# Limit of Absolute Value

## Theorem

Let $x, \xi \in \R$ be real numbers.

Then:

$\size {x - \xi} \to 0$ as $x \to \xi$

where $\size {x - \xi}$ denotes the Absolute Value.

## Proof

Let $\epsilon > 0$.

Let $\delta = \epsilon$.

From the definition of a limit of a function, we need to show that $\size {\map f x - 0} < \epsilon$ provided that $0 < \size {x - \xi} < \delta$, where $\map f x = \size {x - \xi}$.

Thus, provided $0 < \size {x - \xi} < \delta$, we have:

 $\ds \size {x - \xi} - 0$ $=$ $\ds \size {x - \xi}$ $\ds$ $<$ $\ds \delta$ $\ds$ $=$ $\ds \epsilon$

Hence the result.

$\blacksquare$