Let m and n be two positive integers greater than 1. If limα→0(ecos(αn)-eαm)=-(e2), then the value of mn is [2015]
(2)
limα→0ecosαn-eαm=-e2
⇒limα→0e[ecosαn-1-1]cosαn-1×cosαn-1αm=-e2
⇒elimα→0-2sin2αn2(αn2)2×(αn2)2αm=-e2
⇒-e2α2n-m=-e2
⇒mn=2