Question

$\underset{x\to 0}{lim}\frac{x(1+acosx)-bsinx}{x^3}=1$, then

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

$\underset{x\to 0}{lim}\frac{x\{1+a(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\dots \dots )\}-b(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\dots \dots )}{x^3}=1\underset{x\to 0}{lim}\frac{x(1+a-b)-x^3(\frac{a}{2!}-\frac{b}{3!})+x^5(\frac{a}{4!}-\frac{b}{5!})-\dots \dots }{x^3}=1SinceRHSinfinite,\therefore 1+a-b=0and-\frac{a}{2}+\frac{b}{6}=1\therefore b=-\frac{3}{2},a=-\frac{5}{2}$