Question

Find c

$\underset{x\to 0}{lim}\frac{\sin x-x}{x^3}$ is $=-\frac{1}{c}.$

Answer 1

$\underset{x\to 0}{lim}\frac{\sin x-x}{x^3}=\underset{x\to 0}{lim}\frac{(x-\frac{x^3}{3!}+\frac{x^5}{5!}-...)-x}{x^3}=-\frac{1}{3!}=-\frac{1}{6}$