Question
Express $|x+5|<\frac{1}{2}$ in neighborhood form.

Answer

Comparing $|x+5|< \frac{1}{2}$ with $|x - a|<δ,$ we get
$a = -5$ and $δ = \frac{1}{2}$
Neighbourhood form $: N(a; δ)$
Putting $a = -5$ and $= δ =\frac{1}{2}$
$|x+5|<\frac{1}{2}=N\left(-5 ; \frac{1}{2}\right)$

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