IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

toshi

lim    1--5x/1--cos6x
x-->0
   2sin^2 (5/2)x
lim     -------------------------         [since 1-cos2x= 2sin^2x]
x-->0        2sin^2(3x)

now, by applying formala of limits when x-->0

          sin 5/2x
      [ --------------] [5/2x}^2
         5/2x
lim       -----------------------------------------    [since lim     sinmx/mx = 1]
x-->0             sin3x                                     x-->0
            [-------------] [3x]^2
               3x
now,

lim       (25/4x^2 )
x-->0--------------------   = 25/36
            9x^2
hope u got it!

Ask & Discuss Questions with Community & Experts