The value of the limit limx→0{11/sin2x+21/sin2x+……+n1/sin2x}sin2x is
A
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution
The correct option is D limx→0{11sin2x+21sin2x+……+n1sin2x}sin2xPut1sin2x=t≥1∴limt→∞(1t+2t+……+nt)1t=limt→∞(nt)1t[(1n)t+(2n)t+……+1]1t=nlimt→∞[(1n)t+(2n)t+……+1]1t=n[0+0+……+1]0=n