$\displaystyle

\lim_{x\to 0} = \frac{\sqrt{x+5}-\sqrt{5}}{x}$

and what I got was

$\displaystyle

\frac{1}{\sqrt{5}+\sqrt{5}}

$

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- Jul 28th 2009, 06:25 AM #1

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- Jul 28th 2009, 06:42 AM #2
$\displaystyle \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {x + 5} - \sqrt 5 }}

{x} = \mathop {\lim }\limits_{x \to 0} \frac{{\left( {\sqrt {x + 5} - \sqrt 5 } \right)\left( {\sqrt {x + 5} + \sqrt 5 } \right)}}

{{x\left( {\sqrt {x + 5} + \sqrt 5 } \right)}} =$

$\displaystyle = \mathop {\lim }\limits_{x \to 0} \frac{{x + 5 - 5}}

{{x\left( {\sqrt {x + 5} + \sqrt 5 } \right)}} = \mathop {\lim }\limits_{x \to 0} \frac{1}{{\sqrt {x + 5} + \sqrt 5 }} =$

$\displaystyle = \frac{1}{{\sqrt {0 + 5} + \sqrt 5 }} = \frac{1}{{2\sqrt 5 }} = \frac{{\sqrt 5 }}{{10}}.$

- Jul 28th 2009, 06:51 AM #3

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- Jul 28th 2009, 06:55 AM #4

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In the future, when you ask for a check on what you did, it would be better to show how you got the result rather than just the result.

Rationalize the numerator by multiplying numerator and denominator by $\displaystyle \sqrt{x+5}+ \sqrt{5}$:

$\displaystyle \frac{x+ 5- x}{x(\sqrt{x+5}+ \sqrt{x})}= \frac{x}{x(\sqrt{x+5}+ \sqrt{5}}$$\displaystyle \frac{1}{\sqrt{x+5}+ \sqrt{5}}$.

Now you can put x= 0 and get $\displaystyle \frac{1}{\sqrt{5}+\sqrt{5}}$, what you have. But I would recommend that you write as $\displaystyle \frac{1}{2\sqrt{5}}$ or even $\displaystyle \frac{\sqrt{5}}{10}$.

Blast! DeMath got in ahead of me!

- Jul 28th 2009, 07:16 AM #5

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This is anotherone I tried doing today.

$\displaystyle

\lim_{x\to 0} = \frac{\frac{1}{3+x}-\frac{1}{3}}{x}

$

$\displaystyle

\frac{\frac{3-3+x}{3(3+x)}}{x}

$

$\displaystyle

\frac{\frac{x}{9+3x}}{x}

=9?

$

- Jul 28th 2009, 08:43 AM #6