual back example . \begin{aligned}&3. \ The following is about the sequence |x_n| The limit of a The definition of , What is right , What's wrong ？ If it's right , Try to explain why ; If it's wrong ,\\\\&\ \ \ \ Try to give a counterexample .&\end{aligned} ​3.  Next Column Turn off On Count Column ∣xn​∣ Of extremely limit yes a Of set The righteous , which some yes Yes Of , which some yes wrong Of ？ Such as fruit yes Yes Of , try say bright The reason is from ; Such as fruit yes wrong Of ,     try to Out One individual back example .​​

   ( 1 )   Yes On ren It means to set Of ε > 0 , save stay N ∈ N + , When n > N when , No etc. type x n − a < ε become state ;    ( 2 )   Yes On ren It means to set Of ε > 0 , save stay N ∈ N + , When n > N when , Yes nothing poor many term x n , send No etc. type ∣ x n − a ∣ < ε become state ;    ( 3 )   Yes On ren It means to set Of ε > 0 , save stay N ∈ N + , When n > N when , No etc. type ∣ x n − a ∣ < c ε become state , Its in c by some individual just often Count ;    ( 4 )   Yes On ren It means to set Of m ∈ N + , save stay N ∈ N + , When n > N when , No etc. type ∣ x n − a ∣ < 1 m become state . \begin{aligned} &\ \ (1)\ For any given \varepsilon \gt 0, There is N \in \mathbb{N_+}, When n \gt N when , inequality x_n-a \lt \varepsilon establish ;\\\\ &\ \ (2)\ For any given \varepsilon \gt 0, There is N \in \mathbb{N_+}, When n \gt N when , There are infinite numbers x_n, Make inequality |x_n-a| \lt \varepsilon establish ;\\\\ &\ \ (3)\ For any given \varepsilon \gt 0, There is N \in \mathbb{N_+}, When n \gt N when , inequality |x_n-a| \lt c\varepsilon establish , among c Is a normal number ;\\\\ &\ \ (4)\ For any given m \in \mathbb{N_+}, There is N \in \mathbb{N_+}, When n \gt N when , inequality |x_n-a| \lt \frac{1}{m} establish .\\\\ & \end{aligned} ​  (1)  Yes On ren It means to set Of ε>0, save stay N∈N+​, When n>N when , No etc. type xn​−a<ε become state ;  (2)  Yes On ren It means to set Of ε>0, save stay N∈N+​, When n>N when , Yes nothing poor many term xn​, send No etc. type ∣xn​−a∣<ε become state ;  (3)  Yes On ren It means to set Of ε>0, save stay N∈N+​, When n>N when , No etc. type ∣xn​−a∣<cε become state , Its in c by some individual just often Count ;  (4)  Yes On ren It means to set Of m∈N+​, save stay N∈N+​, When n>N when , No etc. type ∣xn​−a∣<m1​ become state .​

