Question

The value of $\underset{x\to 0}{\lim }\frac{(\sin x-x{)}^2+\left(1-\cos x^3\right)}{x^5\sin x}$ is

• A. $7/36$
• B. $31/18$
• C. $19/36$
• D. $11/18$

Answer 1

Answer: (C)

$\underset{x\to 0}{\lim }\frac{(\sin x-x{)}^2+\left(1-\cos x^3\right)}{x^5\sin x}$
$=\underset{x\to 0}{\lim }\frac{(\sin x-x{)}^2+\left(1-\cos x^3\right)}{x^6\frac{\sin x}{x}}$
$=\underset{x\to 0}{\lim }{\left(\frac{\sin x-x}{x^3}\right)}^2+\left(\frac{1-\cos x^3}{x^6}\right)$
$=\underset{x\to 0}{\lim }{\left(\frac{x-\frac{x^3}{3!}-x}{x^3}\right)}^2+\left(\frac{1-\left(1-\frac{x^6}{2!}\right)}{x^6}\right)$
$={\left(\frac{-1}{6}\right)}^2+\frac{1}{2}=\frac{19}{36}$