Prove using the definition of a limit that $\displaystyle\lim_{x \rightarrow 1}|x|=1$ Prove $$\displaystyle\lim_{x \rightarrow 1}|x|=1$$ using the $\varepsilon$-$\delta$ definition of a limit.
I know that we need to choose an $\delta=\min\{\frac{1}{2},\epsilon\}$ however, I don't understand why $\frac{1}{2}$ needs to be chosen and why we can't just choose $\delta=\varepsilon$. Is it possible to also pick $\delta \leq 1$ instead of $\frac{1}{2}$?
 A: Hint: use the relation
$$\big||x|-1\big|\le|x-1|.$$
A: Proposition
Let $|\cdot|:\mathbb{R}\to\mathbb{R}_{\geq 0}$ be the modulus function.
Then for every pair of real numbers $x,y\in\mathbb{R}$, one has that
\begin{align*}
||x| - |y|| \leq |x - y|
\end{align*}
Proof
According to the triangle inequality, we can say that
\begin{align*}
|x| = |x - y + y| \leq |x - y| + |y| \Rightarrow |x| - |y| \leq |x - y|
\end{align*}
Similarly, $|y| - |x| \leq |x - y|$.
Hence $|x - y| \geq \max\{|x| - |y|, |y| - |x|\} = ||x| - |y||$.
Solution
Based on the previous proposition, we can prove a more general result.
That is to say: the function $|\cdot|$ is continuous for every $a\in\mathbb{R}$.
Indeed, let $|x - a| < \delta$. Then we have that
\begin{align*}
|x - a| < \delta \Rightarrow ||x| - |a|| \leq |x - a| < \delta := \varepsilon
\end{align*}
Consequently, we conclude that for every $\varepsilon > 0$, there corresponds $\delta_{\varepsilon} = \varepsilon$ such that
\begin{align*}
|x - a| < \delta_{\varepsilon} \Rightarrow ||x| - |a|| < \varepsilon
\end{align*}
and we are done.
Hopefully this helps!