Explain ：

   ( 1 )   wrong By mistake , Count Column { ( − 1 ) n + 1 n } , a = 1 , ∀   ε > 0 , ∃   N = [ 1 ε ] , When n > N when , ( − 1 ) n + 1 n − 1 ≤ 1 n < ε ,           but { ( − 1 ) n + 1 n } Of extremely limit No save stay .    ( 2 )   wrong By mistake , Count Column x n = { n ,         n = 2 k − 1 , 1 − 1 n , n = 2 k , , k ∈ N + , a = 1 . ∀   ε > 0 , ∃ N = [ 1 ε ] , When n > N And n by accidentally Count when ,          ∣ x n − a ∣ = 1 n < ε become state , but { x n } Of extremely limit No save stay .    ( 3 )   just indeed , ∀ ε > 0 , take 1 c   ε > 0 , Press false set up , ∃ N ∈ N + , When n > N when , No etc. type ∣ x n − a ∣ < c ⋅ 1 c ε = ε become state .    ( 4 )   just indeed , ∀ ε > 0 , take m ∈ N + , send 1 m < ε , Press false set up , ∃ N ∈ N + , When n > N when , No etc. type ∣ x n − a ∣ < 1 m < ε become state . \begin{aligned} &\ \ (1)\ error , The sequence \left\{(-1)^n+\frac{1}{n}\right\},a=1,\forall\ \varepsilon \gt 0,\exists\ N=\left[\frac{1}{\varepsilon}\right], When n\gt N when ,(-1)^n+\frac{1}{n}-1\le \frac{1}{n}\lt \varepsilon,\\\\ &\ \ \ \ \ \ \ \ \ but \left\{(-1)^n+\frac{1}{n}\right\} There is no limit to .\\\\ &\ \ (2)\ error , The sequence x_n=\begin{cases}n,\ \ \ \ \ \ \ n=2k-1,\\\\1-\frac{1}{n},n=2k,\end{cases},k\in N_+,a=1.\forall\ \varepsilon\gt 0,\exists N=\left[\frac{1}{\varepsilon}\right], When n \gt N And n For even when ,\\\\ &\ \ \ \ \ \ \ \ |x_n-a|=\frac{1}{n} \lt \varepsilon establish , but \{x_n\} There is no limit to .\\\\ &\ \ (3)\ correct ,\forall \varepsilon \gt 0, take \frac{1}{c}\ \varepsilon \gt 0, As assumed ,\exists N \in N_+, When n \gt N when , inequality |x_n-a| \lt c \cdot \frac{1}{c}\varepsilon = \varepsilon establish .\\\\ &\ \ (4)\ correct ,\forall \varepsilon \gt 0, take m \in N_+, send \frac{1}{m} \lt \varepsilon, As assumed ,\exists N \in N_+, When n \gt N when , inequality |x_n-a| \lt \frac{1}{m} \lt \varepsilon establish .\\\\ & \end{aligned} ​  (1)  wrong By mistake , Count Column { (−1)n+n1​},a=1,∀ ε>0,∃ N=[ε1​], When n>N when ,(−1)n+n1​−1≤n1​<ε,          but { (−1)n+n1​} Of extremely limit No save stay .  (2)  wrong By mistake , Count Column xn​=⎩⎪⎨⎪⎧​n,       n=2k−1,1−n1​,n=2k,​,k∈N+​,a=1.∀ ε>0,∃N=[ε1​], When n>N And n by accidentally Count when ,        ∣xn​−a∣=n1​<ε become state , but { xn​} Of extremely limit No save stay .  (3)  just indeed ,∀ε>0, take c1​ ε>0, Press false set up ,∃N∈N+​, When n>N when , No etc. type ∣xn​−a∣<c⋅c1​ε=ε become state .  (4)  just indeed ,∀ε>0, take m∈N+​, send m1​<ε, Press false set up ,∃N∈N+​, When n>N when , No etc. type ∣xn​−a∣<m1​<ε become state .​

4.   set up Count Column x n Of One like term x n = 1 n c o s   n π 2 , ask lim ⁡ n → ∞ x n = ? seek Out N , send When n > N when , x n And Its extremely limit And Bad Of most Yes value      Small On just Count ε . When ε = 0.001 when , seek Out Count N . \begin{aligned}&4. \ Set sequence {x_n} Of General items x_n=\frac{1}{n}cos\ \frac{n\pi}{2}, ask \lim_{n \rightarrow \infty}x_n=? Find out N, Make to be n \gt N when ,x_n The absolute value of the difference from its limit \\\\&\ \ \ \ Less than a positive number \varepsilon. When \varepsilon=0.001 when , Find the number N.&\end{aligned} ​4.  set up Count Column xn​ Of One like term xn​=n1​cos 2nπ​, ask n→∞lim​xn​=? seek Out N, send When n>N when ,xn​ And Its extremely limit And Bad Of most Yes value      Small On just Count ε. When ε=0.001 when , seek Out Count N.​​

