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Question #a25bb

1 Answer

Feb 8, 2017

#lim_(x->0^+) 2/x-2/abs(x) = 0#

Explanation:

#x>0 => abs x = x => 2/x-2/abs(x) = 0#

so:

#lim_(x->0^+) 2/x-2/abs(x) = 0#

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