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If $\lim_{n \to \infty} a_{n} = a$ and $\lim_{n \to \infty} b_{n} = b$, how can we show that $\lim_{n \to \infty} \min\{a_{n},b_{n}\} = \min\{a,b\}$?

I say $\min\{a_{n},b_{n}\} $ has two cases: $a_{n}$ and $b_{n}$. So (1) $\lim_{n \to \infty} a_{n} = a$ and (2) $\lim_{n \to \infty} b_{n} = b$. Now I don't know how to imply the $\min\{a,b\}$.

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    ^ The above should be its own answer and not just$a$comment.2012-12-19

2 Answers 2

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Hint: $\max\left\{a_n,b_n\right\}=\frac{1}{2}(a_n+b_n)+\frac{1}{2}\left|a_n-b_n\right|$ and $\min\left\{a_n,b_n\right\}=\frac{1}{2}(a_n+b_n)-\frac{1}{2}\left|a_n-b_n\right|$

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    great @Nameless, great help as usual! :D2012-12-16
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If $a_n\to a$ and $b_n\to b$, then $\min (a_n,b_n)=\frac{1}{2}(a_n+b_n-|a_n-b_n|)\to \frac{1}{2}(a+b-|a-b|)=\min (a,b).$

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    thanks a lot. i am becoming every hour better in maths by walking around here.. great, short and clear!2012-12-16