Limit of Absolute Value

Theorem

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

Then:

$\left\vert{x - \xi}\right\vert \to 0$ as $x \to \xi$

where $\left\vert{x - \xi}\right\vert$ 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 $\left\vert{f \left({x}\right) - 0}\right\vert < \epsilon$ provided that $0 < \left\vert{x - \xi}\right\vert < \delta$, where $f \left({x}\right) = \left\vert{x - \xi}\right\vert$.

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

	\(\displaystyle \)	\(\displaystyle \left\vert{x - \xi}\right\vert - 0\)	\(=\)	\(\displaystyle \left\vert{x - \xi}\right\vert\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(<\)	\(\displaystyle \delta\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \epsilon\)	\(\displaystyle \)

Hence the result.

$\blacksquare$

Sources

• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 8.15 \ (4)$