calculus - Find the value of : $lim_{xtoinfty}sqrt{x+sqrt{x+sqrt{x+sqrt{x}}}}-sqrt{x}$

$\lim_{x\to\infty}\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x}}}}-\sqrt{x}$

I tried conjugating and it didn't lead me anywhere please help guys.

Thanks,

Answer

You can get the following :

$$\begin{align}\sqrt{x+\sqrt{x+\sqrt{x+\sqrt x}}}-\sqrt x&=\frac{\sqrt{x+\sqrt{x+\sqrt x}}}{\sqrt{x+\sqrt{x+\sqrt{x+\sqrt x}}}+\sqrt x}\\&=\frac{\sqrt{1+\left(\sqrt{x+\sqrt x}\right)/x}}{\sqrt{1+\left(\sqrt{x+\sqrt{x+\sqrt x}}\right)/x}+1}\end{align}$$

Now divide both the numerator and the dinominator by $\sqrt x$.