/*! 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 51 Using L'Hôpital's rule (Section... [FREE SOLUTION] | 91Ó°ÊÓ

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Using L'Hôpital's rule (Section \(3.6\) ) one can verify that $$ \lim _{x \rightarrow+\infty} \frac{e^{x}}{x}=+\infty, \quad \lim _{x \rightarrow+\infty} \frac{x}{e^{x}}=0, \quad \lim _{x \rightarrow-\infty} x e^{x}=0 $$ In these exercises: (a) Use these results, as necessary, to find the limits of \(f(x)\) as \(x \rightarrow+\infty\) and as \(x \rightarrow-\infty\). (b) Sketch a graph of \(f(x)\) and identify all relative extrema, inflection points, and asymptotes (as appropriate). Check your work with a graphing utility. $$ f(x)=\frac{e^{2}}{1-x} $$

Short Answer

Expert verified
As \( x \rightarrow \pm \infty, f(x) \rightarrow 0 \). The graph has a vertical asymptote at \( x=1 \).

Step by step solution

01

Determine Limit as x Approaches Infinity

Consider the function \( f(x) = \frac{e^{2}}{1-x} \). As \( x \rightarrow \infty \), the denominator \( 1-x \rightarrow -\infty \). Therefore, \( f(x) = \frac{e^{2}}{1-x} \) will also approach 0 because \( \frac{a}{-\infty} \rightarrow 0 \) for any constant \( a \).
02

Determine Limit as x Approaches Negative Infinity

As \( x \rightarrow -\infty \), the denominator \( 1-x \) becomes a very large positive number since \( x \) is negative and \( 1 - (-x) = 1 + x \rightarrow \infty \). Therefore, as \( x \rightarrow -\infty \), \( f(x) \rightarrow 0 \) because \( \frac{a}{\infty} \rightarrow 0 \).
03

Analyze the Function for Extrema and Inflection Points

The function \( f(x) = \frac{e^{2}}{1-x} \) does not have any critical points where the derivative exists because the numerator is a constant, and the denominator defines the rate of change. Based on the form of the function, there are no places where the derivative equals zero, implying no relative extrema exist within the domain of this function.
04

Identify Asymptotes

The function \( f(x) = \frac{e^{2}}{1-x} \) has a vertical asymptote at \( x = 1 \) since the denominator (\( 1-x \)) becomes zero, causing the function to become undefined. There is no horizontal asymptote since as \( x \rightarrow \pm \infty \), the function approaches 0 which implies the x-axis is also a horizontal asymptote.
05

Graph the Function

Plot \( f(x) = \frac{e^{2}}{1-x} \): it decreases towards 0 as \( x \rightarrow +\infty \) and \( x \rightarrow -\infty \), with a vertical asymptote at \( x = 1 \). The function does not cross the x-axis or y-axis, as \( f(x) eq 0 \) for any \( x \). It is a hyperbolic decay function with its curve approaching but never touching the asymptotes.

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Key Concepts

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

Understanding Limits
Limits are a fundamental concept in calculus that describe the behavior of a function as the input approaches a particular value. In the original exercise, we explore the limits of the function \( f(x) = \frac{e^2}{1-x} \) as \( x \to +\infty \) and \( x \to -\infty \).

- As \( x \to +\infty \), the denominator \( 1-x \) becomes significantly negative, thus making the fraction approach zero. This is because dividing a constant by a very large negative number results in a value very close to zero.- Similarly, as \( x \to -\infty \), the denominator \( 1-x \) nears positive infinity. Once more, dividing \( e^2 \) by an infinite value approaches zero.

In simpler terms, limits allow us to understand the end behavior of functions and how they behave as inputs extend toward infinity or negative infinity. They're crucial in determining the asymptotic nature of functions and are fundamental in calculus for establishing continuity and defining derivatives.
Exploring Asymptotes
Asymptotes give important information about how a function behaves as the input values become very large or very small.

- A **vertical asymptote** occurs when the function approaches infinity as the input gets close to certain values. In the function \( f(x) = \frac{e^2}{1-x} \), there is a vertical asymptote at \( x = 1 \). This is because when \( x \) is 1, the denominator becomes zero, making the function value undefined.- On the other hand, **horizontal asymptotes** describe the value that a function approaches as \( x \) goes to infinity or negative infinity. For our function, as discussed earlier, \( f(x) \) approaches zero as \( x \to \pm \infty \), hence the x-axis (\( y = 0 \)) can be considered a horizontal asymptote.

Understanding asymptotes helps with graphing the function and predicting its long-term behavior. They show where the function stabilizes, and where it explodes, providing valuable insights into the function's growth and decay patterns.
Graphing Functions with Asymptotic Behavior
Graphing functions provides a visual representation of the mathematical relationships described by equations. In the given function \( f(x) = \frac{e^2}{1-x} \), we perceive its behavior based on its limits and asymptotes.

- **Plotting the Function:** Start by determining the key features—notice where it gets close to an axis or undefined. In this function, it approaches zero as \( x \rightarrow \pm \infty \), verifying our previous limit analysis.- **Identifying Asymptotes:** Include the vertical asymptote at \( x = 1 \) on your graph. This line represents where the function values shoot infinitely high or low, as \( x \) nears 1.- **No Crossing Points:** Interestingly, this function does not intercept the x-axis or y-axis, emphasizing its behavior below zero along the entire curve.

Sketching functions impresses upon us the importance of recognizing asymptotic paths and limit behavior. With these observations, predicting function behavior becomes intuitive, providing a thorough understanding of both local and global function characteristics.

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Most popular questions from this chapter

A man is floating in a rowboat 1 mile from the (straight) shoreline of a large lake. A town is located on the shoreline 1 mile from the point on the shoreline closest to the man. As suggested in Figure Ex-58, he intends to row in a straight line to some point \(P\) on the shoreline and then walk the remaining distance to the town. To what point should he row in order to reach his destination in the least time if (a) he can walk \(5 \mathrm{mi} / \mathrm{h}\) and row \(3 \mathrm{mi} / \mathrm{h}\) (b) he can walk \(5 \mathrm{mi} / \mathrm{h}\) and row \(4 \mathrm{mi} / \mathrm{h}\) ?

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