If  $\underset{x\to 0}{Lt}\left(\frac{\cos 4x+a\cos 2x+b}{x^4}\right)$ is finite then the values of  $a,b$ are respectively

It is finite  $1+a+b=0$

$\underset{x\to 0}{Lt}\frac{-4\sin 4x-2a\sin 2x}{+4x^3}$

$=\underset{x\to 0}{Lt}\frac{-16\cos 4x-4a\cos 2x}{12x^2}$

$-16-4a=0$

$\Rightarrow a=-4$

$b=3$