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Please help me find limit of the sequence $$ a_n=\frac{n+1}{n+2} $$ and find index $n_0\in\mathbb{N}$ for $\varepsilon=0,01$.

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1 Answer 1

up vote 3 down vote accepted

$\lim a_n=\lim\frac{n+1}{n+2}=\lim\frac{n(1+\frac{1}{n})}{n(1+\frac{2}{n})}=\frac{1}{1}=1$, because $\lim\frac{1}{n}=0$ and $\lim\frac{2}{n}=0$ if $n\rightarrow\infty$.

Now find $n_0$

$|a_n-a|<\epsilon$

$|\frac{n+1}{n+2}-1|<0,01$

$|\frac{n+1-(n+2)}{n+2}|<\frac{1}{100}$

$|\frac{n+1-n-2}{n+2}|<\frac{1}{100}$

$|\frac{-1}{n+2}|<\frac{1}{100}$

Implement the formula: $|\frac{a}{b}|=\frac{|a|}{|b|}$ we have

$\frac{|-1|}{|n+2|}<\frac{1}{100}$

Because $|-a|=a$, for all $a\in R$, and because $n\in N$ we have:

$\frac{1}{n+2}<\frac{1}{100}$

$100<n+2$

$98<n$

$n>98$ $\Rightarrow$ $n_0=98$

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