Problem 33
Question
The Heaviside Function A generalization of the unit step function or Heaviside
function \(H\) of Example 3 is the function \(H_{c}\) defined by
$$
H_{c}\left(t-t_{0}\right)=\left\\{\begin{array}{ll}
0 & \text { if } t
Step-by-Step Solution
Verified Answer
If \(c \neq 0\), then the limit as \(t\) approaches \(t_0\) of \(H_c(t-t_0)\) does not exist, because the left limit \(\lim_{t \rightarrow t_0^-} H_c(t-t_0) = 0\) and the right limit \(\lim_{t \rightarrow t_0^+} H_c(t-t_0) = c\) are not equal.
1Step 1: Define the left limit
We want to find the limit as \(t\) approaches \(t_0\) from the left, which can be written as \(\lim_{t \rightarrow t_0^-} H_c(t-t_0)\). In this case, when \(t < t_0\), the function is defined as \(H_c(t-t_0) = 0\). Therefore, the left limit is:
\[\lim_{t \rightarrow t_0^-} H_c(t-t_0) = 0\]
2Step 2: Define the right limit
Similarly, we want to find the limit as \(t\) approaches \(t_0\) from the right, which can be written as \(\lim_{t \rightarrow t_0^+} H_c(t-t_0)\). In this case, when \(t \geq t_0\), the function is defined as \(H_c(t-t_0) = c\). Therefore, the right limit is:
\[\lim_{t \rightarrow t_0^+} H_c(t-t_0) = c\]
3Step 3: Determine if the left and right limits are equal
We found that the left limit is 0, and the right limit is c. If \(c \neq 0\), then the left and right limits are not equal, which means the limit as \(t\) approaches \(t_0\) does not exist.
In conclusion, if \(c \neq 0\), then \(\lim_{t \rightarrow t_0} H_c(t-t_0)\) does not exist, because the left and right limits are not equal.
Key Concepts
Unit Step FunctionLeft LimitRight LimitLimits in Calculus
Unit Step Function
The unit step function, often called the Heaviside function, is a mathematical function that jumps from one value to another at a specified point. It's commonly denoted as \(H(t)\) and is defined as follows:
The Heaviside function effectively "turns on" at \(t = 0\), making it instrumental in systems that need a threshold or a starting point. The generalization, denoted by \(H_c(t-t_0)\), adds a level of customization by allowing different jump values \(c\) and shift points \(t_0\). This allows for modeling more complex real-world phenomena where the response may differ in magnitude or timing.
- \(H(t) = 0\) if \(t < 0\)
- \(H(t) = 1\) if \(t \geq 0\)
The Heaviside function effectively "turns on" at \(t = 0\), making it instrumental in systems that need a threshold or a starting point. The generalization, denoted by \(H_c(t-t_0)\), adds a level of customization by allowing different jump values \(c\) and shift points \(t_0\). This allows for modeling more complex real-world phenomena where the response may differ in magnitude or timing.
Left Limit
The concept of the left limit pertained to the behavior of a function as it approaches a particular point from the left side on a number line. In calculus, this is represented as \(\lim_{t \rightarrow t_0^-} f(t)\).
For the Heaviside function \(H_c(t-t_0)\), as \(t\) approaches \(t_0\) from the left (\(t < t_0\)), the function value is \(0\). Therefore, the left limit is denoted as:
For the Heaviside function \(H_c(t-t_0)\), as \(t\) approaches \(t_0\) from the left (\(t < t_0\)), the function value is \(0\). Therefore, the left limit is denoted as:
- \(\lim_{t \rightarrow t_0^-} H_c(t-t_0) = 0\)
Right Limit
The right limit is the behavior of a function as it approaches a point from the right side on the number line. This is written in calculus as \(\lim_{t \rightarrow t_0^+} f(t)\).
For the Heaviside function \(H_c(t-t_0)\), as \(t\) approaches \(t_0\) from the right (\(t \geq t_0\)), the function equals the constant \(c\). Therefore, the right limit is expressed as:
For the Heaviside function \(H_c(t-t_0)\), as \(t\) approaches \(t_0\) from the right (\(t \geq t_0\)), the function equals the constant \(c\). Therefore, the right limit is expressed as:
- \(\lim_{t \rightarrow t_0^+} H_c(t-t_0) = c\)
Limits in Calculus
Limits are foundational in calculus and provide a means to determine the behavior of a function as the input approaches a certain value. When examining limits, we particularly focus on whether the "left limit" and "right limit" at a point are equal.
For a limit to exist at any particular point \(t_0\), the following criteria must be met:
For a limit to exist at any particular point \(t_0\), the following criteria must be met:
- The left limit \(\lim_{t \rightarrow t_0^-} f(t)\) must exist.
- The right limit \(\lim_{t \rightarrow t_0^+} f(t)\) must exist.
- The left and right limits must be equal: \(\lim_{t \rightarrow t_0^-} f(t) = \lim_{t \rightarrow t_0^+} f(t)\).
Other exercises in this chapter
Problem 33
In Exercises 33-36, determine whether the function is continuous on the closed interval. \(f(x)=\sqrt{16-x^{2}}, \quad[-4,4]\)
View solution Problem 33
Determine whether the statement is true or false. If it is true, explain why. If it is false, explain why or give an example that shows it is false. The limit o
View solution Problem 34
Use the technique of Exercise \(33 \mathrm{a}-\mathrm{b}\) to find \(\lim _{h \rightarrow 0} \frac{f(8+h)-f(8)}{h}\) if \(f(x)=\sqrt[3]{x}\), using the viewing
View solution Problem 34
Determine whether the function is continuous on the closed interval. \(g(x)=\ln (x+3)+\sqrt{4-x^{2}}, \quad[-2,1]\)
View solution