/*! 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 73 The notion of an asymptote can b... [FREE SOLUTION] | 91影视

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Chapter 1: Problem 73

The notion of an asymptote can be extended to include curves as well as lines. Specifically, we say that curves \(y=f(x)\) and \(y=g(x)\) are asymptotic as \(x \rightarrow+\infty\) provided $$ \lim _{x \rightarrow+\infty}[f(x)-g(x)]=0 $$ In these exercises, determine a simpler function \(g(x)\) such that \(y=f(x)\) is asymptotic to \(y=g(x)\) as \(x \rightarrow+\infty\) or \(x \rightarrow-\infty\) Use a graphing utility to generate the graphs of \(y=f(x)\) and \(y=g(x)\) and identify all vertical asymptotes. $$ f(x)=\sin x+\frac{1}{x-1} $$

Short Answer

Expert verified
No suitable asymptotic function due to oscillation; vertical asymptote at \( x = 1 \).

Step by step solution

01

Identify the Dominant Terms

For large values of \( x \), the term \( \frac{1}{x-1} \) will dominate the function \( f(x) = \sin x + \frac{1}{x-1} \) since the sine function is bounded between -1 and 1. Therefore, \( g(x) = \frac{1}{x-1} \) could be a simpler function that approximates \( f(x) \) for large values of \( x \).
02

Check Asymptotic Behavior

Calculate \( \lim_{x \rightarrow +\infty} [f(x) - g(x)] = \lim_{x \rightarrow +\infty} \left(\sin x + \frac{1}{x-1} - \frac{1}{x-1} \right) = \lim_{x \rightarrow +\infty} \sin x \). Since \( \sin x \) oscillates and does not approach 0, \( f(x) \) and \( g(x) \) are not asymptotic as \( x \rightarrow +\infty \).
03

Re-Evaluate Strategy

Since the initial choice of \( g(x) = \frac{1}{x-1} \) did not show asymptotic behavior due to the oscillating nature of \( \sin x \), check the periodic nature of the sine function. Identify if there might be a horizontal line or simpler constant function that captures the average or effective nature of \( \,\sin x + \frac{1}{x-1}\, \).
04

Include Oscillation Correction

Despite \( g(x) = \frac{1}{x-1} \) not being suitable alone, \( g(x) = \frac{1}{x-1} + 0 \) (considering 0 as an oscillation correction due to \( \sin x \)) still gives a similar trend for large \( x \). However, note that since the sine function oscillates and does not vanish, finding a suitable \( g(x) \) strictly asymptotic to \( f(x) \) taking into account the smallest common dominant behavior is key.
05

Graphical Representation and Identification

Use a graphing utility to graph \( f(x) = \sin x + \frac{1}{x-1} \) and \( g(x) = \frac{1}{x-1} \). Two observations: the graph of \( f(x) \) will demonstrate vertical asymptotes at \( x = 1 \), because \( \frac{1}{x-1} \) approaches infinity, and sinusoidal behavior in other regions.

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Key Concepts

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

Vertical Asymptotes
Vertical asymptotes occur when a function grows infinitely large as the input (x) approaches a certain point. This is often due to a division by zero in a function, such as what happens in our exercise with the term \( \frac{1}{x-1} \).
When \( x \) approaches 1 from either side, the denominator becomes zero, causing the function to shoot up or down to infinity. This characteristic vertical line at \( x = 1 \) is the vertical asymptote.
  • Vertical asymptotes help identify boundaries for understanding the behavior of a function near specific points.
  • By checking the denominator, especially if it's a rational expression, you can predict where a function might have these asymptotes.
  • Knowing where the function is undefined is key鈥攍ike at \( x = 1 \) in this case.
Dominant Terms
Understanding dominant terms is crucial when analyzing asymptotic behavior. Dominant terms are parts of an expression that disproportionately affect the function鈥檚 value as \( x \) becomes very large. In our given function \( f(x) = \sin x + \frac{1}{x-1} \), the dominant term for large \( x \) is \( \frac{1}{x-1} \).
This is because \( \sin x \) remains between -1 and +1 irrespective of \( x \), while \( \frac{1}{x-1} \) can become quite large as \( x \) approaches the vertical asymptote. Identifying these dominant terms is key to simplifying functions for large \( x \) values, especially when considering asymptotic behavior.
  • Identify all terms in the function and evaluate their behavior as \( x \rightarrow +\infty \) or \( x \rightarrow -\infty \).
  • Focus on terms that grow the largest or cause the greatest variation to understand how they dominate other parts of the function.
  • Knowing which terms dominate allows for simplification when estimating limits and asymptotic behavior.
Graphing Utility
A graphing utility is a tool that aids in visualizing mathematical functions. It can graph complex equations, helping identify features like asymptotes and periods in oscillating functions, such as \( \sin x + \frac{1}{x-1} \).
By inputting the equation into a graphing utility, you can see where vertical asymptotes appear (as lines shooting up and down), and understand the general behavior of the function.
  • Use a graphing utility to visually confirm theoretical analyses鈥攕uch as the existence of vertical asymptotes or bounded oscillations.
  • When graphs show sharp spikes or dips, these typically occur right near vertical asymptotes.
  • It assists in confirming guesses around dominant terms by showing their effects visually.
Graphing is a critical step in bridging the gap between algebraic predictions and actual behavior, making it a vital tool in the study of precalculus and beyond.

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Most popular questions from this chapter

In Example 5 we used the Squeezing Theorem to prove that $$ \lim _{x \rightarrow 0} x \sin \left(\frac{1}{x}\right)=0 $$ Why couldn鈥檛 we have obtained the same result by writing $$ \begin{aligned} \lim _{x \rightarrow 0} x \sin \left(\frac{1}{x}\right) &=\lim _{x \rightarrow 0} x \cdot \lim _{x \rightarrow 0} \sin \left(\frac{1}{x}\right) \\ &=0 \cdot \lim _{x \rightarrow 0} \sin \left(\frac{1}{x}\right)=0 ? \end{aligned} $$

Find the limits. $$ \lim _{x \rightarrow+\infty} \sin \left(\frac{\pi x}{2-3 x}\right) $$

Show that the equation \(x^{3}+x^{2}-2 x=1\) has at least one solution in the interval \([-1,1] .\)

(a) Use the Intermediate-Value Theorem to show that the equation \(x=\cos x\) has at least one solution in the interval \([0, \pi / 2] .\) (b) Show graphically that there is exactly one solution in the interval. (c) Approximate the solution to three decimal places.

Find the limits. $$ \lim _{x \rightarrow 0} e^{\sin x} $$
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