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If $\{a_n\}_{n=1}^\infty$ is convergent with limit $L$ and $\{b_n\}_{n=1}^\infty$ is convergent with limit $M$, $L \gt M$, and $c_n = \max(a_n,b_n)$, then show that $\lim_{n \to \infty} c_n =L.$

Exists epsilon >0 s.t. for n>N1, |an-L| < epsilon & Exists epsilon >0 s.t. for n>N2, |bn-M| < epsilon

Set N= max(N1,N2)

I felt s given L>M, beyond a point an>bn, i.e an dominates b, s.t. beyond this value of n. |cn-L| < epsilon

This was my line of thinking. Unsure of how to mathematically demonstrate

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    $\begingroup$ How have you tried to solve this? Where are you stuck? Please edit your question to include this information. Problems that don't demonstrate some effort are usually downvoted and closed. $\endgroup$ – Robert Shore Apr 13 at 1:07
  • $\begingroup$ If you (@Harsh_Kumar) edit your post to include the info that Robert mentioned, then I would be happy to help. $\endgroup$ – Matt A Pelto Apr 13 at 2:22
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    $\begingroup$ Since $L > M$, $L-M > 0$. Choose $\epsilon < (L-M)/2$ and see what happens. $\endgroup$ – marty cohen Apr 13 at 2:42
  • $\begingroup$ You have a good idea with the set "$N=\max(N_1,N_2)$ " bit and your intuition at the end seems appropriate but you seem to be stating the wrong definition for a convergent sequence. The definition says "for every $\varepsilon>0$, there exists a positive integer $N$ such that ...". Notice how the definition says every $\varepsilon>0$ and also that we have $L-M>0$. $\endgroup$ – Matt A Pelto Apr 13 at 2:46
  • $\begingroup$ Your argument is correct but you should write it out properly. The sequence $a_n-b_n$ tends to a positive number $L-M$ and hence its terms must be positive after a certain point. Thus $\max(a_n, b_n) =a_n$ after a certain value of $n$. $\endgroup$ – Paramanand Singh Apr 13 at 5:52
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Since $L>M$, we take $\varepsilon=\frac{L-M}2$. Since $\lim_{n \to \infty} a_n=L$ and $\lim_{n \to \infty} b_n=M$, we know there are positive integers $N_1$ and $N_2$ such that $$|a_n-L|<\varepsilon \text{ whenever } n \geq N_1 \\\text{and} \\ |b_n-M|<\varepsilon \text{ whenever } n \geq N_2.$$ Select $N=\max\{N_1, N_2\}$. So if $n \geq N$, then we have $$b_n<\frac{L+M}2<a_n.$$ Hence for every $n\geq N$ we will have that $c_n=a_n$.


Same sort of idea ($\varepsilon$) is helpful in some other introductory analysis proofs; for example in showing that the max of two continuous functions is continuous.

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    $\begingroup$ the comment below is exactly the second part of the question, which i had no idea on how to solve $\endgroup$ – Harsh Kumar Apr 13 at 2:52
  • $\begingroup$ Hint: Either the two functions are equal at $a$ (an arbitrary point in $\mathbb R$ at which you show the max of two continuous functions is continuous) or else there is $\delta>0$ (perhaps the min of 2 deltas) so that one of the two functions remains the max for all $x$ such that $|x-a|<\delta$. There is also a simple formula (somewhat related to this choice of $\varepsilon$) for the max of two functions which could be used. Two birds, one stone :) @HarshKumar $\endgroup$ – Matt A Pelto Apr 13 at 3:30
  • $\begingroup$ ah, the sandwich theorem? $\endgroup$ – Harsh Kumar Apr 13 at 11:53
  • $\begingroup$ @HarshKumar No. Let $f,g:\mathbb R \to \mathbb R$ be continuous functions, and define $h(x):=\max\{f(x),g(x)\}$. If $f(a)=g(a)$, then for any $\varepsilon>0$ how can we find $\delta>0$ so that $|h(x)-h(a)|=|h(x)-f(a)|=|h(x)-g(a)|<\varepsilon$? If $f(a) \neq g(a)$, then either $f(a)>g(a)$ or $g(a)>f(a)$. First pick an $\varepsilon'>0$ (similar but not identical to the $\varepsilon$ used in the argument above) and find $\delta'>0$ so that the function which is greater at $a$ remains greater for all $x$ s.t. $|x-a|<\delta'$. Then simply pass the continuity of $f$ or $g$ on to $h$. $\endgroup$ – Matt A Pelto Apr 15 at 20:31
  • $\begingroup$ Or if you want to use a formula for the max of two sequences (as tonychow suggests)/the max of two functions, then you will also need to use theorems saying limit of sum equals sum of limits/sum of continuous functions is continuous. $\endgroup$ – Matt A Pelto Apr 15 at 20:40
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It would be very helpful to know that $$\forall a, b \in \mathbb{R}, \, \max\{a,b\}=\frac{|a+b|+|a-b|}{2}.$$ Then $$c_n = \frac{|a_n+b_n|+|a_n-b_n|}{2} \, \forall n \in \mathbb{N}.$$

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