11D (source code)
= 11D
{scope}
{title2=Analysis I}
The two preliminary <sequence> limit laws follow directly from the <triangle inequality>. For the sum, choose indices so that $|x_n-x|<\varepsilon/2$ and $|y_n-y|<\varepsilon/2$ beyond their respective cutoffs. Beyond the larger cutoff,
$$
|(x_n+y_n)-(x+y)|
\le|x_n-x|+|y_n-y|<\varepsilon.
$$
Thus
$$
\boxed{x_n+y_n\longrightarrow x+y.}
$$
For the reciprocal, convergence to nonzero $x$ gives $|x_n-x|<|x|/2$ eventually. The reverse <triangle inequality> then gives $|x_n|\ge|x|/2$, and consequently
$$
\left|\frac1{x_n}-\frac1x\right|
=\frac{|x_n-x|}{|x_n|\,|x|}
\le\frac{2|x_n-x|}{|x|^2}.
$$
Choosing the eventual error smaller than both $|x|/2$ and $\varepsilon|x|^2/2$ proves
$$
\boxed{\frac1{x_n}\longrightarrow\frac1x.}
$$
\b[The nonzero limit gives the denominator a positive lower bound; nonzero individual terms alone would not do so.] The assumption $x_n\ne0$ ensures the reciprocal sequence is defined even before this eventual bound applies.
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