Jay Mckay

2022-01-24

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

Brenton Pennington

Expert

I solved only the first one
Note that $|\frac{1+n}{1-2n}+\frac{1}{n}|$
$=|\frac{2+2n+1-2n}{2\left(1-2n\right)}|$
$=|\frac{3}{2}×\frac{1}{1-2n}|$
For every $\xi =0$ given we can choose n such that
$|\frac{3}{2}×\frac{1}{1-2}|=\frac{3}{2\left(2N-1\right)}<\xi$
$⇒2N-1>\frac{3}{2\xi }$
$⇒N>\left(1+\frac{3}{2\xi }\right)\frac{1}{2}$
$⇒N>\left(\frac{1}{2}+\frac{3}{4\xi }\right)$
Hence, $\mathrm{\forall }n\ge N$
$|\frac{1+2n}{1-2n}-\left(-\frac{1}{2}\right)|<\xi$
Thus, $\underset{n\to \mathrm{\infty }}{lim}|\frac{1+2n}{1-2n}|=-\frac{1}{2}$
(From def of limit)

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