Question

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

Answer 1

$\underset{x\to 0}{lim}\frac{\sin x-x+\left(1/6\right)x^3}{x^5}$
$=\underset{x\to 0}{lim}\frac{[x-(1/3!)x^3+(1/5!)x^5-\cdots ]-x+\left(1/6\right)x^3}{x^5}$
$=\underset{x\to 0}{lim}\frac{1}{x^5}[\frac{1}{5!}{x}^5+\text{higher powers of x}]$
$=\underset{x\to 0}{lim}[\frac{1}{120}+\text{higher powers of x}]=\frac{1}{120}.$
$\Rightarrow a=120$