Prove using the definition on the limit that limn→∞1+n1−2n=−12

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

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High School
Calculus and Analysis
Calculus 1
Limits and continuity

slaggingV

Answered question

2021-08-19

Prove using the definition on the limit that

lim_{n \to \infty} \frac{1 + n}{1 - 2 n} = \frac{- 1}{2}

Answer & Explanation

wheezym

Skilled2021-08-20Added 103 answers

$| \frac{1 + n}{1 - 2 n} + \frac{1}{2} |$
$= | \frac{2 + 2 n + 1 - 2 n}{2 (1 - 2 n)} |$
$= | \frac{3}{2} \cdot \frac{1}{1 - 2 n} |$
For every $ϵ > 0$ given we can choose N that
$| \frac{3}{2} \cdot \frac{1}{1 - 2} = \frac{3}{2 (2 N - 1)} | < ϵ$

$2 N - 1 > \frac{3}{2 ϵ}$
$N > (1 + \frac{3}{2 ϵ}) \frac{1}{2}$
$N > (\frac{1}{2} + \frac{3}{4 ϵ})$
Hence $\forall n \geq N$
$| \frac{1 + 2 n}{1 - 2 n} - (- \frac{1}{2}) | < ϵ$
Hence
$lim_{n \to \infty} | \frac{1 + 2 n}{1 - 2 n} | = - \frac{1}{2}$

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