# Thread: How to calculate this limit?

1. ## How to calculate this limit?

$\displaystyle \lim_{x \to \infty_+}\left(\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}\right)$

Thank you very much!

2. ## Re: How to calculate this limit?

Hello, gotmejerry!

Nice job with the Latex!

$\displaystyle \lim_{x \to \infty}\left(\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}\right)$

Multiply numerator and denominator by the conjugate:

$\displaystyle \frac{\sqrt{x+\sqrt{x+\sqrt{x}}} - \sqrt{x}}{1} \cdot\frac{\sqrt{x+\sqrt{x + \sqrt{x}}} + \sqrt{x}}{\sqrt{x+\sqrt{x+\sqrt{x}}} + \sqrt{x}}$

. . $\displaystyle =\;\dfrac{x + \sqrt{x+\sqrt{x}} - x}{\sqrt{x+\sqrt{x+\sqrt{x}}} + \sqrt{x}} \;=\;\frac{\sqrt{x+\sqrt{x}}}{\sqrt{x+\sqrt{x + \sqrt{x}}} + \sqrt{x}}$

Divide numerator and denominator by $\displaystyle \sqrt{x}\!:$

$\displaystyle \dfrac{\dfrac{\sqrt{x+\sqrt{x}}}{\sqrt{x}}} {\dfrac{\sqrt{x+\sqrt{x+\sqrt{x}}}}{\sqrt{x}} + \dfrac{\sqrt{x}}{\sqrt{x}}} \;=\;$ .$\displaystyle \dfrac{\sqrt{\dfrac{x+\sqrt{x}}{x}}}{\sqrt{\dfrac{ x+\sqrt{x+\sqrt{x}}}{x}} + 1}}} \;=\;\dfrac{\sqrt{1+\dfrac{1}{\sqrt{x}}}} {\sqrt{1 + \dfrac{\sqrt{x+\sqrt{x}}}{x}} + 1}$

. . $\displaystyle =\; \dfrac{\sqrt{1+\dfrac{1}{\sqrt{x}}}} {\sqrt{1 + \dfrac{\sqrt{x+\sqrt{x}}}{\sqrt{x^2}}} + 1} \;=\;\dfrac{\sqrt{1+\dfrac{1}{\sqrt{x}}}} {\sqrt{1 + \sqrt{\dfrac{x}{x^2} + \dfrac{\sqrt{x}}{x^2}}}+ 1}\;=\;\dfrac{\sqrt{1+\dfrac{1}{\sqrt{x}}}} {\sqrt{1 + \sqrt{\dfrac{1}{x} + \dfrac{1}{x^{\frac{3}{2}}}}}+ 1}$

$\displaystyle \lim_{x\to\infty}\dfrac{\sqrt{1+\dfrac{1}{\sqrt{x} }}} {\sqrt{1 + \sqrt{\dfrac{1}{x} + \dfrac{1}{x^{\frac{3}{2}}}}+ 1}} \;\;=\;\;\frac{\sqrt{1+0}}{\sqrt{1 + \sqrt{0+0}} + 1} \;\;=\;\;\frac{1}{2}$

3. ## Re: How to calculate this limit?

Thank you very much! Your LATEX is very good too