Question

$\lim_{x \to 0} \frac{x - sin(x)}{x^3}=$

• A. 1/2
• B. 1/3
• C. 1/6
• D. -1/6

Answer 1

Answer:

C. 1/6

$\lim_{x \to 0} \frac{x - sin(x)}{x^3}=$

applying L Hospitals rule

$\lim_{x \to 0} \frac{x - sin(x)}{x^3}=\lim_{x \to 0} \frac{1 - cos(x)}{3x^2}= \lim_{x \to 0} \frac{sin(x)}{6x}$

$=\frac{1}{6} \lim_{x \to 0} \frac {sin(x)}{x} = \frac{1}{6} \times 1= \frac{1}{6}$