Question

If $\underset{x \rightarrow 0}{lim} \frac{sin ⁡ 2 x - a sin ⁡ x}{\left(\frac{x}{3}\right)^{3}}=L$ exists finitely, then the absolute value of $L$ is equal to

Answer 1

$L=\underset{x \rightarrow 0}{l i m}27\frac{sin x \left(2 cos ⁡ x - a\right)}{x \cdot x^{2}}=\underset{x \rightarrow 0}{l i m}\left(\frac{sin ⁡ x}{x}\right)\left(\frac{2 cos ⁡ x - a}{x^{2}}\right)\times 27$
This limit to exists finitely
$\underset{x \rightarrow 0}{lim} \frac{2 cos ⁡ x - a}{x^{2}}=$ finite
$\therefore $ It must be $\frac{0}{0}$ form
$\therefore 2cos \left(0\right)-a=0\Rightarrow a=2$
$\Rightarrow L=\underset{x \rightarrow 0}{lim} \left(\frac{sin ⁡ x}{x}\right)\times 2\left(\frac{cos ⁡ x - 1}{x^{2}}\right)\times 27$
$\Rightarrow L=1\times 2\times \left(- \frac{1}{2}\right)\times 27=-27$