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|1+n1−2n+12| =|2+2n+1−2n2(1−2n)| =|32⋅11−2n| For every ϵ>0 given we can choose N that |32⋅11−2=32(2N−1)|<ϵ
2N−1>32ϵ N>(1+32ϵ)12 N>(12+34ϵ) Hence ∀n≥N |1+2n1−2n−(−12)|<ϵ Hence limn→∞|1+2n1−2n|=−12
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