slaggingV

2021-08-19

Prove using the definition on the limit that
$\underset{n\to \mathrm{\infty }}{lim}\frac{1+n}{1-2n}=\frac{-1}{2}$

wheezym

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

$2N-1>\frac{3}{2ϵ}$
$N>\left(1+\frac{3}{2ϵ}\right)\frac{1}{2}$
$N>\left(\frac{1}{2}+\frac{3}{4ϵ}\right)$
Hence $\mathrm{\forall }n\ge N$
$|\frac{1+2n}{1-2n}-\left(-\frac{1}{2}\right)|<ϵ$
Hence
$\underset{n\to \mathrm{\infty }}{lim}|\frac{1+2n}{1-2n}|=-\frac{1}{2}$

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