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Mathematics 7 Online
OpenStudy (anonymous):

when lim x intend to 0 then proove that logcos x/sin2x=1/2

OpenStudy (raden):

because can be 0/0, use L'Hopital rule

OpenStudy (raden):

looks it doesnt be 1/2 :)

OpenStudy (agent0smith):

is it log(cosx/sin2x) or log(cosx)/sin2x? Neither appear to approach a limit of 1/2 just from the graphs...

OpenStudy (raden):

i think the 2nd one, maybe

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