Absolute value of limit

Question

Expain why the following is true:

If $$\lim_ {x\to a}\ f(x) = k$$

then

$$\lim_ {x\to a}\ |f|(x) = |k|$$

Answer 1

Lemma: For any $a,b\in\Bbb R,$ we have $\bigl||a|-|b|\bigr|\le|a-b|.$

Proof: On the one hand, we have by triangle inequality that $$|a|-|b|=|(a-b)+b|-|b|\le|a-b|+|b|-|b|=|a-b|.$$ On the other hand, we likewise have $|b|-|a|\le|b-a|.$ Since $|b|-|a|=-(|a|-|b|)$ and $|b-a|=|a-b|,$ then we have $\pm(|a|-|b|)\le|a-b|,$ and so $\bigl||a|-|b|\bigr|\le|a-b|.$ $\Box$

Hint: See if you can apply the Lemma above, together with the $\epsilon$-$\delta$ definition of function limits. What do you know? What are you trying to show? How can the Lemma bridge the gap?