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. 

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*Way to solve this:
$$ \lim_{x \to 0} \frac{\sqrt{x+9}-3}{x}=\lim_{x \to 0} (\frac{\sqrt{x+9}-3}{x}*\frac{\sqrt{x+9}+3}{\sqrt{x+9}+3})=\lim_{x \to 0} (\frac{x}{x*(\sqrt{x+9}+3)})=\lim_{x \to 0} (\frac{1}{\sqrt{x+9}+3})=\frac{1}{6}$$


*Way to solve this:
Use L'Hospital rule:
$$ \lim_{x \to 0} \frac{\sqrt{x+9}-3}{x}=\lim_{x \to 0} (\frac{\sqrt{x+9}-3)'}{x'}=\lim_{x \to 0} \frac{\frac{1}{2*\sqrt{x+9}}}{1}=\frac{1}{6}$$

