My question concerns the following proof:

> 
> Prove that $$\lim\_{x\to -5} \vert x - 5 \vert=10.$$
> 
> 
> 

We are given $\varepsilon > 0$. Then we have
$$\begin{align\*}
\vert \vert x - 5 \vert - 10 \vert &< \varepsilon \\
\vert -(x-5)-10 \vert &< \varepsilon \qquad \text{($x-5 < 0$)} \\
\vert (-x-5) \vert &< \varepsilon \\
\vert (x-(-5)) \vert &< \varepsilon
\end{align\*}$$

So, let $\delta = \varepsilon$.

So for $\vert x - (-5) \vert < \delta = \varepsilon$, you have

$$\begin{align\*}
\vert-(x+5) \vert &< \varepsilon \\
\vert -(x-5)-10 \vert &< \varepsilon \\
\vert \vert x - 5 \vert - 10 \vert &< \varepsilon \qquad \text{(because $x-5 < 0$)}.
\end{align\*}$$

Now surely I'm missing something very obvious here, but why are we allowed to make the claim that $x-5 < 0$, or is that a restriction we are imposing? It seems if we wanted to make this claim we would need to include in the proof something like: assume $x < 5$?