Limit of $\vert x-2\vert$ as $x \to -2$

Question

I believe that it equals $-4$. In the $\varepsilon$-$\delta$ definition, we can set $\delta=\varepsilon$ and I find this satisfies the definition. The problem is I can't seem to prove, based on this, that $$0 \leq \vert x-(-2)\vert \leq\delta =\varepsilon \implies \big\vert\vert x-2\vert-(-4)\big\vert < \varepsilon.$$ It seems such a basic question, but I keep hitting dead ends. A series of equations or even a hint would help.

Answer 1

Hint: Since $|x - 2|$ is continuous, we have: $$\lim_{x \to -2}|x - 2| = |-2 - 2| = |-4| = 4.$$

By definition, given $\epsilon > 0$, you want to find $\delta > 0$ such that: $$|x - (-2)| < \delta \implies ||x - 2| - 4| < \epsilon.$$ Using that $||a|-|b|| < |a-b|$ implies $||x - 2| - 4| <|x-6|$ and rewriting in simples terms, you want to find $\delta > 0$ such that: $$|x+2| < \delta \implies |x - 6| < \epsilon$$ Can you go on?

Answer 2

Let $\epsilon>0$ be given. We want to find a $\delta>0$ such that if $0<|x-(-2)|<\delta$, then $\lvert|x-2|-4\rvert<\epsilon$.

To do this, you can use $\lvert|x-2|-4\rvert=\lvert|x-2|-|-4|\rvert\le\lvert(x-2-(-4)\rvert=\lvert x+2\rvert$.