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Chapter 2: Problem 16

Find the limits in Exercises \(11-22\) $$\lim _{s \rightarrow 2 / 3}(8-3 s)(2 s-1)$$

Short Answer

Expert verified
The limit is 2.

Step by step solution

01

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 \((8-3s)(2s-1)\).
02

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 \( s = \frac{2}{3} \): \[ 2\left(\frac{2}{3}\right) - 1 = \frac{4}{3} - 1 = \frac{4}{3} - \frac{3}{3} = \frac{1}{3} \]
03

Multiply the Simplified Terms

With the two simplified terms from Step 2, multiply them together to find the limit:\[(8 - 3s)(2s - 1) = 6 \times \frac{1}{3} = 2\]
04

Conclude the Limit Calculation

The result from Step 3 shows the value of the limit:\[\lim_{s \rightarrow \frac{2}{3}} (8 - 3s)(2s - 1) = 2\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.
Begin with substituting to quickly identify any potential simplifications before further manipulation. This step sets the stage for simplifying and calculating the limit without complexity.
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 \( \frac{4}{3} - \frac{3}{3} = \frac{1}{3} \).
Efficient simplification prevents errors and makes subsequent calculations easier. By resolving expressions within parentheses first, you streamline the process of multiplication and achieve a more direct path to the answer.
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 \( 6 \times \frac{1}{3} = 2 \).
Multiplying factors can often reveal the true value of the limit. Ensuring accuracy in multiplication ensures the precision of your final answer. This step is crucial as it brings together all previous simplifications into a conclusive result.
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.
Engage with such a text to deepen your grasp on calculus principles and techniques. Leveraging resources like Thomas Calculus can vastly enhance your problem-solving toolkit and offer diverse strategies to tackle calculus challenges with confidence.

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Most popular questions from this chapter

In Exercises \(47-50,\) graph the function \(f\) to see whether it appears to have a continuous extension to the origin. If it does, use Trace and Zoom to find a good candidate for the extended function's value at \(x=0\) .If the function does not appear to have a continuous extension, can it be extended to be continuous at the origin from the right or from the left? If so, what do you think the extended function's value(s) should be? $$ f(x)=\frac{10^{|x|}-1}{x} $$

For what value of \(b\) is $$ g(x)=\left\\{\begin{array}{ll}{x,} & {x<-2} \\ {b x^{2},} & {x \geq-2}\end{array}\right. $$ continuous at every \(x ?\)

If \(x^{4} \leq f(x) \leq x^{2}\) for \(x\) in \([-1,1]\) and \(x^{2} \leq f(x) \leq x^{4}\) for \(x<-1\) and \(x>1,\) at what points \(c\) do you automatically know \(\lim _{x \rightarrow c} f(x) ?\) What can you say about the value of the limit at these points?

Explain why the equation \(\cos x=x\) has at least one solution.

Removable discontinuity Give an example of a function \(f(x)\) that is continuous for all values of \(x\) except \(x=2,\) where it has a removable discontinuity. Explain how you know that \(f\) is discontinuous at \(x=2,\) and how you know the discontinuity is removable. A function discontinuous at every point a. Use the fact that every nonempty interval of real numbers contains both rational and irrational numbers to show that the function $$ f(x)=\left\\{\begin{array}{ll}{1,} & {\text { if } x \text { is rational }} \\\ {0,} & {\text { if } x \text { is irrational }}\end{array}\right. $$ is discontinuous at every point. b. Is \(f\) right-continuous or left-continuous at any point?
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