Exercise · ZW-UNIQUENESS-OF-LIMITS-009 · ★★★ Advanced

If \((-1)^n\to L\), with \(\varepsilon=\tfrac12\) the terms \(1\) and \(-1\) (which appear for large \(n\)) satisfy \(|1-L|<\tfrac12\) and \(|-1-L|<\tfrac12\). What contradiction follows?

Choose one option.

Show answer

C. \(2=|1-(-1)|<1\)

Explanation

By the triangle inequality, \(2=|1-(-1)|\le|1-L|+|L-(-1)|<1\), which is false.

Review the concept: Uniqueness of limits

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.

\[a_n\to L,\ a_n\to M\ \Rightarrow\ L=M\]

More practice problems: uniqueness of limits

ZW-UNIQUENESS-OF-LIMITS-001 If a_n→2 and also a_n→ M, what is M?
ZW-UNIQUENESS-OF-LIMITS-002 If lim _n→∞(4n+1)/(n)=L, what is L?
ZW-UNIQUENESS-OF-LIMITS-003 To rule out a_n→3 and a_n→7 holding together, one takes ε=(|L-M|)/(2). What is it for L=3, M=7?
ZW-UNIQUENESS-OF-LIMITS-004 In the proof, N_1 works for convergence to L and N_2 for convergence to M. If N_1=12 and N_2=30, wh…
ZW-UNIQUENESS-OF-LIMITS-005 Which statement is true?
ZW-UNIQUENESS-OF-LIMITS-006 If a_n→ L and a_n→ M, then a_n-a_n=0 converges to L-M and also to 0. What follows?
ZW-UNIQUENESS-OF-LIMITS-007 With |L-M|=0.6 and ε=0.3, for n≥ N the triangle inequality gives |L-M|≤|L-a_n|+|a_n-M|< ...
ZW-UNIQUENESS-OF-LIMITS-008 An alternative proof shows |L-M|<ε for every ε>0. Which fact lets us conclude L=M?