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. 

Of course $h(0)$ is undefined as you cannot divide by zero. As you said $\displaystyle\lim_{x\rightarrow 0}\frac{\sqrt{x+9}-3}x=\frac 1 6$, according to L'Hospital's rule. Why do you care about $\displaystyle\frac{\sqrt{x+9-3}}x\ne h(x)$?
