$\lim_{x \to 0 }|x| = 0$

Question

Given $\epsilon > 0$ we want to $\delta > 0$ such as $|x - 0| < \delta \Rightarrow |f(x) - f(0)| < \epsilon $.

How \begin{align} |x| = \begin{cases} x, \text{ if } x > 0\\ -x, \text{ if } x < 0 \end{cases} \end{align}

From desigualities we have

if $x > 0$ : $|x - 0| < \epsilon$

and if $x < 0 : $

$|-x - 0 | < \epsilon \Rightarrow |-1||x - 0| < \epsilon \Rightarrow |x - 0| < \epsilon$

Because this, we take it $\delta = \epsilon$.

So, $|x - 0|<\delta \Rightarrow |x - 0|<\epsilon$.

Therefore, for definition $\lim_{x \to 0 }|x| = 0$

It's right?

With the hints I rewrited like this: Given $\epsilon > 0$ we want to $\delta > 0$ such as $|x - 0| < \delta > \Rightarrow ||x| - 0| < \epsilon $.

As $||x| - 0| = |x|$ we have that $|x| < \epsilon$ so we take it $\delta = \epsilon$.
So, $|x - 0|<\delta \Rightarrow ||x| - 0|<\epsilon$.

Therefore, for definition $\lim_{x \to 0 }|x| = 0$

Answer 1

You are looking at the wrong definition. What you need should be the definition of $\lim_{x\to 0}f(x)=0$, not $\lim_{x\to 0}f(x)=f(0)$. The latter is the continuity of $f$ at $x=0$. The function $f(x)=|x|$ is indeed continuous at $0$ though.

The estimate you want should be $$ ||x|-0|<\epsilon. $$

You don't need to discuss two cases. Simply observe that $$ ||x|-0|=|x| $$

Answer 2

Let us use a standart definition of a limit of a function: $\forall \varepsilon>0\; \exists \delta:\forall x\in\stackrel{\circ}{U}_{\delta}(x_{0})\Rightarrow |f(x) - L|<\varepsilon$. For given function $f(x) = |x|$ we have $x_{0}=0$. Let us take $\delta = \varepsilon/2$. Then indeed $|x - 0 | \leq |\delta| = \varepsilon/2 < \varepsilon$. Because of arbitrary choice of $\varepsilon$, we have that $\lim\limits_{x \to 0} |x| = 0$.