At how many points is this piecewise function discontinuous? 
$$f(x) =
\begin{cases}
\frac{3x}{x^2-16}  & \text{when $x\leq 2$} \\[2ex]
\frac{x-2}{x^2-5x+6} & \text{when $x\gt 2$}
\end{cases}
$$
At what points is this piecewise function discontinuous?

My solution:
$x=-4, 4, 2, 3$ are the points  at which the denominator is zero, but $x=4\not\leq2$ and $x=2\not\gt2$. Therefore, the points are $x=-4$ and $x=3$. But the answer is given as $3$.
What am I doing wrong?
 A: You are correct about $x=-4$ and $x=3$, but you also need to check $x=2$ since that's the point where you jump from one function to the other.
For the piecewise function to be continuous at $x=2$ you need $$lim_{x \to 2} f(x) = f(2).$$  Since $f(x)$ is defined differently from the left and the right of $x=2$ you will need to instead use each one sided limit.

$$lim_{x \to 2^-} f(x) = f(2)$$
$$lim_{x \to 2^-} \frac{3x}{x^2-16} = \frac{3(2)}{(2)^2-16}$$
$$\frac{3(2)}{(2)^2-16} = \frac{3(2)}{(2)^2-16}$$
$$\frac{6}{-12}=\frac{6}{-12}$$
$$-\frac{1}{2}=-\frac{1}{2}$$
So this one works out.  Now we also need to check the right sided limit.

$$lim_{x \to 2^+} f(x) = f(2)$$
$$lim_{x \to 2^+} \frac{x-2}{x^2-5x+6} = \frac{3(2)}{(2)^2-16}$$
$$lim_{x \to 2^+} \frac{x-2}{(x-2)(x-3)} = -\frac{1}{2}$$
$$lim_{x \to 2^+} \frac{1}{x-3} = -\frac{1}{2}$$
$$\frac{1}{(2)-3} = -\frac{1}{2}$$
$$\frac{1}{-1} = -\frac{1}{2}$$
$$-1=-\frac{1}{2}$$
Clearly this is not true, $-1 \neq -\frac{1}{2}$.  Therefore the two sided limit doesn't exist and so we also have a discontinuity at $x=2$.
A: There are 3 points of discontinuity, see the graph bellow:

So you are particularly right. But for $x=2$:
\begin{aligned}
\lim_{x \to 2^-}\frac{3x}{x^2-16} &= -\frac 12, 
&&\text{while} \\[2ex]
\lim_{x \to 2^+}\frac{x - 2}{x^2-5x+6} &= \lim_{x \to 2^+} \frac{1}{x-3}=-1,
&&\text{so} \\[4ex] 
\lim_{x \to 2^-}f(x) &\ne \lim_{x \to 2^+}f(x)
\end{aligned}
