Question

If $l=\underset{x\to 0}{lim}\frac{\sin ax-\sin x-x}{x^3}$ exists, and is finite, then the values of '$l$' and '$a$' are respectively equal to

• A. $\frac{7}{6},2$
• B. $-\frac{4}{3},2$
• C. $-\frac{7}{6},-2$
• D. $-\frac{7}{6},2$

Answer 1

The correct option is D $-\frac{7}{6},2$

$l=\underset{x\to 0}{lim}\frac{\sin ax-\sin x-x}{x^3}$

$=\underset{x\to 0}{lim}\frac{(ax-\frac{a^3{x}^3}{3!}+\frac{a^5{x}^5}{5!}.....)-(x-\frac{x^3}{3!}+\frac{x^5}{5!})-x}{x^3}$

$=\underset{x\to 0}{lim}\frac{(a-2)x+(\frac{1}{6}-\frac{a^3}{6})x^3+.....}{x^3}$

$\underset{x\to 0}{lim}\frac{(a-2)}{x^2}+(\frac{1}{6}-\frac{a^3}{6})+x^2(.....)+......$

For limit to exist, $(a-2)=0$
$a=2$
$l=(\frac{1}{6}-\frac{a^3}{6})=\frac{1-8}{6}=\frac{-7}{6}$.
$l=\frac{-7}{6}$