This is the limit 2

Consider the McLaurin expansion $\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}+\ldots$

Taking the first two terms alone results in the denominator of the limit being 0 for all $x$ , so the third term is also required. As such we will want to take the McLaurin expansion of the $a\sin(x)$ term to the 5th order.

This gives a new fraction of:

$\frac{a(x-\frac{x^3}{6}+\frac{x^5}{120})-bx+cx^2+x^3}{2x^2(x-\frac{x^2}{2}+\frac{x^3}{3})-2x^3+x^4}$

Grouping powers and cancelling we get:

$\frac{(a-b)x+cx^2+(1-\frac{a}{6})x^3+\frac{ax^5}{120}}{\frac{2x^5}{3}}$

Splitting this gives:

$\frac{(a-b)x}{\frac{2x^5}{3}}+\frac{cx^2}{\frac{2x^5}{3}}+\frac{(1-\frac{a}{6})x^3}{\frac{2x^5}{3}}+\frac{\frac{ax^5}{120}}{\frac{2x^5}{3}}=\frac{3(a-b)}{2x^4}+\frac{3c}{2x^3}+\frac{(3-\frac{a}{2})}{2x^2}+\frac{3a}{240}$

In order for the limit to exist, the numerator of all terms with $x$ in the denominator must be 0, which means $a=b$ , $c=0$ , and $a=6$ .

The limit is therefore $\frac{3\times6}{240}=\frac{3}{40}$ which means $x+y+a+b+c=3+40+6+6+0=\boxed{55}$