Question

If $l={lim}_{x\to 0}\frac{x(1+acosx)-bsinx}{x^3}$ is finite,

where $l\in R$, then

• A. (a)$(a,b)=(-1,0)$
• B. (b) a & b are any real numbers
• C. (c)$l=0$
• D. (d)$l=\frac{1}{2}$

Answer 1

Answer: (A), (D)

(a)$(a,b)=(-1,0)$
(d)$l=\frac{1}{2}$

For given limit to exist, it should be of form $\frac{0}{0}$.

Now for limit, using L'Hopital rules (differentiating $3$ time because denominator has power $3$):

$l=\underset{x\to 0}{lim}=\frac{x(1+acosx)-bsinx}{x^3}=\underset{x\to 0}{lim}\frac{1-axsinx+acosx-bcosx}{3x^2}$

For limit to be finite:

$1+(a-b)=0⟹a-b=-1$........[1]

Again differentiating:

$l=\underset{x\to 0}{lim}\frac{-axcosx-asinx-asinx+bsinx}{6x}$

Again differentiating:

$l=\underset{x\to 0}{lim}\frac{-acosx+axsinx-acosx-acosx+bcosx}{6}=\frac{-a-a-a+b}{6}$

Now for $b=0$;

$a=-1+b=-1$

So, $l=\frac{-3a}{6}=\frac{1}{2}$