Solve using the limit's definition

Question

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 !

Answer 1

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} $.

Answer 2

The solution goes as follows: $$\lim_{x\to\infty}\frac{\sqrt{x+1}}{\sqrt{x}+1}$$ $$=\lim_{\frac{1}{x}\to 0}\frac{\sqrt{1+\frac{1}{x}}}{1+\frac{1}{\sqrt{x}}}$$ $$=\frac{1+0}{1+0}$$ $$=1$$