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Chapter 2: Problem 56

Suppose that \lim _{x \rightarrow-2} p(x)=4, \lim _{x \rightarrow-2} r(x)=0, and \lim _{x \rightarrow-2}\( s(x)=-3 .\) Find $$\begin{array}{l}{\text { a. } \lim _{x \rightarrow-2}(p(x)+r(x)+s(x))} \\\ {\text { b. } \lim _{x \rightarrow-2} p(x) \cdot r(x) \cdot s(x)} \\ {\text { c. } \lim _{x \rightarrow-2}(-4 p(x)+5 r(x)) / s(x)}\end{array}$$

Short Answer

Expert verified
a) 1 b) 0 c) \( \frac{16}{3} \)

Step by step solution

01

Understand the Given Limits

We are given three limits: \( \lim_{x \rightarrow -2} p(x) = 4 \), \( \lim_{x \rightarrow -2} r(x) = 0 \), and \( \lim_{x \rightarrow -2} s(x) = -3 \). These limits will be used to solve parts a, b, and c.
02

Solve Part A

We need to find \( \lim_{x \rightarrow -2} (p(x) + r(x) + s(x)) \). Using the property of limits that allows us to add them: \[\lim_{x \rightarrow -2} (p(x) + r(x) + s(x)) = \lim_{x \rightarrow -2} p(x) + \lim_{x \rightarrow -2} r(x) + \lim_{x \rightarrow -2} s(x).\]Substituting the given values, we have:\[\lim_{x \rightarrow -2} p(x) + \lim_{x \rightarrow -2} r(x) + \lim_{x \rightarrow -2} s(x) = 4 + 0 + (-3) = 1.\]
03

Solve Part B

Next, we need to find \( \lim_{x \rightarrow -2} (p(x) \cdot r(x) \cdot s(x)) \). We use the property of limits that allows us to multiply them:\[\lim_{x \rightarrow -2} (p(x) \cdot r(x) \cdot s(x)) = \lim_{x \rightarrow -2} p(x) \cdot \lim_{x \rightarrow -2} r(x) \cdot \lim_{x \rightarrow -2} s(x).\]Substituting the given values, we have:\[4 \cdot 0 \cdot (-3) = 0.\]
04

Solve Part C

Finally, we need to find \( \lim_{x \rightarrow -2} \frac{-4p(x) + 5r(x)}{s(x)} \). Using the property of limits for division:\[\lim_{x \rightarrow -2} \frac{-4p(x) + 5r(x)}{s(x)} = \frac{\lim_{x \rightarrow -2} (-4p(x) + 5r(x))}{\lim_{x \rightarrow -2} s(x)}.\]We calculate the numerator:\[\lim_{x \rightarrow -2} (-4p(x) + 5r(x)) = -4 \cdot \lim_{x \rightarrow -2} p(x) + 5 \cdot \lim_{x \rightarrow -2} r(x) = -4 \cdot 4 + 5 \cdot 0 = -16.\]The denominator is:\[\lim_{x \rightarrow -2} s(x) = -3.\]Thus, the overall limit is:\[\frac{-16}{-3} = \frac{16}{3}.\]
05

Summarize the Solutions

For part a, the limit is 1. For part b, the limit is 0. For part c, the limit is \( \frac{16}{3} \).

