Problem 16
Question
Find the limits in Exercises \(11-22\) $$\lim _{s \rightarrow 2 / 3}(8-3 s)(2 s-1)$$
Step-by-Step Solution
Verified Answer
The limit is 2.
1Step 1: Substitute the Limit Point
Start by substituting the value of the limit point into the function. Here, the limit is as \( s \) approaches \( \frac{2}{3} \). Therefore, replace \( s \) with \( \frac{2}{3} \) in the expression \((8-3s)(2s-1)\).
2Step 2: Evaluate the Expression Inside the Parentheses
Calculate each part of the expression inside the parentheses separately:- For \( 8 - 3s \), substitute \( s = \frac{2}{3} \): \[ 8 - 3\left(\frac{2}{3}\right) = 8 - 2 = 6 \]- For \( 2s - 1 \), again substitute \( s = \frac{2}{3} \): \[ 2\left(\frac{2}{3}\right) - 1 = \frac{4}{3} - 1 = \frac{4}{3} - \frac{3}{3} = \frac{1}{3} \]
3Step 3: Multiply the Simplified Terms
With the two simplified terms from Step 2, multiply them together to find the limit:\[(8 - 3s)(2s - 1) = 6 \times \frac{1}{3} = 2\]
4Step 4: Conclude the Limit Calculation
The result from Step 3 shows the value of the limit:\[\lim_{s \rightarrow \frac{2}{3}} (8 - 3s)(2s - 1) = 2\]
Key Concepts
Substitution MethodSimplifying ExpressionsMultiplying FactorsThomas Calculus
Substitution Method
The substitution method is a straightforward and powerful technique used to evaluate limits. When confronting a limit such as \( \lim_{s \rightarrow \frac{2}{3}} (8-3s)(2s-1) \), the first step is to replace the variable with the limit point. Here, we substitute \( s = \frac{2}{3} \) directly into the expression.
- Substituting the limit point helps us directly tackle the indeterminate forms, if any arise.
- This approach assumes the function is continuous at the point of substitution, allowing for a seamless calculation process.
Simplifying Expressions
Once you have substituted the value into the function, the next step is simplifying the expressions within the parentheses. For the exercise, simplifying means calculating each part separately:
- For \( 8 - 3s \), substituting \( s = \frac{2}{3} \) gives us \( 8 - 2 = 6 \).
- For \( 2s - 1 \), substituting \( s = \frac{2}{3} \) gives us \( \frac{4}{3} - \frac{3}{3} = \frac{1}{3} \).
Multiplying Factors
After simplifying, it's time to multiply the simplified terms. This involves combining terms to find the resultant value of the expression:
- The simplified terms in the exercise are \( 6 \) and \( \frac{1}{3} \).
- Multiplying these gives \( 6 \times \frac{1}{3} = 2 \).
Thomas Calculus
Thomas Calculus offers extensive methods and theories that underpin many mathematical practices, including limits. This resource provides comprehensive insights into concepts like the substitution method itself:
- This text gives a profound understanding of fundamental calculus operations, which are essential for solving limit problems.
- It explores various approaches to handle indeterminate forms and presents rigorous proofs that solidify conceptual understanding.
Other exercises in this chapter
Problem 16
Find the limits in Exercises \(11-18\) $$\lim _{h \rightarrow 0} \frac{\sqrt{6}-\sqrt{5 h^{2}+11 h+6}}{h}$$
View solution Problem 16
Each of Exercises \(15-30\) gives a function \(f(x)\) and numbers \(L, c,\) and \(\epsilon>0 .\) In each case, find an open interval about \(c\) on which the in
View solution Problem 17
At what points are the functions in Exercises 13-30 continuous? $$y=|x-1|+\sin x$$
View solution Problem 17
In Exercises \(13-22,\) find the limit of each rational function (a) as \(x \rightarrow \infty\) and \((b)\) as \(x \rightarrow-\infty\) . $$h(x)=\frac{7 x^{3}}
View solution