\max \{a, b\}=\frac{(a+b)+|a-b|}{2} and \min \{a, b\}=\frac{(a+b)-|a-b|}{2} 3. Show that if an → L and b, → M then min(an, bn) → min{L, M} and max(an, bn)→max{L, M }.(Recall that min(an, bn) is the sequence n H min{an, bn} and max(a,, bn) is the sequence n → max{an, bn}.)
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