How to find continuities with piecewise? $$f(x) = \begin{cases}x &\text{if}& x\le 0 \\ x^2 &\text{if}&  0\lt x\lt 1 \\ 2-x &\text{if}&x \ge 1\end{cases}$$
I don't understand how to begin solving this problem.Aren't continuities usually solved by looking at the denominator, setting it to zero and solving? There is no denominator in a piecewise function. Is this a trick question?
The Stackexchange site is glitching and it won't display the question properly. Here is the pastebin of the problem: http://pastebin.com/U4mAyvwn

Hint. Since the function is continuous on open intervals, potential issues are at endpoints.
$$f(x) = \begin{cases}x &\text{if}& x\le 0 \\ x^2 &\text{if}&  0\lt x\lt 1 \\ 2-x &\text{if}&x \ge 1.\end{cases}$$
Then here you just have to check if
$$
\begin{align}
\lim_{x \to 0^-}f(x)&=\lim_{x \to 0^+}f(x)\\
\lim_{x \to 1^-}f(x)&=\lim_{x \to 1^+}f(x).
\end{align}
$$
Can you take it from here?
