# Simple limit with radical, but I keep getting the wrong answer.

• Jan 26th 2010, 09:24 AM
Sirmio
Hello. I've worked this problem several times, but I keep coming up with the wrong answer and it's driving me mad. Here it is:

http://imgur.com/4cbgL.gif

The answer is 1/6, but I keep getting undefined. I multiply by the conjugate, then reduce, and I get: $\frac{1}{\sqrt{x+7}-3}$ I have no clue where I'm messing up with the algebra.

Thanks for any help.
• Jan 26th 2010, 09:36 AM
Plato
$\frac{{\sqrt {x + 7} - 3}}
{{x - 2}} = \frac{{(x + 7) - 9}}
{{(x - 2)\left( {\sqrt {x + 7} + 3} \right)}}$
• Jan 26th 2010, 09:45 AM
Soroban
Hello, Sirmio!

Your algebra must be off . . .

Quote:

$\lim_{x\to2}\,\frac{\sqrt{x+7} - 3}{x-2}$

Multiply top and bottom by the conjugate:

. . $\frac{\sqrt{x+7}-3}{x-2}\cdot\frac{\sqrt{x+7}+3}{\sqrt{x+7}+3} \;=\;\frac{(x+7) - 9}{(x-2)(\sqrt{x+7} + 3)}$ . $=\;\frac{{\color{red}\rlap{/////}}x-2}{({\color{red}\rlap{/////}}x-2)(\sqrt{x+7}+3)} \;=\;\frac{1}{\sqrt{x+7}+3}$

Therefore: . $\lim_{x\to2}\frac{1}{\sqrt{x+7}+3} \;=\;\frac{1}{\sqrt{9}+3} \;=\;\frac{1}{6}$

• Jan 26th 2010, 09:48 AM
Sirmio
Thanks! I didn't change the sign on the conjugate. :( I knew it was simple. Thanks for the help!