Explain ：

   lim ⁡ n → ∞ x n = 0     ∣ x n − 0 ∣ = ∣ 1 n c o s   n π 2 ∣ ≤ 1 n , want send ∣ x n − 0 ∣ < ε , only want 1 n l t ε , namely n > 1 ε , the With ∀   ε > 0 , take N = [ 1 ε ] , be When n > N when ,     Just Yes ∣ x n − 0 ∣ < ε .     When ε = 0.001 when , take N = [ 1 ε ] = 1000 . namely if ε = 0.001 , only want n > 1000 , Just Yes ∣ x n − 0 ∣ < 0.001 . \begin{aligned} &\ \ \lim_{n \rightarrow \infty}x_n=0\\\\ &\ \ \ |x_n-0|=|\frac{1}{n}cos\ \frac{n\pi}{2}| \le \frac{1}{n}, To make |x_n-0| \lt \varepsilon, as long as \frac{1}{n} lt \varepsilon, namely n \gt \frac{1}{\varepsilon}, therefore \forall \ \varepsilon \gt 0, take N=\left[\frac{1}{\varepsilon}\right], Then when n \gt N when ,\\\\ &\ \ \ There is |x_n-0| \lt \varepsilon.\\\\ &\ \ \ When \varepsilon=0.001 when , take N=\left[\frac{1}{\varepsilon}\right]=1000. I.e \varepsilon=0.001, as long as n \gt 1000, There is |x_n-0| \lt 0.001.\\\\ & \end{aligned} ​  n→∞lim​xn​=0   ∣xn​−0∣=∣n1​cos 2nπ​∣≤n1​, want send ∣xn​−0∣<ε, only want n1​ltε, namely n>ε1​, the With ∀ ε>0, take N=[ε1​], be When n>N when ,    Just Yes ∣xn​−0∣<ε.    When ε=0.001 when , take N=[ε1​]=1000. namely if ε=0.001, only want n>1000, Just Yes ∣xn​−0∣<0.001.​

5.   root According to the Count Column extremely limit Of set The righteous Prove bright ：    ( 1 )   lim ⁡ n → ∞ 1 n 2 = 0                  ( 2 )   lim ⁡ n → ∞ 3 n + 1 2 n + 1 = 3 2    ( 3 )   lim ⁡ n → ∞ n 2 + a 2 n = 1      ( 4 )   lim ⁡ n → ∞ 0. 999 ⋅ ⋅ ⋅ 9 ⏟ n individual = 1 \begin{aligned}&5. \ Prove according to the definition of the limit of sequence ：\\\\&\ \ (1)\ \lim_{n \rightarrow \infty}\frac{1}{n^2}=0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)\ \lim_{n \rightarrow \infty}\frac{3n+1}{2n+1}=\frac{3}{2}\\\\&\ \ (3)\ \lim_{n \rightarrow \infty}\frac{\sqrt{n^2+a^2}}{n}=1\ \ \ \ (4)\ \lim_{n \rightarrow \infty}0.\underbrace{999\cdot\cdot\cdot9}_{n individual }=1&\end{aligned} ​5.  root According to the Count Column extremely limit Of set The righteous Prove bright ：  (1) n→∞lim​n21​=0                (2) n→∞lim​2n+13n+1​=23​  (3) n→∞lim​nn2+a2​​=1    (4) n→∞lim​0.n individual 999⋅⋅⋅9​​=1​​

