lim 2sinx - sin2x

_____________________

x^3

x--> 0

lim 2sinx - sin2x

_____________________   =   1

            x^3

x--> 0

x->0 --------------------          since sin 2x=2 sinx cosx

            x^3  

lim   2sinx    lim   (1- cosx)           lim  sinx /x  =1    

x->0 -------   x->0 ---------            x->0          

          x                 x^2                          

2   lim  ( 2sin^2(x/2))/ x^2        using identity 1-cos2x = 2sin^2(x)     

   x->0 

now ( 2sin^2(x/2))/ x^2    may be written as  2sin^2(x/2)/4 (x^2/4)

lim  ( sin^2(x/2))/ (x/2)^2 = 1 

x->0

thus we have 2*(2/4)=1 

if u r satisfied wid the answer click on 'yes' :)

2sinx-2sinxcosx

2sinx(sinx^2{x\2}

4x^3

like my answer

lim x---0 (2sinx-sin2x)/x^3=lim x---0 (2sinx-2sinxcosx)/x^3=lim x---0 (2sinx{1-cosx})/x^3=lim x---0 (2sinx * 2sin^2 x/2)/x^3=lim x---0 ({2sinx /x}* {2sin^2 x/2}/({4x^2}/4)= 1 * 1 = 1LIKE MY ANSWER IF IT HELPED YOU