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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)?

Cherry_Developer
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3 Answers3

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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} $

marty cohen
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    I can't seem to see the big picture. Meaning, how to apply what you've demonstrated to the actual problem. – Cherry_Developer Nov 15 '15 at 01:46
  • Combine this with what you have and you get $\left| \frac { \sqrt { x+1 } -\sqrt { x } -1 }{ \sqrt { x } +1 } \right| < |\frac{1+1/(2\sqrt{x})}{\sqrt{x}}|$. – marty cohen Nov 15 '15 at 02:51
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Hint for an intuitive solution:

$$\left|\frac{\sqrt{x+1}}{\sqrt{x}+1}-1\right|<\left|\frac{\sqrt{x+1}}{\sqrt{x}}-1\right|=\left|\sqrt{\frac{x+1}{x}}-1\right|=\left|\sqrt{1+\frac{1}{x}}-1\right|.$$ Now by letting that $x\to\infty$ then the result follows.

janmarqz
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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 $M = max${${(1-\sqrt\frac{1}{3})^2},2$}$=2$

lcn
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