Limits: sqrtx - 3 / x - 9 as x approaches 9

[peep1963]

Hi, I got a question on limits on my homework.

The problem is:

Find the limit as x is approaching 9
the square root of x minus 3 over x - 9

sorry I can't do the square root sign, but it looks something like this
/X|-3 / X-9

Anyways, the first thing i did was plug the nine in but that leaves a zero in the bottom. so then I did the conjugate. I got that the limit is -1/6 but the right answer is a postive 1/6. I don't get what I did wrong.

Please help.

 

[stapel]

sorry I can't do the square root sign....

For formatting advice, please follow the links in the "Forum Help" pull-down menu at the very top of the page. Thank you.

To clarify, is your limit as follows?

. . . . .lim_{x -> 9} (sqrt[x] - 3) / (x - 9)

In LaTeX formatting, the above looks like this:

. . . . .\(\displaystyle \large{\begin{array}{c}limit\\x\rightarrow 9\end{array}\,\,\frac{\sqrt{x}\,-\,3}{x\,-\,9}}\)

Thank you.

Eliz.

 

[pka]

\(\displaystyle \L
\frac{{\sqrt x - 3}}{{x - 9}}\left( {\frac{{\sqrt x + 3}}{{\sqrt x + 3}}} \right) = \frac{{x - 9}}{{\left( {x - 9} \right)\left( {\sqrt x + 3} \right)}}\)