Mathematics - Limits Question with Solution | TestHub

${\lim }_{x\to 0}\frac{\cos 2x+a\cos 4x-b}{x^4}=$ finite
$\begin{array}{rl}& L=\frac{\{1-\frac{(2x{)}^2}{\underline{2}}+\frac{(2x{)}^4}{\underline{4}\cdots \}\}+a\left\{1-\frac{(4x{)}^2}{\underline{2}}+\frac{(4x{)}^4}{\lfloor 4}\dots \cdot \right\}-b}}{x^4}\\ & L=\frac{(1+a-b)-x^2(2+8a)+x^4\left(\frac{2}{3}+\frac{32}{3}a\right)+x^6()\dots }{x^4}\end{array}$
$\begin{array}{rl}& \therefore 1+\mathrm{a}-\mathrm{b}=0\text{ and }2+8\mathrm{a}=0\Rightarrow \mathrm{a}=-\frac{1}{4}\\ & \mathrm{b}=\mathrm{a}+1\\ & =-\frac{1}{4}+1=\frac{3}{4}\\ & \therefore \mathrm{a}+\mathrm{b}=-\frac{1}{4}+\frac{3}{4}=\frac{1}{2}\end{array}$