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Suppose $\lim_{x\to a} f(x) = L$. Prove that $\lim_{x\to a} \lvert f(x)\rvert$ exists.

The $\epsilon-\delta$ definition of limits may be helpful. My logic just cannot get there without feeling incomplete. I tried proof by contradiciton and including the squeeze theorem. I really just can't get a grasp for how to think of proofs. Any help is appreciated, thanks!

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If we can prove that $||f(x)| - |L|| \leq |f(x) - L|$ for all $x$, and if we use the assumption, then for every $\varepsilon > 0$ there is some $\delta > 0$ such that $0 < |x-a| < \delta$ implies $||f(x)| - |L|| \leq |f(x)-L| < \varepsilon$; then $|f(x)| \to |L|$ as $x \to a$.

The target inequality is actually not deep, which follows directly from triangle inequality. Let $a,b$ be real numbers. If we use the triangle inequality, then $|a| = |a-b+b| \leq |a-b| + |b|$ and $|b| = |b-a+a| \leq |a-b| + |a|$. So $|a| - |b| \leq |a-b|$ and $|b| - |a| = -(|a| - |b|) \leq |a-b|$; so $||a| - |b|| \leq |a-b|$.

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  • $\begingroup$ Ahh, it was the fact that the inequality is therefore strictly less than epsilon that I wasn't grasping. Also the explanation of the inequality is much appreciated. Thank you very much. $\endgroup$ – LocoSap Feb 23 '18 at 4:11
  • $\begingroup$ Hey glad you overcame it! :) $\endgroup$ – Benicio Feb 23 '18 at 4:37
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$\big||f(x)|-|L|\big|\leq|f(x)-L|$, can you finish from here?

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  • $\begingroup$ I understand how you have the right hand side of the inequality because of what's given. So sorry if this is very elementary but how do you say it is greater than or equal to the left hand side. $\endgroup$ – LocoSap Feb 23 '18 at 3:21
  • $\begingroup$ $|a+b|\leq|a|+|b|$ and realise $a=f(x)-L$ and $b=L$, one gets $|f(x)|-|L|\leq|f(x)-L|$ and realise $a=L-f(x)$ and $b=f(x)$ to get $|L|-|f(x)|\leq|f(x)-L|$. $\endgroup$ – user284331 Feb 23 '18 at 3:23
  • $\begingroup$ Math is great, thank you. $\endgroup$ – LocoSap Feb 23 '18 at 3:30

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