Limit with square roots

lim as x approaches 9 of (√(x)-3)/(x-9)

Im really rusty with my math, and I know I'm missing a step after multiplying by the conjugate....help?

√x -3
-------- =
...x-9

........√x - 3
---------------------- =
...(√x - 3)(√x + 3)

...1
---------
√x + 3

So limit as x approaches 9 is 1/6