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Posted

lim x-->0 [(1/x)+x]

Can I split it into limx-->(1/x) + limx--->0 (x)?

A principle said it can if both of the functions exist.

But does the former exist?

Posted

Indeed. If you know that [imath](a_n) \to a, (b_n) \to b[/imath] then it's fairly easy to prove that [imath](a_n + b_n) \to a+b[/imath]. Doesn't work if one of the limits is undefined though.

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