factoring squareroots in limits

**casesam** · January 21st 2008, 06:13 PM

i have been stuck on this problem for awhile and would love some help.

lim ((1/sqrt(x))-(1/sqrt(x^2+x))
x-->0+

thanks ahead of time

-Dan

**ThePerfectHacker** · January 21st 2008, 06:57 PM

$\frac{1}{\sqrt{x}} - \frac{1}{\sqrt{x^2+x}} = \frac{\sqrt{x^2+x} - \sqrt{x}}{\sqrt{x} \sqrt{x^2+x}} \cdot \frac{\sqrt{x^2+x} + \sqrt{x}}{\sqrt{x^2+x} + \sqrt{x}}$

$\frac{x^2}{\sqrt{x} \cdot \sqrt{x} \cdot \sqrt{x+1} \cdot \sqrt{x} \cdot (\sqrt{x+1} + 1)}$

**mr fantastic** · January 21st 2008, 06:59 PM

Originally Posted by casesam

i have been stuck on this problem for awhile and would love some help.

lim ((1/sqrt(x))-(1/sqrt(x^2+x))
x-->0+

thanks ahead of time

-Dan

Consider:

$\frac{1}{\sqrt{x}} - \frac{1}{\sqrt{x^2 + x}} = \frac{1}{\sqrt{x}} - \frac{1}{\sqrt{x} \sqrt{x + 1}} = \frac{\sqrt{x + 1}}{\sqrt{x} \sqrt{x + 1}} - \frac{1}{\sqrt{x} \sqrt{x + 1}}$

$= \frac{\sqrt{x + 1} - 1}{\sqrt{x} \sqrt{x + 1}} = \frac{(\sqrt{x + 1} - 1)}{\sqrt{x} \sqrt{x + 1}} \times \frac{(\sqrt{x + 1} + 1)}{(\sqrt{x + 1} + 1)} = \frac{(x + 1) - 1}{(\sqrt{x} \sqrt{x + 1} (\sqrt{x + 1} + 1)}$

$= \frac{x}{(\sqrt{x} \sqrt{x + 1} (\sqrt{x + 1} + 1)} = \frac{\sqrt{x}}{\sqrt{x + 1} (\sqrt{x + 1} + 1)}$ .

Therefore ......

**Soroban** · January 21st 2008, 07:05 PM

Hello, Dan!

Wow . . . This is an ugly one!

$\lim_{x\to0^+} \left(\frac{1}{\sqrt{x}} - \frac{1}{\sqrt{x^2+x}}\right)$

The second fraction is: . $\frac{1}{\sqrt{x(x+1)}} \;=\;\frac{1}{\sqrt{x}}\cdot\frac{1}{\sqrt{x+1}}$

So we have: . $\frac{1}{\sqrt{x}} - \frac{1}{\sqrt{x}}\cdot\frac{1}{\sqrt{x+1}} \;=\;\frac{1}{\sqrt{x}}\left(1 - \frac{1}{\sqrt{x+1}}\right) \;=\;\frac{1}{\sqrt{x}}\left(\frac{\sqrt{x+1} - 1}{\sqrt{x+1}}\right)$

Multiply top and bottom by $(\sqrt{x+1} + 1)$

. . $\frac{1}{\sqrt{x}}\cdot\frac{\sqrt{x+1}-1}{\sqrt{x+1}}\cdot\frac{\sqrt{x+1} + 1}{\sqrt{x+1} + 1} \;=\;\frac{1}{\sqrt{x}}\cdot\frac{(x+1)-1}{\sqrt{x+1}(\sqrt{x+1} + 1)}$

. . . . $= \;\frac{1}{\sqrt{x}}\cdot\frac{x}{\sqrt{x+1}(\sqrt {x+1} + 1)} \;=\;\frac{\sqrt{x}}{\sqrt{x+1}(\sqrt{x+1} + 1)}$

Hence: . $\lim_{x\to0^+}\left[\frac{\sqrt{x}}{\sqrt{x+1}(\sqrt{x+1} + 1)}\right] \;=\;\frac{\sqrt{0}}{\sqrt{1}(\sqrt{1} + 1)} \;=\;\frac{0}{2} \;=\;0$

**topsquark** · January 21st 2008, 09:14 PM

Rationalizing the numerator. Gets 'em every time!

-Dan

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