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Let $(u_n)$ and $(v_n)$ be two real sequences with limits L and M respectively. If $x_n$= max$({u_n,v_n})$ and $y_n$=min$(u_n,v_n)$, prove that the sequence $x_n$ and $y_n$ converges to max$(L,M)$ and min$(L,M)$ respectively.

My attempt: It is given that $u_n$ and $v_n$ converges to L and M respectively. So $u_n$+$v_n$ =$L+M$.

Now, taking limits,

$x_n$=max$(u_n,v_n)$=$1/2{(a+b+|a-b|)}$=$1/2{(L+M+|L-M|)}$=max$(L,M)$

Similarly,

$x_n$=min$(u_n,v_n)$=$1/2{(a+b-|a-b|)}$=$1/2{(L+M-|L-M|)}$=min$(L,M)$

Is this correct??

Natasha J
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  • Yes. ${[{}{[}[}[{[}[}}$ – hamam_Abdallah Apr 19 '21 at 02:23
  • You introduce $a,b$ into an equation without defining them. ANd the sentence you wrote is unreadable. "$x_n = \max(u_n, v_n)$" are you defining $x_n$? $\max(u_n,v_n) = \frac 12(a+b+|a-b|)$. What is $a$ and $b$. Are they supposed to be $u_n, v_n$. Then $\frac 12(a+b +|a-b|)= \frac 12(L+M+|L-M|)$. Have $a$ and $b$ switched values. Were they $u_n,v_n$ in the beginning and changed* to $L,M$ mid sentence? That's just not how you make statements. – fleablood Apr 19 '21 at 03:32
  • I think what you are trying to write is: Let $x_n = \max(u_n, v_n)$ then $x_n = \frac 12(u_n + v_n + |u_n-v_n|)$. SO $\lim_{n\to \infty} x_n =\lim_{n\to \infty} \frac 12(u_n+v_n+|u_n+v_n|)= \frac 12(L + M + |L-M|) = \max (L,M)$. But what you actually wrote: "xn=max(un,vn)=1/2(a+b+|a−b|)=1/2(L+M+|L−M|)=max(L,M)" is actually just gibberish. – fleablood Apr 19 '21 at 03:35
  • a and b are reals. It is a rough sketch of the proof. – Natasha J Apr 19 '21 at 03:37
  • @Fleablood. Will keep in mind. And yes that's what I meant to write. But I just forgot to type it that way. Thanks! – Natasha J Apr 19 '21 at 03:44

1 Answers1

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The idea is correct. Few steps are not.

For example $u_n + v_n \neq L + M$ but $u_n + v_n \rightarrow L + M$.

Similarly $x_n = \frac{u_n + v_n + |u_n - v_n|}{2} \rightarrow \frac{L + M + |L - M|}{2} = \max(L,M)$

Anon
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