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Chapter 2: Problem 53

Suppose \(\lim _{x \rightarrow c} f(x)=5\) and \(\lim _{x \rightarrow g} g(x)=-2 .\) Find \begin{equation} \text { a. }\lim _{x \rightarrow c} f(x) g(x) \quad \text { b. } \lim _{x \rightarrow c} 2 f(x) g(x) \\\\\quad\quad\quad \text { c. }\lim _{x \rightarrow c}(f(x)+3 g(x)) \quad \text { d. } \lim _{x \rightarrow c} \frac{f(x)}{f(x)-g(x)} \end{equation}

Short Answer

Expert verified
a. \(-10\), b. \(-20\), c. \(-1\), d. \(\frac{5}{7}\).

Step by step solution

01

Understanding the Problem

We are given limits of two functions, \( \lim_{x \rightarrow c} f(x) = 5 \) and \( \lim_{x \rightarrow c} g(x) = -2 \), and asked to find the limits of expressions involving these functions as \( x \) approaches \( c \).
02

Multiplying Limits for Part a

For \( \lim_{x \rightarrow c} f(x)g(x) \), we can use the property that the limit of a product is the product of the limits, provided the limits exist. Thus, \( \lim_{x \rightarrow c} f(x)g(x) = \lim_{x \rightarrow c} f(x) \cdot \lim_{x \rightarrow c} g(x) = 5 \cdot (-2) = -10 \).
03

Multiplicative Constant for Part b

For \( \lim_{x \rightarrow c} 2f(x)g(x) \), apply the limit of a product and constant rule: \( \lim_{x \rightarrow c} (2f(x)g(x)) = 2 \cdot \lim_{x \rightarrow c} f(x) \cdot \lim_{x \rightarrow c} g(x) = 2 \cdot 5 \cdot (-2) = -20 \).
04

Adding Limits for Part c

For \( \lim_{x \rightarrow c} (f(x)+3g(x)) \), use the limit of a sum rule: \( \lim_{x \rightarrow c} (f(x) + 3g(x)) = \lim_{x \rightarrow c} f(x) + 3 \cdot \lim_{x \rightarrow c} g(x) = 5 + 3 \cdot (-2) = 5 - 6 = -1 \).
05

Simplifying Expression for Part d

For \( \lim_{x \rightarrow c} \frac{f(x)}{f(x) - g(x)} \), use the quotient rule. The limit of the quotient of two functions is the quotient of their limits, as long as the limit of the denominator is not zero. \( \lim_{x \rightarrow c} f(x) = 5 \) and \( \lim_{x \rightarrow c} g(x) = -2 \), so \( \lim_{x \rightarrow c} (f(x) - g(x)) = 5 - (-2) = 7 \) and thus, \( \lim_{x \rightarrow c} \frac{f(x)}{f(x) - g(x)} = \frac{5}{7} \).

