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Expain why the following is true:

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

then

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

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

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  • 1
    $\begingroup$ What is the definition of $|f|$? What do you know about continuity. $\endgroup$ – Sigur Nov 11 '13 at 23:50
  • $\begingroup$ I only know that the limit is finite. $\endgroup$ –  ShadowHero Nov 11 '13 at 23:52
  • $\begingroup$ |f| is the absolute value of f $\endgroup$ –  ShadowHero Nov 11 '13 at 23:53
  • $\begingroup$ Do you think that it is a continuous function? If yes, what can you say about the limit? $\endgroup$ – Sigur Nov 11 '13 at 23:57
  • $\begingroup$ So try to learn about limits of continuous functions. $\endgroup$ – Sigur Nov 12 '13 at 0:00
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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?

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  • $\begingroup$ I really appreciate your help, I will try my best. $\endgroup$ –  ShadowHero Nov 12 '13 at 1:09
  • $\begingroup$ If you get stuck, or aren't sure if you've done something right, feel free to ask me. (I may be eating dinner, but I'll reply when I can.) $\endgroup$ – Cameron Buie Nov 12 '13 at 1:13
  • $\begingroup$ Thanks, I don't want to bug people a lot with my problems and the fact I diden't study enough to solve them on my own... $\endgroup$ –  ShadowHero Nov 12 '13 at 1:20

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