Question

$\underset{x\to 0}{\lim }\frac{{\sin }^{-1}x-x}{x^3\cos x}$ is equal to

• A. $1/2$
• B. $1/3$
• C. $1/6$
• D. $1/12$

Answer 1

Answer: (C)

$=\underset{x\to 0}{\lim }\frac{{\sin }^{-1}x-x}{x^3\cos x}$
$\left(\frac{0}{0}form\right)$
$=\underset{x\to 0}{\lim }\frac{\frac{1}{\sqrt{1-x^2}}-1}{x^3(-\sin x)+3x^2\cos x}$
(using L- Hospital's rule)
$=\underset{x\to 0}{\lim }\frac{1-\sqrt{1-x^2}}{\sqrt{1-x^2}.x^2(-x\sin x+3\cos x)}$
$\times \frac{1+\sqrt{1-x^2}}{1+\sqrt{1-x^2}}$
$=\underset{x\to 0}{\lim }\frac{1-1+x^2}{\left[\begin{array}{rl}& x^2\sqrt{1-x^2}(1+\sqrt{1-x^2})\\ & (-x\sin x+3\cos x)\end{array}\right]}$
$=\underset{x\to 0}{\lim }\frac{1}{\left[\begin{array}{rl}& \sqrt{1-x^2}(1+\sqrt{1-x^2})\\ & (-x\sin x+3\cos x)\end{array}\right]}$
$=\frac{1}{1(1+1)(3)}$
$=\frac{1}{6}$