L=limx→0 sin⁡x+aex+be−x+cln⁡(1+x)x3=∞The value of L isEquation ax2 + bx + c = 0 hasThe solution set of ||x+c|−2a|<4b is

L=limx→0 sin⁡x+aex+be−x+cln⁡(1+x)x3=limx→0 x−x33!+a1+x1!+x22!+x33!+b1−x1!+x22!−x33!+cx−x22+x33x3=limx→0 (a+b)+(1+a−b+c)x+a2+b2−c2x2+−13!+a3!−b3!+c3x3x3or a+b=0,1+a−b+c=0,a2+b2−c2=0and L=−13!+a3!−b3!+c3Solving the first three equations, we get c = 0, a = -1/2, b = 1/2.Then, L = - 1/3.Equation ax2 + bx + c = 0 reduces to x2 -x=0 or x=0, 1.||x+c|−2a|<4b reduces to ∥x|+1|<2 or     −2<|x|+1<2 or     0≤|x|<1 or     x∈[−1,1]