Exercise · ZW-UNIQUENESS-OF-LIMITS-007 · ★★☆ Intermediate
With \(|L-M|=0.6\) and \(\varepsilon=0.3\), for \(n\ge N\) the triangle inequality gives \(|L-M|\le|L-a_n|+|a_n-M|<\) ...
Choose one option.
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- D. \(0.6\)
Exercise · ZW-UNIQUENESS-OF-LIMITS-007 · ★★☆ Intermediate
Choose one option.
Each summand is less than \(0.3\), so the sum is less than \(0.6\). We get \(0.6<0.6\), absurd.
Uniqueness of limits says that a sequence \((a_n)\) cannot converge to two different numbers \(L\) and \(M\); that is why it makes sense to speak of “the” limit.