I'm not sure if I got this one right

**drkidd22** · July 28th 2009, 06:25 AM

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

and what I got was

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

**DeMath** · July 28th 2009, 06:42 AM

Originally Posted by drkidd22

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

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

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

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

**drkidd22** · July 28th 2009, 06:51 AM

Oh ok, so I did do it right, just didn't complete the whole thing.

**HallsofIvy** · July 28th 2009, 06:55 AM

Originally Posted by drkidd22

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

and what I got was

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

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

Blast! DeMath got in ahead of me!

**drkidd22** · July 28th 2009, 07:16 AM

This is anotherone I tried doing today.

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

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

$ \frac{\frac{x}{9+3x}}{x} =9? $

**skeeter** · July 28th 2009, 08:43 AM

Originally Posted by drkidd22

This is anotherone I tried doing today.

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

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

$ \frac{\frac{x}{9+3x}}{x} =9? $

$\frac{\frac{3-(3+x)}{3(3+x)}}{x} = \frac{-x}{3x(3+x)} = \frac{-1}{3(3+x)}$

limit is $-\frac{1}{9}$

in future, start a new problem with a new thread.

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I'm not sure if I got this one right