Answer:
Option A
Explanation:
Given limα→0(ecos(αn)−eαm)=−(e2)
⇒ limα→0e[ecos(αn)−1−1].cos(αn)−1cos(αn)−1αm=−e2
⇒ limα→0e{ecos(αn)−1−1cos(αn)−1} .
limα→0−2sin2αn2αm
=-e/2
⇒ e×1×(−2)limα→0sin2(αn2)α2n4.α2n4αm=−e2
⇒ e×1×(−2)×1limα→0α2n−m4=−e2
For this to exist.
2n-m=0 ⇒ mn=2