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If $\lim _{n\to \infty} a_n = A$, $\lim _{n\to \infty} b_n = B$ and $A<B$, prove $\lim _{n\to \infty}\max\{a_n,b_n\} = B$

Side question: Does the $A<B$ imply $a_n < b_n$?

Since we know $a_n$ converges to $A$, doesn't the problem kind of suggest $\max{a_n,b_n} = b_n$ (as we also know $A\not = B$, so if the max were $a_n$ the limit would have to be A, but then the problem is futile)?

[Not homework, just practice for learnings sake :-)]

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  • $\begingroup$ No. $a_n<b_n$ may not hold for all $n$, but it will hold "eventually", which is what the problem wants you to check. $\endgroup$ – Prahlad Vaidyanathan Oct 1 '13 at 7:18
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Hint: For sufficiently large $n$ we have $a_n < A + \dfrac{B-A}{2} = \dfrac{A+B}{2} = B - \dfrac{B-A}{2}< b_n$.

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The function $(x,y) \mapsto \max(x,y)$ is continuous. Hence $\lim_n \max(a_n,b_n) = \max(A,B) = B$.

If $A<B$, then eventually you will have $a_n < b_n$. This follows because $\lim_n (b_n-a_n) = B-A > 0$. Hence for some large $N$, you have $b_n-a_n > 0$ for $n \ge N$.

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  • $\begingroup$ Can't use anything involving functions or continuity. $\endgroup$ – JohanLiebert Oct 1 '13 at 8:21
  • $\begingroup$ Well, you don't need to. The second conclusion implies the first... $\endgroup$ – copper.hat Oct 1 '13 at 8:28
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Hint: Let $\epsilon=(B-A)/2$. There is an $N_1$ such that if $n\gt N_1$ then $|a_n-A|\lt \epsilon$. There is an $N_2$ such that if $n\gt N_2$ then $|b_n-B|\lt \epsilon$.

Let $N=\max(N_1,N_2)$. If $n\gt N$, then by the Triangle Inequality, $b_n\gt a_n$. Thus if $n\gt N$, then $\max(a_n,b_n)=b_n$, and the result follows.

Remark: In answer to your question, it is quite possible for $a_n$ to be bigger than $b_n$ for some $n$. But as we have shown, after a while this cannot be the case.

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