1. We need the limit

$$\lim_{x\to 0} \frac{\cos(2x)+a\cos(4x)-b}{x^4}$$

to be finite.

2. Expand the cosine terms near $x=0$ using Taylor series:

$$\cos t = 1-\frac{t^2}{2!}+\frac{t^4}{4!}+\cdots$$

So,

$$\cos(2x)=1-\frac{(2x)^2}{2}+\frac{(2x)^4}{24}+\cdots =1-2x^2+\frac{16x^4}{24}+\cdots =1-2x^2+\frac{2}{3}x^4+\cdots$$

and

$$\cos(4x)=1-\frac{(4x)^2}{2}+\frac{(4x)^4}{24}+\cdots =1-8x^2+\frac{256x^4}{24}+\cdots =1-8x^2+\frac{32}{3}x^4+\cdots$$

3. Substitute into the numerator:

$$\cos(2x)+a\cos(4x)-b$$ $$=\left(1-2x^2+\frac{2}{3}x^4+\cdots\right) +a\left(1-8x^2+\frac{32}{3}x^4+\cdots\right)-b$$ $$=(1+a-b)+(-2-8a)x^2+\left(\frac{2}{3}+\frac{32a}{3}\right)x^4+\cdots$$

4. For

$$\frac{\cos(2x)+a\cos(4x)-b}{x^4}$$

to have a finite limit, the constant term and the $x^2$ term in the numerator must vanish.

So,

• Constant term:

$$1+a-b=0$$

• Coefficient of $x^2$:

$$-2-8a=0$$

5. Solve for $a$:

$$-2-8a=0 \implies 8a=-2 \implies a=-\frac14$$

Then from $1+a-b=0$,

$$b=1+a=1-\frac14=\frac34$$

6. Now compute $a+b$:

$$a+b=-\frac14+\frac34=\frac12$$

7. Therefore, the correct option is

$$\boxed{\frac12}$$

which is option D.