Question on delta-epsilon proof nested absolute value

Question

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 $x < 5$?

Answer 1

You're right, that is a restriction we are imposing. We want $$|x-(-5)| < \delta \Rightarrow ||x-5| - 10| < \varepsilon.$$ Note that if you find some $\delta$ that works, than any $\delta'\leq\delta$ will also do. What we could do is take $\delta = \min\{\varepsilon, 10\}$. Then,

• $|x + 5| < \delta \leq 10$, so $x-5<0$, and
• $|x + 5| < \delta \leq \varepsilon$,

which are the two restrictions on $x$ you needed. Now the rest of your argument follows.

Answer 2

Let $|x+5|<1$ and for given $\varepsilon > 0$, there is $\delta>0$ such that if $|x+5|<\delta$ then $$||x - 5|- 10 = |\sqrt{(x-5)^2}-10|=\dfrac{|x+5|~|x-15|}{\sqrt{(x-5)^2}+10}< \dfrac{21|x+5|}{10}<\dfrac{21}{10}\delta$$ take $\delta = \min\{1,\dfrac{10}{21}\varepsilon \}$.