Question

Find the values of $a,b$ such that $\underset{x\to 0}{lim}\frac{x(1+a\cos x)-b\sin x}{x^3}=1.$

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

Answer 1

Answer: (A), (D)

The correct options are
A $a=\frac{-5}{2}$
D $b=\frac{-3}{2}$

Here we use the expansions $\sin x=x-x^3/3!+x^5/5!-\cdots$ and $\cos x=1-x^2/2!+x^4/4!-\cdots$ Then we have $=\underset{x\to 0}{lim}\frac{x(1+a\cos x)-b\sin x}{x^3}$

$=\underset{x\to 0}{lim}\frac{x+ax(1-x^2/2!+x^4/4!-\cdots )}{x^3}-\frac{b(x-x^3\ast 3!+x^5/5!-\cdots )}{x^3}$

$=\underset{x\to 0}{lim}\frac{(1+a-b)x+(b/6-a/2)x^3+(a/24-b/120)x^5+\cdots }{x^3}$

$=\underset{x\to 0}{lim}\frac{(1+a-b)+(b/6-a/2)x^2+(a/24-b/120)x^4\cdots }{x^2}$ Since the limit is given as 1, a finite quantity, we must have

$1+a-b=0...\left(1\right)$ and $b/6-a/2=1..\left(2\right)$
Solving
(1) and (2),we have $a=-5/2,b=-3/2.$