Explain ：

   ( 1 )   ∣ x n − 0 ∣ = 1 n 2 , ∀   ε > 0 , by 了 send ∣ x n − 0 ∣ < ε , only want 1 n 2 < ε or n > 1 ε , take N = [ 1 ε ] ,           be When n > N when , Just Yes ∣ 1 n 2 − 0 ∣ < ε , namely lim ⁡ n → ∞ 1 n 2 = 0 .    ( 2 )   ∣ x n − 3 2 ∣ = 1 4 n + 2 < 1 4 n , ∀   ε > 0 , by 了 send ∣ x n − 3 2 ∣ < ε , only want 1 4 n < ε or n > 1 4 ε ,           take N = [ 1 4 ε ] , be When n > N when , Just Yes ∣ 3 n + 1 2 n + 1 − 3 2 ∣ < ε , namely lim ⁡ n → ∞ 3 n + 1 2 n + 1 = 3 2 .    ( 3 )   ∣ x n − 1 ∣ = ∣ n 2 + a 2 n − 1 ∣ = a 2 n ( n 2 + a 2 + n ) = a 2 n 2 + n n 2 + a 2 < a 2 2 n 2 , ∀   ε > 0 ,           by 了 send ∣ x n − 1 ∣ < ε , only want a 2 2 n 2 < ε or n > ∣ a ∣ 2 ε , take N = [ ∣ a ∣ 2 ε ] , be When n > N when , Just Yes           ∣ n 2 + a 2 n − 1 ∣ < ε , namely lim ⁡ n → ∞ n 2 + a 2 n = 1 .    ( 4 )   ∣ x n − 1 ∣ = ∣ 0. 999 ⋅ ⋅ ⋅ 9 ⏟ n individual − 1 ∣ = ∣ − 0. 111 ⋅ ⋅ ⋅ 1 ⏟ n individual ∣ = 1 1 0 n , ∀   ε > 0 , by 了 send ∣ x n − 1 ∣ < ε ,          only want 1 1 0 n < ε or n > l g   1 ε , take N = [ l g   1 ε ] , be When n > N when , Just Yes ∣ x n − 1 ∣ < ε , namely lim ⁡ n → ∞ 0. 999 ⋅ ⋅ ⋅ 9 ⏟ n individual = 1 . \begin{aligned} &\ \ (1)\ |x_n-0|=\frac{1}{n^2},\forall\ \varepsilon \gt 0, In order to make |x_n-0| \lt \varepsilon, as long as \frac{1}{n^2} \lt \varepsilon or n \gt \frac{1}{\sqrt{\varepsilon}}, take N=\left[\frac{1}{\sqrt{\varepsilon}}\right],\\\\&\ \ \ \ \ \ \ \ \ Then when n \gt N when , There is |\frac{1}{n^2}-0| \lt \varepsilon, namely \lim_{n \rightarrow \infty}\frac{1}{n^2}=0.\\\\ &\ \ (2)\ \left|x_n-\frac{3}{2}\right|=\frac{1}{4n+2} \lt \frac{1}{4n},\forall\ \varepsilon \gt 0, In order to make \left|x_n-\frac{3}{2}\right| \lt \varepsilon, as long as \frac{1}{4n} \lt \varepsilon or n \gt \frac{1}{4\varepsilon},\\\\&\ \ \ \ \ \ \ \ \ take N=\left[\frac{1}{4\varepsilon}\right], Then when n \gt N when , There is \left|\frac{3n+1}{2n+1}-\frac{3}{2}\right| \lt \varepsilon, namely \lim_{n \rightarrow \infty}\frac{3n+1}{2n+1}=\frac{3}{2}.