Decide whether or not a function is continuous Is the piecewise function 
$$f(x)=\begin{cases}
      \frac{x^2-x-6}{x-3} & x>3 \\
      \frac{\sqrt{x}-3}{x-9} & x \leq 3 
\end{cases}
$$
continuous?
I've done the following:
$\lim_{x\rightarrow3^+}\frac{\sqrt{x}-3}{x-9}=\frac{\sqrt{3}-3}{3^+ -9}=\frac{\sqrt{3}-3}{6}$
$\lim_{x\rightarrow3^-}\frac{\sqrt{x}-3}{x-9}=\lim_{x\rightarrow3^-}\frac{(x-3)(x+2)}{(x-3)}=\lim_{x\rightarrow3^-}(x+2)=3^- +2=5$
Since $\lim_{x\rightarrow3^+} f(x)\neq \lim_{x\rightarrow3^-} f(x)$, the function is not continuous.
Any thoughts whether this is correct?

Let us find the value of $f(3)$ first (i.e. $x=3$).
$$f(3)=\frac{\sqrt{3}-3}{3-9}=-\frac{\sqrt{3}-3}{6}$$
Now, let us find the left-side limit as $x\to3$ (i.e. $x\lt3$).
$$\lim\limits_{x\to3^-}(f(x))=\frac{\sqrt{3}-3}{3-9}=-\frac{\sqrt{3}-3}{6}$$
The left-side limit as $x\to3$ is equal to $f(3)$. Let us find the right-side limit as $x\to3$ (i.e. $x\gt3$).
$$\lim\limits_{x\to3^+}(f(x))=\lim\limits_{x\to3^+}\frac{x^2-x-6}{x-3}=\lim\limits_{x\to3^+}\frac{(x-3)(x+2)}{(x-3)}=\lim\limits_{x\to3^+}(x+2)=5$$
The left-side and right-side limits as $x\to3$ are not equal. Thus, the ordinary limit does not exist and the function is not continuous. There exists a jump discontinuity at $x=3$.
