Problem 23
Question
Construct your own \(f(x)\) with these discontinuities at \(x=1\). Limit for \(x \rightarrow 1^{+},\) no limit for \(x \rightarrow 1^{-}\)
Step-by-Step Solution
Verified Answer
Function: \( f(x) = \frac{1}{x-1} \) for \( x \leq 1 \) and \( f(x) = 2 \) for \( x > 1 \). Right limit exists, left does not.
1Step 1: Understand the Requirements
To fulfill the requirements, we need to create a function \( f(x) \) with a discontinuity exactly at \( x = 1 \). Specifically, the function should have a limit from the right as \( x \to 1^+ \) and no limit from the left as \( x \to 1^- \).
2Step 2: Design the Function Piecewise
Since we need different behaviors from the left and right of \( x = 1 \), constructing a piecewise function is ideal. For \( x > 1 \), choose a simple function like a constant function \( f(x) = 2 \) to ensure a limit exists as \( x \to 1^+ \). For \( x \leq 1 \), choose a function that oscillates or is undefined at \( x = 1 \), such as \( f(x) = \frac{1}{x-1} \).
3Step 3: Define the Function
The function can be defined as: \[f(x) = \begin{cases} \frac{1}{x-1}, & \text{if } x \leq 1 \2, & \text{if } x > 1 \end{cases}\]In this piecewise function, as \( x \to 1^{+} \), the value of \( f(x) = 2 \), providing a clear limit from the right. As \( x \to 1^{-} \), \( f(x) = \frac{1}{x-1} \) becomes undefined.
4Step 4: Verify the Right-Sided Limit
Evaluate the limit from the right. Consider \( \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} 2 = 2 \). The limit exists and equals 2.
5Step 5: Verify the Left-Sided Behavior
Check the behavior from the left. As \( x \to 1^- \), \( f(x) = \frac{1}{x-1} \) approaches negative or positive infinity depending on the direction from which \( x \) approaches 1. Hence, no finite limit exists.
Key Concepts
Piecewise FunctionRight-Hand LimitUndefined LimitBehavior of Functions Around Points
Piecewise Function
A piecewise function is a type of function that is defined by multiple sub-functions, each of which applies to a specific interval of the domain. This means that the function can take different expressions based on the input value of the variable.
For example, consider the function we constructed: \[ f(x) = \begin{cases} \frac{1}{x-1}, & \text{if } x \leq 1 \ 2, & \text{if } x > 1 \end{cases} \]In this case, the function behaves differently depending on whether \(x\) is less than or equal to 1, or if \(x\) is greater than 1. Such a structure allows us to create specific behaviors around points of interest, such as discontinuities or abrupt changes in the function's values.
Piecewise functions are particularly useful in modeling real-world situations where a process might change behavior between different states or conditions.
For example, consider the function we constructed: \[ f(x) = \begin{cases} \frac{1}{x-1}, & \text{if } x \leq 1 \ 2, & \text{if } x > 1 \end{cases} \]In this case, the function behaves differently depending on whether \(x\) is less than or equal to 1, or if \(x\) is greater than 1. Such a structure allows us to create specific behaviors around points of interest, such as discontinuities or abrupt changes in the function's values.
Piecewise functions are particularly useful in modeling real-world situations where a process might change behavior between different states or conditions.
Right-Hand Limit
The right-hand limit of a function at a specific point refers to the value that the function approaches as the variable gets close to that point from the positive side (i.e., values greater than the point). Essentially, it involves examining the function's behavior strictly from one side.
In our constructed piecewise function, we are interested in the right-hand limit at \(x = 1\).
In our constructed piecewise function, we are interested in the right-hand limit at \(x = 1\).
- For \(x > 1\), \(f(x) = 2\), so \(\lim_{x \to 1^+} f(x) = 2\).
Undefined Limit
An undefined limit at a point occurs when a function does not approach a specific value as the variable approaches that point. This can happen for several reasons, such as the function oscillating indefinitely or becoming unbounded.
In our exercise, the function \(f(x) = \frac{1}{x-1}\) for \(x \leq 1\) is of particular interest for values as \(x\) approaches 1 from the left. As \(x\) approaches 1 from the negative side (or numbers less than 1), the function's behavior becomes undefined because it leads to division by zero.
In our exercise, the function \(f(x) = \frac{1}{x-1}\) for \(x \leq 1\) is of particular interest for values as \(x\) approaches 1 from the left. As \(x\) approaches 1 from the negative side (or numbers less than 1), the function's behavior becomes undefined because it leads to division by zero.
- As \(x\) nears 1 from the left, \(f(x)\) does not settle to a finite number; instead, it tends to either very large positive or negative values.
Behavior of Functions Around Points
The behavior of functions around specific points is crucial for understanding their overall continuity and forms a foundation for many calculus concepts. When assessing this behavior, we often look at both limits and actual function values at and around the points of interest.
In the context of our function, the point \(x = 1\) is critical. This point represents where different behaviors from each piece of our piecewise function meet.
In the context of our function, the point \(x = 1\) is critical. This point represents where different behaviors from each piece of our piecewise function meet.
- For \(x \leq 1\): The function \(f(x) = \frac{1}{x-1}\) indicates drastic behavior changes as \(x\) nears 1.
- For \(x > 1\): The function \(f(x) = 2\) denotes consistent behavior leading to a constant limit.
Other exercises in this chapter
Problem 23
Complete Pascal's triangle for \(n=5\) and \(n=6 .\) Why do the numbers across each row add to \(2^{n}\) ?
View solution Problem 23
Find the limits if they exist. An \(\varepsilon-\delta\) test is not required. $$ \lim _{x \rightarrow 0} \frac{\sin x}{\sin x / 2} $$
View solution Problem 23
Draw the graph of \(y=\sin \frac{1}{2} x .\) State its slope at \(x=0, \pi / 2\), \(\pi,\) and \(2 \pi .\) Does \(\frac{1}{2} \sin x\) have the same slopes?
View solution Problem 24
Complete \((x+h)^{5}=x^{5}+\) ____. What are the binomial coefficients \(\left(\begin{array}{l}5 \\ 1\end{array}\right)\) and \(\left(\begin{array}{l}5 \\\ 2\en
View solution