If $\underset{x\to 0}{\lim }\frac{a\sin x-bx+c{x}^2+x^3}{2x^2{\log }_e(1+x)-2x^3+x^4}$ exists and is  finite then  $a+b+c=\_\_\_\_$

$\underset{x\to 0}{lim} \frac{a\sin x-bx+c{x}^2+x^3}{2x^2\log (1+x)-2x^3+x^4}$

$=\underset{x\to 0}{lim} \frac{a\left\{x-\frac{x^3}{3!}+\frac{x^5}{5!}-\dots \right\}-bx+c{x}^2+x^3}{2x^2\left(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}\cdots \right)-2x^3+x^4}$

$=\underset{x\to 0}{lim} \frac{(a-b)x+c{x}^2+\left(-\frac{a}{3!}+1\right)x^3+\frac{a}{5!}{x}^5+\dots \dots }{\frac{2}{3}{x}^5-\frac{1}{2}{x}^6+\dots \dots .}$

$=\underset{x\to 0}{lim} \frac{(a-b)+cx+\left(-\frac{a}{3!}+1\right)x^2+\frac{a}{5!}{x}^4+\dots \dots .}{\frac{2}{3}{x}^4-\frac{1}{2}{x}^5+\dots \dots .}$ 

If this limit exists, then  

$a-b=0,c=0,-\frac{a}{3!}+1=0\Rightarrow a=b=6$and $c=0$

For this values of$a, b, c$, the limit is given by 

$\underset{x\to 0}{lim} \frac{\frac{6}{5!}{x}^4+\dots \dots }{\frac{2}{3}{x}^4-\frac{1}{2}{x}^5+\dots \dots }=\frac{6}{5!}\times \frac{3}{2}=\frac{3}{40}$