Find the value of : $\lim_{x\to\infty}\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x}}}}-\sqrt{x}$

Question

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

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

Thanks,

Answer 1

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 $\sqrt x$.

Answer 2

I think the best way to do it is to note that

$$\sqrt{x + \sqrt{x}} \sim \sqrt{x}$$

So you can write that $\lim_{x\to\infty}\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x}}}}-\sqrt{x}$ = $\lim_{x\to\infty}\sqrt{x+\sqrt{x}}-\sqrt{x}$

(We can't simplify any further cause that would bring us to $\sqrt{x} - \sqrt{x}$ which doesn't make sense.)

Then write as

$$\sqrt{x} \cdot (\sqrt{1 + \frac{1}{\sqrt{x}}} - 1) \sim \sqrt{x} \cdot (1 + \frac{1}{2\sqrt{x}} - 1) = \frac{1}{2}$$