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Does limit of absolute value of something always equal absolute value of limit of something? Specifically, can I just say the below equality is true or do I need to prove it? I'm not sure how to prove it. Could you help me?

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    $\begingroup$ If $a_n=(-1)^n$ then $\lim a_n$ doesn't exist (so neither does $|\lim a_n|$, yet $\lim |a_n|=1.$ $\endgroup$
    – coffeemath
    Aug 9, 2013 at 5:59
  • $\begingroup$ @AWertheim: Your proposed counterexample looks suspicious; you've brought a summation outside the absolute value, while the OP only brought out a limit. :) $\endgroup$ Aug 9, 2013 at 11:34
  • $\begingroup$ @AndrewD.Hwang goodness, how did I miss that? Tired eyes! Thanks. :) $\endgroup$ Aug 9, 2013 at 14:58

2 Answers 2

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Hint: $f(x) = |x|$ is continuous.

(Of course, the limits must exist for this to apply. For a counterexample, see the comment above made by coffeemath).

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Assuming both of your limits exist, then the absolute value of the limit will be the limit of the absolute value, by continuity of absolute value. However, it's entirely possible that taking the absolute value will cause a limit to exist where it hadn't before.

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