Solve using the limit's definition The question is : 
I am stuck here :
$$| \sqrt{x+1} - \sqrt{x} -1 | / \sqrt{x} + 1$$
i know that the numerator is negetive so i must change it in order to delete the absolute value yet i still don't know how to proceed.
Thanks in advance !
 A: Let $\epsilon >0$, we are seeking for  $\delta >0$, such that , for all  $x\geq  0$ with $x>\delta$, we have  $$ \Bigg| \frac{\sqrt{x+1}}{\sqrt{x}+1} -1\Bigg| < 
\epsilon$$
 Indeed,   $$ \Bigg| \frac{\sqrt{x+1}}{\sqrt{x}+1} -1\Bigg|  = \Bigg| \frac{\sqrt{x+1} -\sqrt{x}-1}{\sqrt{x}+1} \Bigg|  $$
Note that , for all $x\geq 0$ we have  $\sqrt{x+1} \leq \sqrt{x}+ 1  $ (you can see it by squaring both sides). Thus $$ \Bigg| \frac{\sqrt{x+1} -\sqrt{x}-1}{\sqrt{x}+1} \Bigg|=   \frac{\sqrt{x}+1-\sqrt{x+1} }{\sqrt{x}+1} $$
But  $ \sqrt{x+1} \geq \sqrt{x}$ since the square root is an increasing function, and so  $-\sqrt{x+1} \leq -\sqrt{x} $  and  $\frac{1}{\sqrt{x+1}} \leq  \frac{1}{\sqrt{x}} $, hence 
$$ \Bigg| \frac{\sqrt{x+1}}{\sqrt{x}+1} -1\Bigg| = \frac{\sqrt{x}+1-\sqrt{x+1} }{\sqrt{x}+1}  \leq  \frac{\sqrt{x}+1-\sqrt{x} }{\sqrt{x}+1}= \frac{1}{\sqrt{x}+1} \leq \frac{1}{\sqrt{x}}  < \frac{1}{\sqrt{\delta}} $$
So if we choose  $\delta $ such that $ \frac{1}{\sqrt{\delta}} < \epsilon   $ , then we are done. So enough to tkae  $\delta > \frac{1}{\epsilon ^2} $.
A: The solution goes as follows:
$$\lim_{x\to\infty}\frac{\sqrt{x+1}}{\sqrt{x}+1}$$
$$=\lim_{\frac{1}{x}\to 0}\frac{\sqrt{1+\frac{1}{x}}}{1+\frac{1}{\sqrt{x}}}$$
$$=\frac{1+0}{1+0}$$
$$=1$$
