If \(\lim _{\mathrm{x}\to 0}\frac{\cos (2\mathrm{x})+\mathrm{acos}(4\mathrm{x})-\mathrm{b}}{{\mathrm{x}}^{4}}\) is finit…

Use the Taylor expansions around \(x=0\):

\(\cos ⁡(2x)=1−2{x}^{2}+\frac{2}{3}{x}^{4}+O({x}^{6})\)

\(\cos ⁡(4x)=1−8{x}^{2}+\frac{32}{3}{x}^{4}+O({x}^{6})\)

\(\cos ⁡(2x)+a\cos ⁡(4x)−b=(1+a−b)+(−2−8a){x}^{2}+(\frac{2}{3}+\frac{32a}{3}){x}^{4}+O({x}^{6})\)

For

\(\underset{x\to 0}{\lim ⁡}\frac{\cos ⁡(2x)+a\cos ⁡(4x)−b}{{x}^{4}}\)

to be finite, the numerator must be \(O({x}^{4})\). Thus the constant and \({x}^{2}\) terms must vanish:

\(1+a−b=0\ \text{(i)}\)

\(−2−8a=0\ \text{(ii)}\)

From (ii): \(a=−\frac{1}{4}\).

Substitute into (i):

\(1−\frac{1}{4}−b=0\Rightarrow b=\frac{3}{4}\)

So

\(a+b=−\frac{1}{4}+\frac{3}{4}=\frac{1}{2}\mathrm{.}\)