How can I manipulate $\frac { \sqrt { x+1 } }{ \sqrt { x } +1 } $ to find $M>0$ to prove a limit? Given the following limit, find such an $M>0$ that for every $x>M$, the expression is $\frac { 1 }{ 3 }$ close to the limit. In other words find $M>0$ that for every $x>M:\left| f(x)-L \right| <\frac { 1 }{ 3 }$ for the following function:
$$\lim _{ x\rightarrow \infty  }{ \frac { \sqrt { x+1 }  }{ \sqrt { x } +1 } =1 } $$
Steps I took:
$$\left| \frac { \sqrt { x+1 }  }{ \sqrt { x } +1 } -1 \right| <\frac { 1 }{ 3 } $$
$$\Longrightarrow \left| \frac { \sqrt { x+1 }  }{ \sqrt { x } +1 } -\frac { \sqrt { x } +1 }{ \sqrt { x } +1 }  \right| <\frac { 1 }{ 3 } $$
$$\Longrightarrow \left| \frac { \sqrt { x+1 } -\sqrt { x } -1 }{ \sqrt { x } +1 }  \right| <\frac { 1 }{ 3 } $$
How can I manipulate the function inside of the absolute value in order to simplify this and find a lower bound $x$ (the M)?

First rationalization
i.e. $\left|\frac{\sqrt{x+1}-\sqrt{x}-1}{\sqrt{x}+1}\right|$
= $\left|\frac{(\sqrt{x+1}-\sqrt{x}-1)(\sqrt{x}-1)}{x-1}\right|$
=$\left|\frac{\sqrt{x(x+1)}-\sqrt{x+1}-x+1}{x-1}\right|$
Then, we could choose M to be a particular value to solve the problem(by taking maximum of M at last). In this case, take $M=2$
$\because x > M = 2$
$\frac{x}{2} > 1$
$\therefore$ $<\left|\frac{\sqrt{x+1}(\sqrt{x}-1)-x+1}{x-\frac{x}{2}}\right|$ 
=$\left|(\frac{2}{x})(\sqrt{x+1}(\sqrt{x}-1)-x+1)\right|$
<$\left|(\frac{2}{x})(\sqrt{x+x}(\sqrt{x}-1)-x+1)\right|,x>2$
=$\left|(\frac{2}{x})(2x-2\sqrt{x}-x+1)\right|$
=$\left|(\frac{2}{x})(1-\sqrt{x})^2\right|$
<$\left|(1-\sqrt{x})^2\right|$<$\frac{1}{3},x>2$
$\because (1-\sqrt{x})^2 < \frac{1}{3}$
$ x > (1-\sqrt\frac{1}{3})^2$
Therefore, take $M = max${${(1-\sqrt\frac{1}{3})^2},2$}$=2$
