# Continuous Functions

It is pretty easy to think about what we mean by a continuous function. Intuitively, it is a function whose graph we can draw without picking up our pencil. We can see intuitively that \(f(x)=x^2-3x+1\) is continuous, and that \[g(x)=\begin{cases} 0 & x\lt 0\\ 1 & x\ge 0 \end{cases} \] is not. On the other hand, if we think about the situation a little longer we can see that if we consider only positive \(x\) values then \(g\) does seem continuous. It turns out that the concept of the limit gives us a way of discussing continuity precisely.

Note that we were slightly cagey about specifying the interval. It could be open, as with \((0,1)\), it could be closed as for \([0,1]\), or it could be half open. We can say a little more about functions that are continuous on closed intervals, as we'll discuss later.

We have already seen several classes of functions that are continuous over their entire domains. In particular, [PolynomialLimit] shows that every polynomial is continuous over the entire real line. In the same way, [CosineLimit] shows that every cosine function is continuous. Since the sine function is just the cosine function shifted by \(\pi/2\), the sine function is also continuous over the real line.

In the case of the function \(g\) given earlier, we can see immediately that \(\lim_{x\to x_0}\) exists and is equal to 0 for every \(x_0\lt 0\), and it exists and is equal to 1 for every \(x_0>1.\). We conclude that \(g\) is continuous on \( (-\infty,0) \) and is continuous on \( (0,\infty) \). On the other hand, we have seen before that while \(g(0)\) is defined, the limit at \(x_0=0\) does not exist, so \(g\) is not continuous at zero or on any interval containing zero.

Because the definition of continuity is tied to limits, it follows that the algebra of continuous functions follows the algebra of limits. Here is a summary result we can refer to.

- \(f+g\) is continuous on \(I\);
- \(fg\) is continuous on \(I\); and in particular
- \(cf\) is continuous on \(I\).

*Proof:*We can prove the part about addition, just to see the flavor of the arguments. Suppose that for any point \(x_0\in I\) that \(f\) and \(g\) are both continuous at \(x_0\). Then by [LimitAddition] we have \[\lim_{x\to x_0} (f+g)(x) = \lim_{x\to x_0} f(x)+\lim_{x\to x_0} g(x) = f(x_0)+g(x_0) = (f+g)(x_0).\] We conclude that \(f+g\) is continuous everywhere that \(f\) and \(g\) are. The other results are equally simple.