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Key Concepts

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

Limit Properties
Understanding limit properties is crucial when evaluating limits in calculus. Limit properties allow us to simplify and calculate limits efficiently. Here are some basic limit properties that are extremely helpful in solving problems:
  • Sum/Difference Property: The limit of a sum or difference is equal to the sum or difference of the limits. Symbolically, if \( \lim_{x \to c} f(x) = L \) and \( \lim_{x \to c} g(x) = M \), then \( \lim_{x \to c} (f(x) \pm g(x)) = L \pm M \).
  • Product Property: The limit of a product is the product of the limits. That is, \( \lim_{x \to c} (f(x) \cdot g(x)) = L \cdot M \).
  • Quotient Property: When dealing with limits of quotients, \( \lim_{x \to c} \frac{f(x)}{g(x)} = \frac{L}{M} \), as long as \( M eq 0 \).
These properties were used in solving parts (a), (b), and (c) of the original exercise, helping us handle each component separately and then combine the results to find the overall limits.
Limit Calculation
Calculating limits often involves simple arithmetic once the properties of limits are applied. For example, in the given exercise, we used these calculations effectively:
  • Addition: For part (a), we found \( \lim_{x \to -2} (p(x) + r(x) + s(x)) \) by adding \( \lim_{x \to -2} p(x) \), \( \lim_{x \to -2} r(x) \), and \( \lim_{x \to -2} s(x) \). The arithmetic here is straightforward: \( 4 + 0 + (-3) = 1 \).
  • Multiplication: Part (b) required multiplying the given limits: \( 4 \cdot 0 \cdot (-3) = 0 \).
  • Division: In part (c), we divided the limit of the expression \( -4p(x) + 5r(x) \) by \( s(x) \). After calculating, this yielded \( \frac{-16}{-3} = \frac{16}{3} \).
By methodically applying these operations, solving each part becomes clear and manageable.
Limit Substitution
Limit substitution is often the first method applied when evaluating limits, especially if the function is well-behaved at the target point. In cases where direct substitution might not work due to indeterminacy (like \( \frac{0}{0} \)), we use limit properties or simplification techniques.
  • Substituting directly into expressions helps seamlessly evaluate the function as in parts (a) and (b) of the exercise. For instance, substituting \( -2 \) into the expressions involved straightforward arithmetic calculations.
  • When facing a limit expression involving division, such as in (c), ensure that the denominator does not approach zero, unless special techniques like L'H么pital's Rule are applied to resolve indeterminacies.
In the exercise, direct substitution allowed us to evaluate each component of the limits effectively, ensuring clear solutions free from indeterminate forms. Labeling components clearly during substitution simplifies tracing errors and confirming results.

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Most popular questions from this chapter

Suppose that \(f\) is an even function of \(x .\) Does knowing that \(\lim _{x \rightarrow 2^{-}} f(x)=7\) tell you anything about either \(\lim _{x \rightarrow-2^{-}} f(x)\) or \(\lim _{x \rightarrow-2^{+}} f(x) ?\) Give reasons for your answer.

Explain why the following five statements ask for the same information. a. Find the roots of \(f(x)=x^{3}-3 x-1\) . b. Find the \(x\) -coordinates of the points where the curve \(y=x^{3}\) crosses the line \(y=3 x+1 .\) c. Find all the values of \(x\) for which \(x^{3}-3 x=1\) . d. Find the \(x\) -coordinates of the points where the cubic curve \(y=x^{3}-3 x\) crosses the line \(y=1 .\) e. Solve the equation \(x^{3}-3 x-1=0\)

Find the limits in Exercises \(84-90 .\) (Hint: Try multiplying and dividing by the conjugate.) $$ \lim _{x \rightarrow-\infty}\left(\sqrt{x^{2}+3}+x\right) $$

In Exercises \(73-76,\) sketch the graph of a function \(y=f(x)\) that satisfies the given conditions. No formulas are required- just label the coordinate axes and sketch an appropriate graph. (The answers are not unique, so your graphs may not be exactly like those in the answer section.) \begin{equation} f(0)=0, \lim _{x \rightarrow \pm \infty} f(x)=0, \lim _{x \rightarrow 0^{+}} f(x)=2, \text { and } \lim _{x \rightarrow 0^{-}} f(x)=-2 \end{equation}

$$ \begin{array}{l}{\text { a. If } \lim _{x \rightarrow 2} \frac{f(x)-5}{x-2}=3, \text { find } \lim _{x \rightarrow 2} f(x)} \\ {\text { b. If } \lim _{x \rightarrow 2} \frac{f(x)-5}{x-2}=4, \text { find } \lim _{x \rightarrow 2} f(x)}\end{array} $$
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