Evaluating $\lim_{x \to 0} \frac{\sqrt{x+9}-3}{x}$ The question is this. 

In $h(x) = \dfrac{\sqrt{x+9}-3}{x}$, show that $\displaystyle \lim_{x \to 0} \ h(x) = \frac{1}{6}$, but that $h(0)$ is undefinied. 

In my opinion if I use this expression $\displaystyle \lim_{x \to 0} \dfrac{\sqrt{x+9}-3}{x}$  above with the $-3$ inside the square root I got an undefinied expression, but if I put the $-3$ out of the square and I use this expression to calculate the limit $\displaystyle \lim_{x \to 0} \dfrac{\sqrt{x+9}-3}{x}$ I will get $\frac{1}{6}$.
Here a print screen of the original question. 
If needed i can post the Pdf of the homework. 

You cannot pull the negative 3 out of the square root.  For example:
$$\sqrt{1-3} = \sqrt{2}i \ne \sqrt{1} - 3 = -2$$
