Why the root , Dude ?

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

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

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

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

$= \displaystyle \lim_{x \to \infty} \frac{\sqrt{\sqrt{x}+1}}{\sqrt{\sqrt{x}+\sqrt{\sqrt{x}+1}}+1}$

$= \displaystyle \lim_{x \to \infty} \frac{\frac{\sqrt{\sqrt{x}+1}}{\sqrt{\sqrt{x}}}}{\frac{\sqrt{\sqrt{x}+\sqrt{\sqrt{x}+1}}+1}{\sqrt{\sqrt{x}}}}$

$= \displaystyle \lim_{x \to \infty} \frac{1+\frac{1}{\sqrt{x}}}{\sqrt{1+\sqrt{1+\frac{1}{\sqrt{x}}}}+\frac{1}{\sqrt{\sqrt{x}}}}$

$= \displaystyle \frac{1+0}{1+1+0+0} = \boxed{\displaystyle \frac{1}{2}}$ ~~~

We can easily solve this problem by using rule#3 and rule#2 .

$\lim _{ x\rightarrow \infty }{ \sqrt { x+\sqrt { x+\sqrt { x } } } } -\sqrt { x }$ if we see this kind of problems ,the first thing we need to do is to move the root and the best way to do that is by making it a^2-b^2[rule #3] so,

$\lim _{ x\rightarrow \infty }{ \frac { (\sqrt { x+\sqrt { x+\sqrt { x } } } -\sqrt { x } )(\sqrt { x+\sqrt { x+\sqrt { x } } } +\sqrt { x } ) }{ \sqrt { x+\sqrt { x+\sqrt { x } } } +\sqrt { x } } }$

$\lim _{ x\rightarrow \infty }{ \frac { (\sqrt { x+\sqrt { x+\sqrt { x } } } )^{ 2 }-(\sqrt { x } )^{ 2 } }{ \sqrt { x+\sqrt { x+\sqrt { x } } } +\sqrt { x } } }$

so we get,

$\lim _{ x\rightarrow \infty }{ \frac { x+\sqrt { x+\sqrt { x } } -x }{ \sqrt { x+\sqrt { x+\sqrt { x } } } +\sqrt { x } } }$

now is the time to apply rule#2. Which is to divide both numerator and denominator by x^n [where x^n is the highest exponent of x containing in the fraction]

$\lim _{ x\rightarrow \infty }{ \frac { \frac { \sqrt { x+\sqrt { x } } }{ \sqrt { x } } }{ \frac { \sqrt { x+\sqrt { x+\sqrt { x } } } +\sqrt { x } }{ \sqrt { x } } } }$

$\lim _{ x\rightarrow \infty }{ \frac { \frac { \sqrt { x } }{ \sqrt { x } } +\frac { \sqrt { \sqrt { x } } }{ \sqrt { x } } }{ \frac { \sqrt { x } }{ \sqrt { x } } +\frac { \sqrt { \sqrt { x } } }{ \sqrt { x } } +\frac { \sqrt { \sqrt { \sqrt { x } } } }{ \sqrt { x } } +\frac { \sqrt { x } }{ \sqrt { x } } } }$

$\lim _{ x\rightarrow \infty }{ \frac { 1+\frac { \sqrt { \sqrt { x } } }{ \sqrt { x } } }{ 2+\frac { \sqrt { \sqrt { x } } }{ \sqrt { x } } +\frac { \sqrt { \sqrt { \sqrt { x } } } }{ \sqrt { x } } } }$

$\lim _{ x\rightarrow \infty }{ \frac { 1+\frac { \sqrt { \sqrt { x } } }{ \sqrt { x } } }{ 2+\frac { \sqrt { \sqrt { x } } }{ (\sqrt { \sqrt { x } } )^{ 2 } } +\frac { \sqrt { \sqrt { \sqrt { x } } } }{ (\sqrt { \sqrt { \sqrt { x } } } )^{ 4 } } } }$

As x tends to infinity we get

$\frac { 1+0 }{ 2+0+0 }$

hence we get the 1/2 which is also 0.5

take x common, apply approximation .

just rationalize and divide both numerator and denominator by x ^ 1/2