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Key Concepts

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

Limit Laws
Limit laws are fundamental rules in calculus that allow us to evaluate the limits of functions and their combinations. These laws simplify the process of finding the limit when several operations like addition, multiplication, or division are involved. It is important to note that limit laws are only applicable when the individual limits involved are finite and exist.Some of the most common limit laws include:
  • Limit of a Constant: The limit of a constant function is simply the constant itself. If \( k \) is a constant, then \( \lim_{x \to c} k = k \).
  • Constant Multiple Law: If \( \lim_{x \to c} f(x) = L \), then \( \lim_{x \to c} [k \cdot f(x)] = k \cdot L \), where \( k \) is a constant.
  • Sum/Difference Law: \( \lim_{x \to c} [f(x) \pm g(x)] = \lim_{x \to c} f(x) \pm \lim_{x \to c} g(x) \).
  • Product Law: \( \lim_{x \to c} [f(x) \cdot g(x)] = \lim_{x \to c} f(x) \cdot \lim_{x \to c} g(x) \).
  • Quotient Law: \( \lim_{x \to c} \frac{f(x)}{g(x)} = \frac{\lim_{x \to c} f(x)}{\lim_{x \to c} g(x)} \), provided \( \lim_{x \to c} g(x) eq 0 \).
These laws provide a framework for solving complex calculus problems by breaking them down into smaller parts.
Product Rule for Limits
The Product Rule for Limits is a powerful tool in calculus used when dealing with products of functions. If you know the limits of the individual functions, this rule allows you to find the limit of their product. The formal statement is: If \( \lim_{x \to c} f(x) = L \) and \( \lim_{x \to c} g(x) = M \), then \( \lim_{x \to c} [f(x) \cdot g(x)] = L \times M \). Here's why this is useful:
  • It simplifies the process of finding limits where multiplication is involved.
  • It saves time as we don't need to calculate the product under the limit separately for each case.
  • For example, in the exercise, \( \lim_{x \to c} f(x) \cdot g(x) = 5 \cdot (-2) = -10 \).
By understanding this rule, you can tackle problems that seem tangled at first glance by simplifying the operations into easily manageable parts.
Quotient Rule for Limits
The Quotient Rule for Limits is essential whenever you're faced with dividing two functions. This rule tells us how to find the limit of a fraction. The rule states: If \( \lim_{x \to c} f(x) = L \) and \( \lim_{x \to c} g(x) = M \), then \( \lim_{x \to c} \frac{f(x)}{g(x)} = \frac{L}{M} \), provided that \( M eq 0 \).A few things to keep in mind:
  • Always check that the limit of the denominator is not zero. If it is, the equation might not apply directly, and further analysis is needed.
  • This rule enables you to divide limits without solving the division directly from scratch.
  • In the provided exercise, the solution handles \( \lim_{x \to c} \frac{f(x)}{f(x) - g(x)} \), careful to note that \( \lim_{x \to c} [f(x) - g(x)] \) is not zero.
Understanding the quotient rule helps you handle complex calculus problems ensuring you're working efficiently and accurately.
Sum Rule for Limits
The Sum Rule for Limits helps us handle addition and subtraction of functions by separating combined operations into simpler parts. This rule states that the limit of the sum of functions is the sum of their limits.The formal notation: If \( \lim_{x \to c} f(x) = L \) and \( \lim_{x \to c} g(x) = M \), then \( \lim_{x \to c} [f(x) + g(x)] = L + M \).This is useful in numerous cases:
  • This rule allows us to add or subtract limits directly.
  • It considerably simplifies analysis, letting us evaluate each function separately.
  • In the exercise, \( \lim_{x \to c} (f(x) + 3g(x)) = 5 + 3(-2) = -1 \).
The Sum Rule makes it easy to tackle complex expressions by transforming them into simpler tasks, leading to faster and more efficient calculations.

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Most popular questions from this chapter

Solving an equation If \(f(x)=x^{3}-8 x+10,\) show that there are values \(c\) for which \(f(c)\) equals (a) \(\pi ;\) (b) \(-\sqrt{3}\) (c) \(5,000,000\) .

You will find a graphing calculator useful. Let $$g(x)=\left(x^{2}-2\right) /(x-\sqrt{2}).$$ a. Make a table of the values of \(g\) at the points \(x=1.4,1.41\) , \(1.414,\) and so on through successive decimal approximations of \(\sqrt{2} .\) Estimate \(\lim _{x \rightarrow \sqrt{2}} \sqrt{2} g(x).\) b. Support your conclusion in part (a) by graphing \(g\) near \(x_{0}=\sqrt{2}\) and using Zoom and Trace to estimate \(y\) -values on the graph as \(x \rightarrow \sqrt{2}\) . c. Find \(\lim _{x \rightarrow \sqrt{2}} g(x)\) algcbraically.

Define \(h(2)\) in a way that extends \(h(t)=\left(t^{2}+3 t-10\right) /(t-2)\) to be continuous at \(t=2\) .

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 \(x ?\)

You will find a graphing calculator useful. Let $$f(x)=\left(x^{2}-1\right) /(|x|-1).$$ a. Make tables of the values of \(f\) at values of \(x\) that approach \(x_{0}=-1\) from above and below. Then estimate \(\quad \lim _{x \rightarrow-1} f(x).\) b. Support your conclusion in part (a) by graphing \(f\) near \(x_{0}=-1\) and using Zoom and Trace to estimate \(y\) -values on the graph as \(x \rightarrow-1\) . c. Find \(\lim _{x \rightarrow-1} f(x)\) algebraically.
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