We have the function $f:D \rightarrow R$ defined on the domain $D = [0,1] \cup (2,3]$.
The function is defined as
$f(x) = \left\{\begin{matrix} x&0 \leq x \leq 1 \\ x-1 & 2 < x \leq 3 \end{matrix}\right.$
At first, I though that maybe we had to prove that the limits as x approached 1 and x approached 2 had to be continuous, but I think I was confusing some of the stuff I learned a while back in introductory calculus.
Since our definition of continuity at a point only means that given any convergent sequence to some $x_0$ in our domain $D$ that the function of $x_n$ also converges to $f(x_n)$, does that just mean I have to show that the function behaves this way at all the points correct?
Well by previous assignments and investigations in the book, it's already been proven that polynomials are continuous over their whole domain. So in this case, our function is defined as two polynomials over continuous intervals, so does that mean that it follows that the function is continuous at every point?
Every sequence in our domain that converges to 1 must be in the interval [0,1] for if it was in (2,3] it could not get arbitrarily close to 1. The same argument applies for any sequence converging to 2 in the domain D.
Is this reasoning correct?
Thanks
Suppose $x_n \to x \in D$. Then for $N$ sufficiently large, you must have $x_n \in [0,1]$ for all $n \ge N$ or $x_n \in (2,3]$ for all $n \ge N$.
It is easy to see that $f$ is continuous when restricted to $[0,1]$ and similarly for the restriction to $(2,3]$.