Values of $a,b,c$ such that $\lim_{x \to 0}\frac{x(a+b-\cos x)-c\sin x}{x^5}=1$ Find the values of $a,b,c$ such that $$\lim_{x \to 0}\frac{x(a+b-\cos x)-c\sin x}{x^5}=1$$
Here's what I have got so far
Using L'Hospital's rule,
$$\lim_{x \to 0}\frac{a+b-\cos x+x\sin x-c\cos x}{5x^4}=1$$
So,$a+b-1=0$
Again,
$$\lim_{x \to 0}\frac{\sin x+\sin x+x\cos x+c\sin x}{20x^3}=1$$
But the solution is given as $a=120,b=60,c=180$
There is no way that $a+b=1$

Since $$\lim_{x \to 0}\frac{x(a+b-\cos x)-c\sin x}{x^5}=1$$ is given, we need
$$(x(a+b-\cos x)-c\sin x)'_{x=0}=0$$ or
$$(a+b-\cos{x}+x\sin{x}-c\cos{x})_{x=0}=0$$ or
$$a+b-c-1=0.$$ 
Thus, $$\lim_{x \to 0}\frac{a+b-(1+c)\cos{x}+x\sin{x}}{5x^4}=1.$$
Hence, we need $$(a+b-(1+c)\cos{x}+x\sin{x})'_{x=0}=0$$ or
$$((1+c)\sin{x}+\sin{x}+x\cos{x})_{x=0}=0,$$ which is true, which says
$$\lim_{x \to 0}\frac{(2+c)\sin{x}+x\cos{x}}{20x^3}=1$$ and from here
$$((2+c)\sin{x}+x\cos{x})'_{x=0}=0$$ or
$$\lim_{x\rightarrow0}\frac{(3+c)\cos{x}-x\sin{x}}{60x^2}=1,$$ where $c=-3$ and we get $-\frac{1}{60}=1$, which is wrong.
