Question

If $\underset{x\to 0}{lim}\frac{\tan x-\sin x}{{\sin }^{3}x}=k$ then the value of $\frac{1}{k^2}+\frac{1}{k}+1$ is

Answer 1

$L=\underset{x\to 0}{lim}\frac{\tan x-\sin x}{{\sin }^{3}x}$
This is $\frac{0}{0}$ form

$\Rightarrow L=\underset{x\to 0}{lim}\frac{\tan x(1-\cos x)}{\sin x(1-{\cos }^{2}x)}\Rightarrow L=\underset{x\to 0}{lim}\frac{1}{\cos x}\times \frac{1}{(1+\cos x)}\Rightarrow L=\frac{1}{2}\Rightarrow k=\frac{1}{2}\therefore \frac{1}{k^2}+\frac{1}{k}+1=4+2+1=7$