<s>[INST] 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)?
 [/INST] The usual rationalization:
$\begin{array}\\
\sqrt{x+1}-\sqrt{x}
&=(\sqrt{x+1}-\sqrt{x})\frac{\sqrt{x+1}+\sqrt{x}}{\sqrt{x+1}+\sqrt{x}}\\
&=\frac{1}{\sqrt{x+1}+\sqrt{x}}\\
&< \frac1{2\sqrt{x}}
\end{array}
$
</s>