If $\underset{x\to 0}{lim} \frac{\cos 4x+a\cos 2x+b}{x^4}$ is finite, then$(a,b)=$

$\underset{x\to 0}{lim} \frac{\cos 4x+a\cos 2x+b}{x^4}$ is finite.

Therefore $\frac{\cos 4x+a\cos 2x+b}{x^4}$ should be of the form $\frac{0}{0}atx=0$.

Consequently, the value $\cos 4x+a\cos 2x+b$ must be zero at $x=0.$

i.e.,

$1+a+b=0$                  …(i)

Using L' Hospital's rule the given limit is 

$\underset{x\to 0}{lim} \frac{-4\sin 4x-2a\sin 2x}{4x^3}\left(\frac{0}{0}\text{ form }\right)$

$=\underset{x\to 0}{lim} \frac{-16\cos 4x-4a\cos 2x}{12x^2}$ [Using L Hospitals rule] 

This should be of the form $\frac{0}{0}$.

$\therefore -16-4a=0$                                …(ii)

Solving (i) and (ii), we get 

$a=-4$ and $b=3$.