Continuity over an Interval
Now that we have explored the concept of continuity at a point, we extend that idea to continuity over an interval. As we develop this idea for different types of intervals, it may be useful to keep in mind the intuitive idea that a function is continuous over an interval if we can use a pencil to trace the function between any two points in the interval without lifting the pencil from the paper. In preparation for defining continuity on an interval, we begin by looking at the definition of what it means for a function to be continuous from the right at a point and continuous from the left at a point.
Continuity from the Right and from the Left
A function \(f \left(\right. x \left.\right)\) is said to be continuous from the right at a if \(\underset{x \rightarrow a^{+}}{\text{lim}} f \left(\right. x \left.\right) = f \left(\right. a \left.\right) .\)
A function \(f \left(\right. x \left.\right)\) is said to be continuous from the left at a if \(\underset{x \rightarrow a^{-}}{\text{lim}} f \left(\right. x \left.\right) = f \left(\right. a \left.\right) .\)
A function is continuous over an open interval if it is continuous at every point in the interval. A function \(f \left(\right. x \left.\right)\) is continuous over a closed interval of the form \(\left[\right. a , b \left]\right.\) if it is continuous at every point in \(\left(\right. a , b \left.\right)\) and is continuous from the right at a and is continuous from the left at b. Analogously, a function \(f \left(\right. x \left.\right)\) is continuous over an interval of the form \(\left(\right. a , b \left]\right.\) if it is continuous over \(\left(\right. a , b \left.\right)\) and is continuous from the left at b. Continuity over other types of intervals are defined in a similar fashion.
Requiring that \(\underset{x \rightarrow a^{+}}{\text{lim}} f \left(\right. x \left.\right) = f \left(\right. a \left.\right)\) and \(\underset{x \rightarrow b^{-}}{\text{lim}} f \left(\right. x \left.\right) = f \left(\right. b \left.\right)\) ensures that we can trace the graph of the function from the point \(\left(\right. a , f \left(\right. a \left.\right) \left.\right)\) to the point \(\left(\right. b , f \left(\right. b \left.\right) \left.\right)\) without lifting the pencil. If, for example, \(\underset{x \rightarrow a^{+}}{\text{lim}} f \left(\right. x \left.\right) \neq f \left(\right. a \left.\right) ,\) we would need to lift our pencil to jump from \(f \left(\right. a \left.\right)\) to the graph of the rest of the function over \(\left(\right. a , b \left]\right. .\)
Example 2.33
Continuity on an Interval
State the interval(s) over which the function \(f \left(\right. x \left.\right) = \frac{x - 1}{x^{2} + 2 x}\) is continuous.
Solution
Since \(f \left(\right. x \left.\right) = \frac{x - 1}{x^{2} + 2 x}\) is a rational function, it is continuous at every point in its domain. The domain of \(f \left(\right. x \left.\right)\) is the set \(\left(\right. − \infty , −2 \left.\right) \cup \left(\right. −2 , 0 \left.\right) \cup \left(\right. 0 , + \infty \left.\right) .\) Thus, \(f \left(\right. x \left.\right)\) is continuous over each of the intervals \(\left(\right. − \infty , −2 \left.\right) , \left(\right. −2 , 0 \left.\right) ,\) and \(\left(\right. 0 , + \infty \left.\right) .\)
Example 2.34
Continuity over an Interval
State the interval(s) over which the function \(f \left(\right. x \left.\right) = \sqrt{4 - x^{2}}\) is continuous.
Solution
From the limit laws, we know that \(\underset{x \rightarrow a}{\text{lim}} \sqrt[]{4 - x^{2}} = \sqrt{4 - a^{2}}\) for all values of a in \(\left(\right. −2 , 2 \left.\right) .\) We also know that \(\underset{x \rightarrow −2^{+}}{\text{lim}} \sqrt{4 - x^{2}} = 0\) exists and \(\underset{x \rightarrow 2^{-}}{\text{lim}} \sqrt{4 - x^{2}} = 0\) exists. Therefore, \(f \left(\right. x \left.\right)\) is continuous over the interval \(\left[\right. −2 , 2 \left]\right. .\)
Checkpoint 2.24
State the interval(s) over which the function \(f \left(\right. x \left.\right) = \sqrt{x + 3}\) is continuous.
The Composite Function Theorem allows us to expand our ability to compute limits. In particular, this theorem ultimately allows us to demonstrate that trigonometric functions are continuous over their domains.
Theorem 2.9
Composite Function Theorem
If \(f \left(\right. x \left.\right)\) is continuous at L and \(\underset{x \rightarrow a}{\text{lim}} g \left(\right. x \left.\right) = L ,\) then
Before we move on to Example 2.35, recall that earlier, in the section on limit laws, we showed \(\underset{x \rightarrow 0}{\text{lim}} \text{cos} x = 1 = \text{cos} \left(\right. 0 \left.\right) .\) Consequently, we know that \(f \left(\right. x \left.\right) = \text{cos} x\) is continuous at 0. In Example 2.35 we see how to combine this result with the composite function theorem.
Example 2.35
Limit of a Composite Cosine Function
Evaluate \(\underset{x \rightarrow \pi / 2}{\text{lim}} \text{cos} \left(\right. x - \frac{\pi}{2} \left.\right) .\)
Solution
The given function is a composite of \(\text{cos} x\) and \(x - \frac{\pi}{2} .\) Since \(\underset{x \rightarrow \pi / 2}{\text{lim}} \left(\right. x - \frac{\pi}{2} \left.\right) = 0\) and \(\text{cos} x\) is continuous at 0, we may apply the composite function theorem. Thus,
Checkpoint 2.25
Evaluate \(\underset{x \rightarrow \pi}{\text{lim}} \text{sin} \left(\right. x - \pi \left.\right) .\)
The proof of the next theorem uses the composite function theorem as well as the continuity of \(f \left(\right. x \left.\right) = \text{sin} x\) and \(g \left(\right. x \left.\right) = \text{cos} x\) at the point 0 to show that trigonometric functions are continuous over their entire domains.
Theorem 2.10
Continuity of Trigonometric Functions
Trigonometric functions are continuous over their entire domains.
This lesson is part of:
Limits