# Absolute value of limit [closed]

Expain why the following is true:

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

then

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

## closed as off-topic by Antonio Vargas, Nick Peterson, Bruno Joyal, egreg, Lord_FarinNov 12 '13 at 21:59

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• What is the definition of $|f|$? What do you know about continuity. – Sigur Nov 11 '13 at 23:50
• I only know that the limit is finite. –  ShadowHero Nov 11 '13 at 23:52
• |f| is the absolute value of f –  ShadowHero Nov 11 '13 at 23:53
• Do you think that it is a continuous function? If yes, what can you say about the limit? – Sigur Nov 11 '13 at 23:57
• So try to learn about limits of continuous functions. – Sigur Nov 12 '13 at 0:00

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?