\\\\ &\ \ (3)\ |x_n-1|=\left|\frac{\sqrt{n^2+a^2}}{n}-1\right|=\frac{a^2}{n(\sqrt{n^2+a^2}+n)}=\frac{a^2}{n^2+n\sqrt{n^2+a^2}}\lt \frac{a^2}{2n^2},\forall\ \varepsilon \gt 0,\\\\&\ \ \ \ \ \ \ \ \ In order to make |x_n-1| \lt \varepsilon, as long as \frac{a^2}{2n^2} \lt \varepsilon or n \gt \frac{|a|}{\sqrt{2\varepsilon}}, take N=\left[\frac{|a|}{\sqrt{2\varepsilon}}\right], Then when n \gt N when , There is \\\\&\ \ \ \ \ \ \ \ \ \left|\frac{\sqrt{n^2+a^2}}{n}-1\right|\lt \varepsilon, namely \lim_{n \rightarrow \infty}\frac{\sqrt{n^2+a^2}}{n}=1.\\\\ &\ \ (4)\ |x_n-1|=|0.\underbrace{999\cdot\cdot\cdot9}_{n individual }-1|=|-0.\underbrace{111\cdot\cdot\cdot1}_{n individual }|=\frac{1}{10^n},\forall\ \varepsilon \gt 0, In order to make |x_n-1| \lt \varepsilon,\\\\&\ \ \ \ \ \ \ \ as long as \frac{1}{10^n} \lt \varepsilon or n \gt lg\ \frac{1}{\varepsilon}, take N=\left[lg\ \frac{1}{\varepsilon}\right], Then when n \gt N when , There is |x_n-1| \lt \varepsilon, namely \lim_{n \rightarrow \infty}0.\underbrace{999\cdot\cdot\cdot9}_{n individual }=1.\\\\ & \end{aligned} ​  (1) ∣xn​−0∣=n21​,∀ ε>0, by 了 send ∣xn​−0∣<ε, only want n21​<ε or n>ε​1​, take N=[ε​1​],          be When n>N when , Just Yes ∣n21​−0∣<ε, namely n→∞lim​n21​=0.  (2) ∣∣∣∣​xn​−23​∣∣∣∣​=4n+21​<4n1​,∀ ε>0, by 了 send ∣∣∣∣​xn​−23​∣∣∣∣​<ε, only want 4n1​<ε or n>4ε1​,          take N=[4ε1​], be When n>N when , Just Yes ∣∣∣∣​2n+13n+1​−23​∣∣∣∣​<ε, namely n→∞lim​2n+13n+1​=23​.  (3) ∣xn​−1∣=∣∣∣∣∣​nn2+a2​​−1∣∣∣∣∣​=n(n2+a2​+n)a2​=n2+nn2+a2​a2​<2n2a2​,∀ ε>0,          by 了 send ∣xn​−1∣<ε, only want 2n2a2​<ε or n>2ε​∣a∣​, take N=[2ε​∣a∣​], be When n>N when , Just Yes          ∣∣∣∣∣​nn2+a2​​−1∣∣∣∣∣​<ε, namely n→∞lim​nn2+a2​​=1.  (4) ∣xn​−1∣=∣0.n individual 999⋅⋅⋅9​​−1∣=∣−0.n individual 111⋅⋅⋅1​​∣=10n1​,∀ ε>0, by 了 send ∣xn​−1∣<ε,         only want 10n1​<ε or n>lg ε1​, take N=[lg ε1​], be When n>N when , Just Yes ∣xn​−1∣<ε, namely n→∞lim​0.n individual 999⋅⋅⋅9​​=1.​

