If  L=limx→0  sinx+aex+be-x+c  loge1+xx3 exists finitely, thenThe value of L isEquation ax2+bx+c=0 hasThe solutions set of x+c-2a<4b is

L=limx→0 sinx+aex+be-x+c  loge1+xx3     =limx→0  x-x33!+a1+x1!+x22!+x33!+b1-x1!+x22!-x33!+cx-x22+x33x3     =limx→0  a+b+1+a-b+c+a2+b2-c2x2+-13!+a3!-b3!+c3x3x3⇒     a+b=0,  1+a-b+c=0,   a2+b2-c2=0and   L=-13!+a3!-b3!+c3Solving   first   three   equations  we  get  c=0,    a=-12,   b=12.∴   L=13Equation  ax2+bx+c=0   reduces  to   x2-x=0⇒x=0,1x+c-2a<4b   reduces   to   x+1<2⇒-2