Piece-wise Continuous
A function \( f(t) \) is called piecewise continuous (or sectionally continuous) on the finite interval \( [a,b] \) if the interval can be split into a finite number of smaller parts such that:
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The function \( f \) remains continuous within the interior of each subinterval.
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As \( t \) nears either end of a subinterval from the inside, the value \( f(t) \) approaches a finite limit.
Suppose \( f \) is piecewise continuous on the interval \( a \le t \le b \). This means we can divide the interval into a few parts where \( f \) behaves nicely. Let \( t_0 \) be a point inside the interval (with \( a \lt t_0 \lt b \)) that is an end of one of these parts.
The value that \( f(t) \) gets close to as \( t \) approaches \( t_0 \) from the left (smaller values) is called the left-hand limit. We write it as \( \lim_{t \to t_0^-} f(t) \) or \( f(t_0^-) \). For example, consider the function:
Example-1
\[ f(t) = \begin{cases}
t, & \text{if } t \lt 2 \\
3, & \text{if } t \ge 2
\end{cases} \]
In this case, when \( t \) gets close to 2 from values less than 2, the left-hand limit is \( \lim_{t \to 2^-} f(t) = 2 \).
Similarly, the value that \( f(t) \) gets close to as \( t \) approaches \( t_0 \) from the right (larger values) is called the right-hand limit. It is written as \( \lim_{t \to t_0^+} f(t) \) or \( f(t_0^+) \). In our example, as \( t \) approaches 2 from values greater than 2, the right-hand limit is \( \lim_{t \to 2^+} f(t) = 3 \).
Notice that at \( t_0 = 2 \), both the left-hand and right-hand limits exist and are finite. However, they are not equal in this example. This shows that the limits from both sides do not have to be the same.
If a function \( f \) is continuous on the entire interval \( a \le t \le b \), it is automatically piecewise continuous. Moreover, if \( f \) is piecewise continuous on \( a \le t \le b \), then it can be integrated over that interval.
For another example, consider the function:
Example-2
\( g(t) \) defined by:
\[ g(t) = \begin{cases}
t^2, & \text{if } t \lt 1 \\
2t, & \text{if } t \ge 1
\end{cases} \]
At \( t_0 = 1 \), the left-hand limit is \( \lim_{t \to 1^-} g(t) = 1^2 = 1 \) and the right-hand limit is \( \lim_{t \to 1^+} g(t) = 2(1) = 2 \). This example again shows that the two limits at the point \( t_0 \) can be different.
Below are three simple examples that show how a function can be piecewise continuous with clear left-hand and right-hand limits.
Example 3
Next, look at the function:
\[
g(t) = \begin{cases}
\cos(t), & \text{if } t \lt \frac{\pi}{4} \\[6pt]
0, & \text{if } t \ge \frac{\pi}{4}
\end{cases}
\]
At \( t_0 = \frac{\pi}{4} \), the left-hand limit is \(\lim_{t \to (\frac{\pi}{4})^-} g(t) = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\). This is the value when \( t \) approaches from the left.
The right-hand limit is \(\lim_{t \to (\frac{\pi}{4})^+} g(t) = 0\), showing the value when \( t \) approaches from the right.
Example 4: Oscillating Function
Consider the function:
\[
f(t) = \sin\left(\frac{1}{t}\right)
\]
This function is defined for \( t \) in the interval \( 0 \lt t \le 1 \). Even if we define \( f(0) \) as 0, the values of \( f(t) \) change too rapidly near \( t = 0 \). The limit as \( t \) approaches 0 does not exist. Thus, \( f(t) \) is not piecewise continuous on [0,1].
Example 5: Dirichlet Function
Consider the function that takes the value 1 for rational numbers and 0 for irrational numbers on the interval [0,1]:
\[
f(t) = \begin{cases}
1, & \text{if } t \text{ is rational} \\[6pt]
0, & \text{if } t \text{ is irrational}
\end{cases}
\]
This function jumps between 1 and 0 at every point. Since there is no way to divide the interval [0,1] into a finite number of parts where the function is continuous, \( f(t) \) is not piecewise continuous.
Example 6: Function with a Vertical Asymptote
Consider the function:
\[
f(t) = \frac{1}{t-1}
\]
on the interval [0,2]. The function has a vertical asymptote at \( t = 1 \), which means the values of \( f(t) \) go to infinity as \( t \) nears 1 from either side. Since the one-sided limits at \( t = 1 \) are not finite, the function \( f(t) \) is not piecewise continuous on [0,2].
