Question

If $\underset{x\to 0}{\lim }\frac{sin2x-asinx}{{\left(\frac{x}{3}\right)}^3}=L$ exists finitely, then the absolute value of $L$ is equal to

Answer 1

Answer: 27

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