Question

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

• A. $b-a=\frac{5}{6}$
• B. $a=\frac{-5}{2}$
• C. $a-b=-1$
• D. $b=\frac{3}{2}$

Answer 1

Answer: (B), (C)

The correct options are
B $a=\frac{-5}{2}$
C $a-b=-1$

We have
$\underset{x\to 0}{lim}\frac{x(1+a\cos x)-b\sin x}{x^3}=1$
$\underset{x\to 0}{lim}\frac{x(1+a(1-\frac{x^2}{2!}+\frac{x^4}{4!}-...))-b(x-\frac{x^3}{3!}+\frac{x^5}{5!}-...)}{x^3}$
$\underset{x\to 0}{lim}\frac{x(1+a-b)+x^3[\frac{-a}{2!}+\frac{b}{3!}]+0\left(x^4\right)}{x^3}$
This limit exist if $1+a-b=0$ and is equal to 1 and if $-a/2+b/6=1$. Solving these two equations, we get $a=-5/2$ and $b=-3/2$