Question

If $\underset{x\to 0}{lim}\frac{x(1+a\cos x)-b\sin x}{x^3}=1$ then the value of $|a+b|$ is

Answer 1

$\underset{x\to 0}{lim}\frac{x+ax(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+...)-b(x-\frac{x^3}{3!}+\frac{x^5}{5!}-....)}{x^3}=1$
$\underset{x\to 0}{lim}\frac{1+a(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+...)-b(1-\frac{x^2}{3!}+\frac{x^4}{5!}-....)}{x^2}=1$
For limit to exist
$1+a-b=0$
And $\frac{-a}{2!}+\frac{b}{3!}=1$
$\Rightarrow a=\frac{-5}{2},b=\frac{-3}{2}$