P l e a s e s e n d q n o 1 Q F i n d t h e d e r i v a t i v e o f f ( x ) 1 - cos 4 x x 2 i f x < 0 a 2 16 + x - 4 i f l i m f ( x ) = f ( 0 ) , f i n d a x → 0 Share with your friends Share 0 Aarushi Mishra answered this Dear student Your question cannot be understood. What needs to be done remain uncertain. I am assuming your question to befx=1-cos 4xx2, if x<0 a, if x=0x16+x-4, if x>0Find value of if it exist such that fx is continuous at x=0For fx to be continuous at x=b, limx→bfx=fbFor limit to exist, left hand limit and right hand limit should exist finitely and they should be equallimh→0+fb+h=limh→0+fb-hlimh→0+f0+h=limh→0+fh=limh→0+ h16+h-4as for x>0 fx=x16+x-4=limh→0+ h16+h-4×16+h+416+h+4Use a+ba-b=a2-b2=limh→0+ h16+h+416+h2-42=limh→0+ h16+h+416+h-16=limh→0+ h16+h+4hlimh→0+fh=limh→0+ 16+h+4=8limh→0+f0-h=limh→0+f0-h=limh→0+1-cos -4h-h2=1-cos 4hh2=limh→0+2sin2 2hh2as for x<0 fx=1-cos 4xx2=limh→0+8sin2 2h4h2=limh→0+8sin 2h2h2limh→0+f0-h=8limh→0+sin 2h2h2=8Hence right hand limit is equal to left hand limitHence limx→0fx exist and limx→0fx=8For fx to be continuouslimx→0fx=f0a=8 1 View Full Answer