6.   if lim ⁡ n → ∞ μ n = a , Prove bright lim ⁡ n → ∞ ∣ μ n ∣ = ∣ a ∣ , and for example say bright ： Such as fruit Count Column { ∣ x n ∣ } Yes extremely limit , but Count Column { x n } not have to Yes extremely limit . \begin{aligned}&6. \ if \lim_{n \rightarrow \infty}\mu_n=a, prove \lim_{n \rightarrow \infty}|\mu_n|=|a|, And give an example ： If the sequence \{|x_n|\} There are limits , But the sequence \{x_n\} There is no limit .&\end{aligned} ​6.  if n→∞lim​μn​=a, Prove bright n→∞lim​∣μn​∣=∣a∣, and for example say bright ： Such as fruit Count Column { ∣xn​∣} Yes extremely limit , but Count Column { xn​} not have to Yes extremely limit .​​

Explain ：

   because by lim ⁡ n → ∞ μ n = a , the With ∀   ε > 0 , ∃   N , When n > N when , Yes ∣ μ n − a ∣ < ε , because ∣ ∣ μ n ∣ − ∣ a ∣ ∣ ≤ ∣ μ n − a ∣ < ε , the With lim ⁡ n → ∞ ∣ μ n ∣ = ∣ a ∣ .    from lim ⁡ n → ∞ ∣ μ n ∣ = ∣ a ∣ , No can PUSH have to lim ⁡ n → ∞ μ n = a , example Such as Count Column { ( − 1 ) n } , lim ⁡ n → ∞ ∣ ( − 1 ) n ∣ = 1 , but yes { ( − 1 ) n } no Yes extremely limit . \begin{aligned} &\ \ because \lim_{n \rightarrow \infty}\mu_n=a, therefore \forall\ \varepsilon \gt 0,\exists\ N, When n \gt N when , Yes |\mu_n-a| \lt \varepsilon, because ||\mu_n|-|a|| \le |\mu_n-a|\lt \varepsilon, therefore \lim_{n \rightarrow \infty}|\mu_n|=|a|.\\\\ &\ \ from \lim_{n \rightarrow \infty}|\mu_n|=|a|, Can't push \lim_{n \rightarrow \infty}\mu_n=a, For example, sequence \{(-1)^n\},\lim_{n \rightarrow \infty}|(-1)^n|=1, however \{(-1)^n\} There are no limits . & \end{aligned} ​   because by n→∞lim​μn​=a, the With ∀ ε>0,∃ N, When n>N when , Yes ∣μn​−a∣<ε, because ∣∣μn​∣−∣a∣∣≤∣μn​−a∣<ε, the With n→∞lim​∣μn​∣=∣a∣.   from n→∞lim​∣μn​∣=∣a∣, No can PUSH have to n→∞lim​μn​=a, example Such as Count Column { (−1)n},n→∞lim​∣(−1)n∣=1, but yes { (−1)n} no Yes extremely limit .​​

7.   set up Count Column { x n } Yes world , also lim ⁡ n → ∞ y n = 0 , Prove bright ： lim ⁡ n → ∞ x n y n = 0 . \begin{aligned}&7. \ Set sequence \{x_n\} bounded , also \lim_{n \rightarrow \infty}y_n=0, prove ：\lim_{n \rightarrow \infty}x_ny_n=0.&\end{aligned} ​7.  set up Count Column { xn​} Yes world , also n→∞lim​yn​=0, Prove bright ：n→∞lim​xn​yn​=0.​​

Explain ：

   because Count Column { x n } Yes world , so ∃   M > 0 , send have to Yes On One cut n Yes ∣ x n ∣ ≤ M . ∀   ε > 0 , from On lim ⁡ n → ∞ y n = 0 , so Yes ε 1 = ε M > 0 ,    ∃   N , When n > N when , Just Yes ∣ y n ∣ < ε 1 = ε M , from and Yes ∣ x n y n − 0 ∣ = ∣ x n ∣ ⋅ ∣ y n ∣ < M   ⋅   ε M = ε , the With lim ⁡ n → ∞ ∣ x n y n ∣ = 0 . \begin{aligned} &\ \ Because of the sequence \{x_n\} bounded , so \exists\ M \gt 0, Make for everything n Yes |x_n| \le M.\forall\ \varepsilon \gt 0, because \lim_{n \rightarrow \infty}y_n=0, So yes \varepsilon_1=\frac{\varepsilon}{M}\gt 0,\\\\&\ \ \exists\ N, When n \gt N when , There is |y_n| \lt \varepsilon_1=\frac{\varepsilon}{M}, Thus there are |x_ny_n-0|=|x_n|\cdot|y_n| \lt M\ \cdot\ \frac{\varepsilon}{M}=\varepsilon, therefore \lim_{n \rightarrow \infty}|x_ny_n|=0.\\\\ & \end{aligned} ​   because Count Column { xn​} Yes world , so ∃ M>0, send have to Yes On One cut n Yes ∣xn​∣≤M.∀ ε>0, from On n→∞lim​yn​=0, so Yes ε1​=Mε​>0,  ∃ N, When n>N when , Just Yes ∣yn​∣<ε1​=Mε​, from and Yes ∣xn​yn​−0∣=∣xn​∣⋅∣yn​∣<M ⋅ Mε​=ε, the With n→∞lim​∣xn​yn​∣=0.​

