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.

$\blacksquare$

Sources

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