What is the Limit of a Radical Expression with a Variable in the Denominator?

[Bear_B]

Homework Statement

lim as h->0 of (f(x(sub 0)+h)-f(x(sub 0))/h

f(x)=3\sqrt{x}+2
x(sub 0)=9

Homework Equations

limit laws and factoring (my first post, not sure what I need to write here)

The Attempt at a Solution

=lim as h->0 of (f(9+h)-f(9)0/h
=lim as h->0 of (3\sqrt{9+h}+2-(3\sqrt{9}+2)/h
=lim as h->0 of (3\sqrt{9+h}+2-3\sqrt{9}-2)/h
=lim as h->0 of (3\sqrt{9+h}-9)/h

I am stuck here. I know I need to somehow move h out of the denominator(most likely by factoring) but am stuck on how to deal with the radical in the numerator.

 

[vela]

Rationalize the numerator by multiplying both the top and bottom by 3\sqrt{9+h}+9.

 

[Bear_B]

Ok, so I multiplied the numerator and denominator by 3\sqrt{9+h}+9 and am still having problems, here's what I got:

=lim as h->0 of (3\sqrt{9+h}-9)/h * (3\sqrt{9+h}+9)/(3\sqrt{9+h}+9)

=lim as h->0 of (9(9+h)+27\sqrt{9+h}-27\sqrt{9+h}-81)/(3h\sqrt{9+h}+9h)

=lim as h->0 of (81+h-81)/(3h\sqrt{9+h}+9h)

=lim as h->0 of h/(3h\sqrt{9+h}+9h)

ok, I am stuck here again and don't see what I did wrong or what I still need to do

 

[Bohrok]

You didn't distribute the 9 in the second line of your work; that will give you 9h in the numerator. Also, don't distribute the h in the denominator.
This is what you should have now without distributing it:

\lim_{h\rightarrow 0} \frac{9h}{h(3\sqrt{9 + h} + 9)}

The h's cancel and then you can let h=0 in the expression.

 

[Bear_B]

Thanks for the help vela, you got me on the right track, and thanks Bohrok, you caught my mistake. I started over and worked it out and got the right answer.