Prove using the definition of the limit that limn→∞1+n1−2n=−12 Prove using the definition of the limit that limn→∞n2+12n−1=∞

I solved only the first one Note that |1+n1−2n+1n| =|2+2n+1−2n2(1−2n)| =|32×11−2n| For every ξ=0 given we can choose n such that |32×11−2|=32(2N−1)&lt;ξ ⇒2N−1&gt;32ξ ⇒N&gt;(1+32ξ)12 ⇒N&gt;(12+34ξ) Hence, ∀n≥N |1+2n1−2n−(−12)|&lt;ξ Thus, limn→∞|1+2n1−2n|=−12 (From def of limit)

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Jay Mckay

Answered question

2022-01-24

Prove using the definition of the limit that

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

Prove using the definition of the limit that

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

Answer & Explanation

Brenton Pennington

Beginner2022-01-25Added 15 answers

I solved only the first one
Note that

| \frac{1 + n}{1 - 2 n} + \frac{1}{n} |

= | \frac{2 + 2 n + 1 - 2 n}{2 (1 - 2 n)} |

= | \frac{3}{2} \times \frac{1}{1 - 2 n} |

For every

ξ = 0

given we can choose n such that

| \frac{3}{2} \times \frac{1}{1 - 2} | = \frac{3}{2 (2 N - 1)} < ξ

\Rightarrow 2 N - 1 > \frac{3}{2 ξ}

\Rightarrow N > (1 + \frac{3}{2 ξ}) \frac{1}{2}

\Rightarrow N > (\frac{1}{2} + \frac{3}{4 ξ})

Hence,

\forall n \geq N

| \frac{1 + 2 n}{1 - 2 n} - (- \frac{1}{2}) | < ξ

Thus,

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

(From def of limit)

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