<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] $$L=\lim_{x\to 0} \frac{1+a\cos 2x + b\cos 4x}{x^4}$$
$$L=\lim_{x\to 0} \frac{1+a(1-\frac{4x^2}{2!} + \frac{16x^{4}}{4!}\cdots)+b(1-\frac{16x^2}{2!} + \frac{256x^4}{4!}\cdots)}{x^4}$$
$$\implies L=\lim_{x\to 0} \frac{(1+a+b)+(-2a-8b)x^2+(2a/3+32b/3)x^4}{x^4}$$
for the limit to exist $(1+a+b)=0, (-2a-8b)=0, (2a/3+32b/3=c$ then
$a=-4/3,b=1/3, c=8/3; [1/a+1/b+1/c]=[21/8]=2.
OP's answer is right.
</s>