$\displaystyle\lim_{x\to\infty}\sqrt{x+\sqrt{x+\sqrt{x}}} - \sqrt{x} =?$

Question

lim  x sonsuza giderken $\sqrt{x+\sqrt{x+\sqrt{x}}}$ - $\sqrt{x}$ =?

Answer 1

$$\lim_{x\to \infty}{\sqrt{x+\sqrt{x+\sqrt{x}}}}-\sqrt{x}=\lim_{x\to \infty}\frac{{x+\sqrt{x+\sqrt{x}}}-x}{{\sqrt{x+\sqrt{x+\sqrt{x}}}}+\sqrt{x}}$$
$$=$$
$$\lim_{x\to \infty}\frac{{\sqrt{x+\sqrt{x}}}}{{\sqrt{x+\sqrt{x+\sqrt{x}}}}+\sqrt{x}}$$
$$=$$ $$\ldots$$

Answer 2

hocanın kaldığı yerden devam edelim

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

$\sqrt {x}=y$ olsun , $x\rightarrow \infty \Rightarrow y\rightarrow \infty$

$\lim _{y\rightarrow \infty }\dfrac {\sqrt {y^{2}+y}}{\sqrt {y^{2}+\sqrt {y^{2}+y}}+y}$

$\lim _{y\rightarrow \infty }\dfrac {y\cdot \sqrt {1+\dfrac {1}{y}}}{y\sqrt {( 1+\dfrac {\sqrt {1+\dfrac {1}{y}}}{y}}+y}$

$\lim _{y\rightarrow \infty }\dfrac {y\sqrt {1+\dfrac {1}{y}}}{y\left( 1+\sqrt {1+\dfrac {\sqrt {1+\dfrac {1}{y}}}{y}}\right) }$

$=\dfrac {1}{2}$