<s>[INST] 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 $a+b=1$
 [/INST] Expand with Taylor:
$$
x(a+b-\cos x)-c\sin x=
x\left(a+b-1+\frac{x^2}{2!}-\frac{x^4}{4!}\right)
-c\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}\right)+o(x^5)
$$
Thus you want
$$
\begin{cases}
a+b-1-c=0 \\[4px]
1/2+c/6=0 \\[4px]
-1/24-c/120=1
\end{cases}
$$
The second equation becomes $c=-3$, the third becomes $-5-c=120$.
No solution.
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