Prove with $\epsilon$ and $\delta$ that $\lim_ {x\to 0} |x|=0$

Question

Prove with $\epsilon$ and $\delta$ that $\lim_{x\to 0} |x|=0$. I just have to find the $\delta$ that works with $\left||x| -0\right|< \epsilon$. Following the absolute value definition I found that when $x>0$, then $\epsilon=\delta$. But I´m stuck when $x<0$, what should I do in that case?.

Answer 1

Put aside the definition of absolute value for a moment and focus on the definition of a limit. Can you convince yourself that what you want is to find a $\delta$ such that if $|x|<\delta$, then $|x|<\epsilon$?

Answer 2

The definition of $\lim_{x\to a} f(x) = L$ is that $$\forall \epsilon \gt 0\ \exists \delta \gt 0 \text{ s.t. } 0\lt|x-a|\lt \delta \implies |f(x)-L|\lt \epsilon$$

In this case

$$||x|-0|=|x|=|x-0|$$ So just choose $\delta = \epsilon$, then $|x-0|\lt \delta \implies ||x|-0|\lt \epsilon$ and you're done.