/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 5 Find the limit of each function ... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 5

Find the limit of each function (a) as \(x \rightarrow \infty\) and (b) as \(x \rightarrow-\infty\). (You may wish to visualize your answer with a graphing calculator or computer.) $$g(x)=\frac{1}{2+(1 / x)}$$

Short Answer

Expert verified
As \( x \to \pm \infty \), \( g(x) \to \frac{1}{2} \).

Step by step solution

01

Understanding the Function

The given function is \( g(x) = \frac{1}{2 + (1/x)} \). As \( x \to \infty \) and \( x \to -\infty \), the \( 1/x \) term becomes negligible, simplifying our calculations.
02

Simplifying as x approaches infinity

As \( x \to \infty \), the term \( \frac{1}{x} \) approaches 0. This means the expression simplifies to \( g(x) = \frac{1}{2 + 0} = \frac{1}{2} \).
03

Renaming x indefinitely for clarity

For both positive and negative infinity, recalling that \( \frac{1}{x} \) approaches 0 regardless of the sign of x helps treat the denominator as \( 2 + 0 \), making \( g(x) = \frac{1}{2} \).
04

Simplifying as x approaches negative infinity

As \( x \to -\infty \), the term \( \frac{1}{x} \) also approaches 0. Hence, the function simplifies to \( g(x) = \frac{1}{2 + 0} = \frac{1}{2} \).
05

Graph Analysis Recommendation

Visualizing \( g(x) \) on a graphing calculator will show that the function approaches a horizontal asymptote at \( y = \frac{1}{2} \) as \( x \) tends towards both positive and negative infinity.

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Key Concepts

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

Infinity in Calculus
Infinity in calculus is an essential concept that plays a vital role in understanding functions and their behavior. Key Ideas To Remember:
  • When we say a variable approaches infinity, we're talking about what happens as the variable grows larger and larger. For example, as \( x \to \infty \), \( x \) becomes a very large positive number.
  • Similarly, \( x \to -\infty \) indicates \( x \) becoming a very large negative number.
  • Infinity itself is not a number, but rather a way to express that values can grow without bound in either the positive or negative direction.
  • In calculus, considering limits as \( x \) approaches infinity helps us determine the long-term behavior of a function.
By considering \( x \to \infty \) and \( x \to -\infty \) for a function like \( g(x) = \frac{1}{2 + (1/x)} \), we see how certain terms become insignificant. In this example, \( 1/x \to 0 \) when \( x \) grows large, either positive or negative, making it much easier to analyze the function's behavior.
Horizontal Asymptotes
Horizontal asymptotes are lines that a graph approaches as \( x \) goes towards plus or minus infinity but never quite touches.Important Aspects:
  • If a function has a horizontal asymptote, it's like having a boundary that the function nears but doesn't cross.
  • In our example, as \( x \) moves towards infinity in both directions, \( g(x) = \frac{1}{2 + (1/x)} \) tends to the horizontal line \( y = \frac{1}{2} \).
  • This happens because the term \( 1/x \) approaches zero when \( x \) is very large, effectively simplifying the function to \( g(x) = \frac{1}{2} \) for both positive and negative infinity.
  • Graphically, plotting shows that \( g(x) \) rounds off at \( y = \frac{1}{2} \), demonstrating the horizontal asymptote's graphical aspect.
Horizontal asymptotes are useful for identifying what happens to a function at its extremes鈥攖hey give us a concise understanding of boundaries within the function's behavior.
Function Simplification
Function simplification is about making functions easier to analyze by reducing them to their core components.Steps and Tricks:
  • In calculus, we often simplify functions as a strategy to find their limits or asymptotes, especially at infinity.
  • With the function \( g(x) = \frac{1}{2 + (1/x)} \), as \( x \to \pm \infty \), \( \frac{1}{x} \) becomes so small that it's negligible.
  • This simplification turns the complex expression into a simple fraction \( \frac{1}{2} \), offering a clearer view of the function's long-term behavior.
  • The simplification removes minor influences (like \( 1/x \)) to focus on significant behaviors and limits.
Understanding how to effectively simplify functions is crucial for calculating limits and comprehending the nature of these expressions as they interact with infinity.

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Most popular questions from this chapter

Prove the limit statements. $$\lim _{x \rightarrow 4}(9-x)=5$$

If the product function \(h(x)=f(x) \cdot g(x)\) is continuous at \(x=0\) must \(f(x)\) and \(g(x)\) be continuous at \(x=0 ?\) Give reasons for your answer.

You will find a graphing calculator useful for Exercise. Let \(f(x)=\left(x^{2}-1\right) /(|x|-1).\) a. Make tables of the values of \(f\) at values of \(x\) that approach \(c=-1\) from above and below. Then estimate \(\lim _{x \rightarrow-1} f(x).\) b. Support your conclusion in part (a) by graphing \(f\) near \(c=-1\) 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.

Another wrong statement about limits Show by example that the following statement is wrong. The number \(L\) is the limit of \(f(x)\) as \(x\) approaches \(c\) if, given any \(\epsilon>0,\) there exists a value of \(x\) for which \(|f(x)-L|<\epsilon\) Explain why the function in your example does not have the given value of \(L\) as a limit as \(x \rightarrow c\)

Gives a function \(f(x)\) and numbers \(L, c,\) and \(\epsilon \geq 0 .\) In each case, find an open interval about \(c\) on which the inequality \(|f(x)-L|<\epsilon\) holds. Then give a value for \(\delta>0\) such that for all \(x\) satisfying \(0<|x-c|<\delta\) the inequality \(|f(x)-L|<\epsilon\) holds. $$f(x)=x^{2}-5, \quad L=11, \quad c=4, \quad \epsilon=1$$
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