Finding $\lim\limits_{h \to 0}\frac{(\sqrt{9+h} -3)}{h}$ I have about 10 of these problems I could post, can't figure any of them out...not sure what to do.
$$\lim_{h \to 0}\dfrac{(\sqrt{9+h} -3)}{h}$$
I know that I need to rationalize it so I multiply by the numerator which gives me 
$9 + h - 9$ (not sure if positive or negative since it is $-3$ squared it should be $9$ but it could also be minus square of $3$ which would be -9
/h sqr(9+h) -3 
From that I can divide by h and get 1/3-3
which I know is wrong. What am I doing wrong?
 A: In response to a comment by Jordan Carlyon. We start by rationalizing the numerator. To do that we multiply and divide $\sqrt{9+h}-3$ by $\sqrt{9+h}+3$. In general $\sqrt{a}-b$ would be rationalized by multiplying and dividing by $\sqrt{a}+b$.
$$\begin{eqnarray*}
\lim_{h\rightarrow 0}\frac{\sqrt{9+h}-3}{h} &=&\lim_{h\rightarrow 0}\frac{
\left( \sqrt{9+h}-3\right) \left( \sqrt{9+h}+3\right) }{h\left( \sqrt{9+h}
+3\right) } \\
&=&\lim_{h\rightarrow 0}\frac{\left( \sqrt{9+h}\right) ^{2}-3^{2}}{h\left( 
\sqrt{9+h}+3\right) } \\
&=&\lim_{h\rightarrow 0}\frac{9+h-9}{h\left( \sqrt{9+h}+3\right) } \\
&=&\lim_{h\rightarrow 0}\frac{h}{h\left( \sqrt{9+h}+3\right) } \\
&=&\lim_{h\rightarrow 0}\frac{1}{\sqrt{9+h}+3} \\
&=&\frac{1}{\displaystyle\lim_{h\rightarrow 0}\sqrt{9+h}+3} \\
&=&\frac{1}{\sqrt{9}+3} \\
&=&\frac{1}{6}
\end{eqnarray*}$$
Alternatively, you may use use L'Hôpital's rule, as I wrote in my reply to your 1st question.
Added: Since you have "not learned about derivatives yet" and this rule uses the evaluation of derivatives, it is to be learned later. 
A: Consider $f:[0,\infty[\to \mathbb{R}:x\mapsto \sqrt{x}$. Then
$$\begin{align*}
f'(9)&=\lim_{h\to 0} \frac{f(9+h)-f(9)}{h}\\
&= \lim_{h\to 0} \frac{\sqrt{9+h}-3}{h},\\
f'(x)&= \frac{1}{2\sqrt{x}},
\end{align*}$$
so $$f'(9)=\frac{1}{6}.$$
A: Let $t=\sqrt{9+h}$. Then $h=t^2-9$ and $h \to 0$ iff $t \to 3$, so your limit is equal to $\lim_{t \to 3} \frac{t-3}{t^2 - 9}$. Since $t^2 -9 = (t-3)(t+3)$, this is just $\lim_{t \to 3} \frac{1}{t+3} = \frac{1}{6}$.
A: Posting as I havent seen anyone suggest Taylor expansion yet.
$$\lim_{h \to 0}\dfrac{\sqrt{9+h} -3}{h} = \lim_{h \to 0}\dfrac{3\sqrt{1+h/9} -3}{h}$$
Taylor expanding the numerator becomes  $3(1+ \frac{1}{2}\cdot\frac{h}{9}) -3 = \frac{h}{6} + O(h^2)$
$$\lim_{h\to 0}
\dfrac{h/6+O(h^2)}{h} = \frac{1}{6}$$
A: If $\rm\ f(x)\: = \ f_0 + f_1\ x +\:\cdots\:+f_n\ x^n\:$ and $\rm\: f_0 \ne 0\:$ then rationalizing the numerator below yields
$$\rm \lim_{x\:\to\: 0}\ \dfrac{\sqrt{f(x)}-\sqrt{f_0}}{x}\ = \ \lim_{x\:\to\: 0}\ \dfrac{f(x)-f_0}{x\ (\sqrt{f(x)}+\sqrt{f_0})}\ =\ \dfrac{f_1}{2\ \sqrt{f_0}}$$
Your problem is the special case  $\rm\ f(x) = 9 + x\ $ with $\rm\ f_0 =9,\ f_1 = 1\:,\:$ so the limit equals $\:1/6\:.\:$  
When you study derivatives you'll see how they mechanize this process in a very general way. Namely the above limit is $\rm\:g'(0)\ $ for $\rm\:g(x) = \sqrt{f(x)}\:,\:$ so applying general rules for calculating derivatives we easily mechanically calculate that $\rm\:g'(x)\: =\: f\:\:'(x)/(2\:\sqrt{f(x)})\:.\:$ Evaluating it at $\rm\:x=0\:$ we conclude that $\rm\: g'(0)\: =\: f\:\:'(0)/(2\:\sqrt{f(0)})\: =\: f_1/(2\:\sqrt{f_0})\:,\:$ exactly as above.