8.   Yes On Count Column { x n } , if x 2 k − 1 → a   ( k → ∞ ) , x 2 k → a   ( k → ∞ ) , Prove bright ： x n → a   ( n → ∞ ) . \begin{aligned}&8. \ For sequence \{x_n\}, if x_{2k-1}\rightarrow a\ (k \rightarrow \infty),x_{2k}\rightarrow a\ (k\rightarrow \infty), prove ：x_n\rightarrow a\ (n\rightarrow \infty).&\end{aligned} ​8.  Yes On Count Column { xn​}, if x2k−1​→a (k→∞),x2k​→a (k→∞), Prove bright ：xn​→a (n→∞).​​

Explain ：

   because by x 2 k − 1 → a   ( k → ∞ ) , the With ∀   ε > 0 , ∃   k 1 , When k > k 1 when , Yes ∣ x 2 k − 1 − a ∣ < ε ; also because by x 2 k → a   ( k → ∞ ) ,    the With ∀   ε > 0 , ∃   k 2 , When k > k 2 when , Yes ∣ x 2 k − a ∣ < ε . remember K = m a x { k 1 ,   k 2 } , take N = 2 K , be When n > N when ,    if n = 2 k − 1 , 2 k − 1 > 2 K , be k > K + 1 2 > k 1 ⇒ ∣ x n − a ∣ = ∣ x 2 k − 1 − a ∣ < ε , if n = 2 k ,    be k > K ≥ k 2 ⇒ ∣ x 2 k − a ∣ < ε . from and only want n > N , Just Yes ∣ x n − a ∣ < ε , namely lim ⁡ x → ∞ x n = a . \begin{aligned} &\ \ because x_{2k-1}\rightarrow a\ (k\rightarrow \infty), therefore \forall\ \varepsilon \gt 0,\exists\ k_1, When k\gt k_1 when , Yes |x_{2k-1}-a|\lt \varepsilon; Again because x_{2k}\rightarrow a\ (k\rightarrow \infty),\\\\ &\ \ therefore \forall\ \varepsilon \gt 0,\exists\ k_2, When k\gt k_2 when , Yes |x_{2k}-a|\lt \varepsilon. remember K=max\{k_1, \ k_2\}, take N=2K, Then when n\gt N when ,\\\\ &\ \ if n=2k-1,2k-1>2K, be k\gt K+\frac{1}{2}\gt k_1 \Rightarrow |x_n-a|=|x_{2k-1}-a|\lt \varepsilon, if n=2k,\\\\ &\ \ be k\gt K\ge k_2\Rightarrow |x_{2k}-a|\lt \varepsilon. So as long as n\gt N, There is |x_n-a|\lt \varepsilon, namely \lim_{x\rightarrow \infty} x_n=a.\\\\ & \end{aligned} ​   because by x2k−1​→a (k→∞), the With ∀ ε>0,∃ k1​, When k>k1​ when , Yes ∣x2k−1​−a∣<ε; also because by x2k​→a (k→∞),   the With ∀ ε>0,∃ k2​, When k>k2​ when , Yes ∣x2k​−a∣<ε. remember K=max{ k1​, k2​}, take N=2K, be When n>N when ,   if n=2k−1,2k−1>2K, be k>K+21​>k1​⇒∣xn​−a∣=∣x2k−1​−a∣<ε, if n=2k,   be k>K≥k2​⇒∣x2k​−a∣<ε. from and only want n>N, Just Yes ∣xn​−a∣<ε, namely x→∞lim​xn​=a.​

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