<s>[INST] If $\lim_{x\to 0} \frac{1+a\cos 2x + b\cos 4x}{x^4}$ exists for all $x\in\mathbb R$ and is equal to $c$, find $\lfloor a^{-1} +b^{-1} + c^{-1}\rfloor$ $$\lim_{x\to 0} \frac{1+a(1-\frac{4x^2}{2!} + \frac{16x^4}{4!}-\cdots)+b(1-\frac{16x^2}{2!} + \frac{256x^4}{4!}-\cdots)}{x^4}$$
$$=\lim_{x\to 0} \frac{ (1+a+b) -\frac{x^2}{2!} (4a+16b) + \frac{x^4}{4!} (16a+256b)}{x^4}$$
For limit to exist, $1+a+b=0$ and $4a+16b=0$
So $a=-\frac 43$ and $b=\frac 13$
Now $$c=\lim_{x\to 0} \frac{\frac{x^4}{4!} (16a+256b)\cdots}{x^4}$$
$$c=\frac{16a+256b}{24}$$
$$c=\frac{2a}{3}+\frac{32}{3}b=\frac{24}{9}$$
Then $$\lfloor a^{-1} + b^{-1} + c^{-1} \rfloor = \left\lfloor 3-\frac 34 + \frac{9}{24}\right\rfloor = \lfloor 2.625\rfloor=2$$
The given answer is $8.$
What is wrong in my solution?
 [/INST] Your result is correct
If we put $X=2x$ then it is equivalent to say that
$$\lim_0\frac{1+a\cos(X)+b\cos(2X)}{X^4}=\frac{c}{16}$$
$$=-\lim_0\frac{a\sin(X)+2b\sin(2X)}{4X^3}$$
$$=-\lim_0\frac{a\cos(X)+4b\cos(2X)}{12X^2}$$
$$=\lim_0\frac{a\cos(X)+16b\cos(2X)}{24}$$
So, we must have
$$a+b+1=0$$
$$a+4b=0$$
thus
$$b=\frac 13,\;\; a=-\frac 43$$
and
$$c=\frac 23(a+16b)=\frac 29(-4+16)=\frac{24}{9}$$
</s>