Question

If $\underset{x\to 0}{lim}\frac{-1+\sqrt{(\tan x-\sin x)+\sqrt{(\tan x-\sin x)+\dots \infty }}}{-1+\sqrt{x^3+\sqrt{x^3+\sqrt{x^3+\dots \infty }}}}$ is $\frac{1}{k},$ then $k$ is equal to

Answer 1

$L=\underset{x\to 0}{lim}\frac{-1+\sqrt{(\tan x-\sin x)+\sqrt{(\tan x-\sin x)+\dots \infty }}}{-1+\sqrt{x^3+\sqrt{x^3+\sqrt{x^3+\dots \infty }}}}\dots \left(1\right)$

$y=\sqrt{(\tan x-\sin x)+y}$
$y^2=(\tan x-\sin x)+y$
$\Rightarrow y=\frac{1\pm \sqrt{1+4(\tan x-\sin x)}}{2}$
As $y>0,$
$y=\frac{1+\sqrt{1+4(\tan x-\sin x)}}{2}$

Now, $z=\sqrt{x^3+z}$
$z^2=x^3+z$
$\Rightarrow z=\frac{1\pm \sqrt{1+4x^3}}{2}$
As $z>0,$
$z=\frac{1+\sqrt{1+4x^3}}{2}$

$\therefore$ $\left(1\right)$ becomes
$L=\underset{x\to 0}{lim}\frac{-1+\frac{1+\sqrt{1+4(\tan x-\sin x)}}{2}}{-1+\frac{1+\sqrt{1+4x^3}}{2}}$
$=\underset{x\to 0}{lim}\frac{-1+\sqrt{1+4(\tan x-\sin x)}}{-1+\sqrt{1+4x^3}}$
Rationalizing, we get
$L=\underset{x\to 0}{lim}\frac{4(\tan x-\sin x)}{4x^3}\times \frac{1+\sqrt{1+4x^3}}{1+\sqrt{1+4(\tan x-\sin x)}}$
$=\underset{x\to 0}{lim}\frac{\tan x-\sin x}{x^3}\times \frac{2}{2}$
$=\underset{x\to 0}{lim}(\frac{\sin x}{x}\times \frac{1}{\cos x}\times \frac{1-\cos x}{x^2})$
$=1\times 1\times \frac{1}{2}=\frac{1}{2}$