lim (x+y) sec (x+y) -x sec x / y y--0 Share with your friends Share 20 Manbar Singh answered this We have, limy→0 x + y sec x + y - x sec xy= limy→0 x sec x + y + y sec x + y - x sec xy= limy→0 xsec x + y - sec x + y sec x + yy= limy→0 x sec x + y - sec xy + limy→0 y sec x + yy= limy→0 x 1cos x + y - 1cos xy + limy→0 sec x +y= limy→0 xcos x - cos x + yy cos x cos x + y + sec x= limy→0 cos x - cos x + yy . xcos x cos x + y + sec x= limy→0 2 sin x + y2 sin y22y2 . xcos x cos x + y + sec x using, cos x - cos y = -2 sin x + y2 sin x - y2= limy→0 sin x + y2 . limy→0 sin y2y2 . limy→0 xcos x cos x + y + sec x= sin x . 1 . xcos x cos x + sec x = xsin xcos x. 1cos x + sec x = x tan x sec x + sec x 51 View Full Answer
