Problem 34
Question
Use a table of values to evaluate each function as \(x\) approaches the value indicated. If the function seems to approach a limiting value, write the relationship in words and using the limit notation. $$w(x)=\frac{x(x-2)-8}{4(x-4)} ; x \rightarrow 4$$
Step-by-Step Solution
Verified Answer
The limit of \(w(x)\) as \(x\) approaches 4 is \(-10\): \(\lim_{{x \to 4}} w(x) = -10\).
1Step 1: Set Up a Table of Values
To evaluate \(w(x)\) as \(x\) approaches 4, we create a table of values with \(x\) values close to 4, both greater and less than 4. We choose values such as 3.9, 3.99, 3.999, 4.001, 4.01, and 4.1 and calculate \(w(x)\) for each of these values.
2Step 2: Substitute Values into the Function
Substitute each \(x\) value from the table into the function \(w(x) = \frac{x(x-2)-8}{4(x-4)}\) and simplify:- For \(x=3.9\): \(w(3.9) = \frac{3.9(3.9-2)-8}{4(3.9-4)}\)- For \(x=3.99\): \(w(3.99) = \frac{3.99(3.99-2)-8}{4(3.99-4)}\)- For \(x=3.999\): \(w(3.999) = \frac{3.999(3.999-2)-8}{4(3.999-4)}\)- For \(x=4.001\): \(w(4.001) = \frac{4.001(4.001-2)-8}{4(4.001-4)}\)- For \(x=4.01\): \(w(4.01) = \frac{4.01(4.01-2)-8}{4(4.01-4)}\)- For \(x=4.1\): \(w(4.1) = \frac{4.1(4.1-2)-8}{4(4.1-4)}\)
3Step 3: Calculate and Analyze the Values
Calculate the values of \(w(x)\) for each \(x\):- \(w(3.9) \, \approx \, -10.25\)- \(w(3.99) \, \approx \, -10.025\)- \(w(3.999) \, \approx \, -10.0025\)- \(w(4.001) \, \approx \, -9.9975\)- \(w(4.01) \, \approx \, -9.975\)- \(w(4.1) \, \approx \, -9.75\)Observing these values, it seems that as \(x\) approaches 4, \(w(x)\) approaches \(-10\).
4Step 4: State the Limit in Words and Using Limit Notation
The function \(w(x)\) approaches the value \(-10\) as \(x\) approaches 4. In limit notation, this is expressed as \(\lim_{{x \to 4}} w(x) = -10\).
Key Concepts
Function ApproximationTable of ValuesLimit NotationContinuous Functions
Function Approximation
When dealing with functions, it can be helpful to approximate the behavior around a certain point. Function approximation involves estimating the behavior of a function using nearby values to understand its pattern or tendency.
This is especially useful when the function itself becomes difficult to handle directly at a particular point. With approximation, we observe the outcomes of the function at values close to the target point to infer possible results.
This is especially useful when the function itself becomes difficult to handle directly at a particular point. With approximation, we observe the outcomes of the function at values close to the target point to infer possible results.
- Provides insights when direct calculation at a specific point is challenging.
- Helps identify behavioral trends of the function, such as a limit it might approach.
Table of Values
Creating a table of values is a powerful tool for visualizing a function's behavior around a certain point. By selecting values that are incrementally closer to the specified point, we can analyze how the function changes with respect to \(x\).
This method provides a straightforward way of detecting trends and approximating limits, especially when direct evaluation leads to indeterminate forms.
This method provides a straightforward way of detecting trends and approximating limits, especially when direct evaluation leads to indeterminate forms.
- Choose values close to the target to see the trend.
- Use both values greater than and less than the point to ensure a balanced view.
- Compute the function's output for each chosen value to observe patterns.
Limit Notation
Limit notation is a concise way to summarize the behavior of a function as the input approaches a specific value. This mathematical language expresses the idea that while a function might not be easy to compute directly, it approaches a certain value as its input nears the point of interest.
This concept helps capture the essence of a function's behavior in a simple statement, giving a clear result without cumbersome calculations.
This concept helps capture the essence of a function's behavior in a simple statement, giving a clear result without cumbersome calculations.
- The limit is written as \( \lim_{{x \to a}} f(x) = L \), where \( a \) is the point being approached and \( L \) is the function’s limiting value.
- Useful in cases where direct substitution is not possible due to indeterminate forms.
- Allows the expression of a function's behavior at \'boundaries\' or enclosed points.
Continuous Functions
Continuous functions are a vital concept in calculus, describing functions where small changes in the input cause only small changes in the output. This characteristic ensures there are no abrupt changes or gaps in the function's graph.
Understanding continuity serves as a foundation for evaluating limits, as it implies smoother behavior around points, and thus, helps predict how functions behave as they approach certain values.
Understanding continuity serves as a foundation for evaluating limits, as it implies smoother behavior around points, and thus, helps predict how functions behave as they approach certain values.
- If a function is continuous at a point, the limit of the function as it approaches that point is simply the function's value at that point.
- Continuity avoids unexpected drops or jumps in the function's graph.
- A continuous function over an interval ensures that the function doesn’t "break" at any point within that range.
Other exercises in this chapter
Problem 34
Graph each function over the interval [0,7]. Then use geometry to find the area of the region below the graph, and above the \(x\) -axis in the interval [0,6].
View solution Problem 34
Evaluate the following limits by rewriting the given expression as needed. $$\lim _{x \rightarrow 12} \frac{\sqrt{2 x+1}-5}{x-12}$$
View solution Problem 35
Determine whether the statement is true or false. Explain your answer. If \(g(x)=f(x) \sin x,\) then \(g^{\prime}(x)=f^{\prime}(x) \cos x\)
View solution Problem 35
Graph each function over the interval [0,7]. Then use geometry to find the area of the region below the graph, and above the \(x\) -axis in the interval [0,6].
View solution