Question

If the value of $\underset{x\to 0}{lim}\frac{\tan x-\sin x}{x^3}$ is equal to $\frac{1}{b}$. Then ,what will be the value of $b?$.

Answer 1

$\underset{x\to 0}{lim}\frac{\tan x-\sin x}{x^3}$

$=\underset{x\to 0}{lim}\frac{\sin x(1-\cos x)}{x^3\cos x}$..................................$\left(taking\tan xascommonfrombothnumeratoranddenominator\right)$
$=\underset{x\to 0}{lim}\frac{2\sin \left(x/2\right)\cos \left(x/2\right)2{\sin }^2\left(x/2\right)}{x^3\cos x}$..........................$(\sin x=2\cdot \sin \frac{x}{2}\cdot \cos \frac{x}{2}and1-\cos x=2\cdot {\sin }^2\frac{x}{2})$

$=\underset{x\to 0}{lim}\frac{4{\sin }^3\left(x/2\right)}{x^3}.\frac{\cos \left(x/2\right)}{\cos x}$

$=\underset{x\to 0}{lim}\frac{1}{2}{\left(\frac{\sin \left(x/2\right)}{x/2}\right)}^3\frac{\cos \left(x/2\right)}{\cos x}=\frac{1}{2}\times 1=\frac{1}{2}$

$\therefore \frac{1}{b}=\frac{1}{2}$

$\therefore b=2$