If $\lim _{x \rightarrow 0} \frac{\cos (2 x)+a \cos (4 x)-b}{x^4}$ is finite, then $(a+b)$ is equal to :

Question

Aicharya · Accepted Answer

$\lim _{x ightarrow 0} \frac{\cos 2 x+a \cos 4 x-b}{x^4}=$ finite $\begin{aligned} & L=\frac{\left\{1-\frac{(2 x)^2}{2!}+\frac{(2 x)^4}{4! \cdots\}\}+a\left\{1-\frac{(4 x)^2}{2!}+\frac{(4 x)^4}{4!} \ldots \cdotight\}-b}ight.}{x^4} \ & L=\frac{(1+a-b)-x^2(2+8 a)+x^4\left(\frac{2}{3}+\frac{32}{3} aight)+x^6() \ldots}{x^4}\end{aligned}$ $\begin{aligned} & 	herefore 1+\mathrm{a}-\mathrm{b}=0 	ext { and } 2+8 \mathrm{a}=0 \Rightarrow \mathrm{a}=-\frac{1}{4} \ & \mathrm{~b}=\mathrm{a}+1 \ & =-\frac{1}{4}+1=\frac{3}{4} \ & 	herefore \mathrm{a}+\mathrm{b}=-\frac{1}{4}+\frac{3}{4}=\frac{1}{2}\end{aligned}$

If $\lim _{x \rightarrow 0} \frac{\cos (2 x)+a \cos (4 x)-b}{x^4}$ is finite, then $(a+b)$ is equal to :

Solution