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(i) Approximate the \(y\) -coordinates of all horizontal asymptotes of \(y=f(x)\) by evaluating \(f\) at the points \(\pm 10 . \pm 100 . \pm 1000 .\pm 100.000,\) and \(\pm 100.000 .000\) (ii) Confirm your conclusions by graphing \(y=f(x)\) over an appropriate interval. (iii) If you have a CAS, then use it to find the horizontal asymptotes. (a) \(f(x)=\frac{x^{2}-1}{5 x^{2}+1}\) (b) \(f(x)=\left(2+\frac{1}{x}\right)^{x}\) (c) \(f(x)=\frac{\sin x}{x}\)

Short Answer

Expert verified
(a) HA: \( y=\frac{1}{5} \); (b) No HA; (c) HA: \( y=0 \).

Step by step solution

01

Understanding Horizontal Asymptotes

A horizontal asymptote (HA) of a function is a horizontal line that the graph of the function approaches as \( x \) approaches \( \pm \infty \). For the function \( f(x) \), we need to evaluate the behavior as \( x \) goes to \( +\infty \) and \( -\infty \).
02

Evaluate \( f(x) \) for \( f(x)=\frac{x^{2}-1}{5x^{2}+1} \)

Calculate \( f(x) \) at \( x = \pm 10, \pm 100, \pm 1000, \pm 100000, \pm 100000000 \):- For large \( x \), the terms \( -1 \) and \(+1 \) in the numerator and denominator become negligible, so \( f(x)\approx\frac{x^2}{5x^2}=\frac{1}{5} \).- Evaluating \( f(x) \): - \( f(10) \approx \frac{99}{501} \approx 0.1976 \) - \( f(100) \approx \frac{9999}{50001} \approx 0.19996 \) - \( f(1000) \approx \frac{999999}{5000001} \approx 0.199996 \)- As \( x \to \pm \infty \), \( f(x) \to \frac{1}{5} \). So the horizontal asymptotes are \( y = \frac{1}{5} \).
03

Evaluate \( f(x) \) for \( f(x)=\left(2+\frac{1}{x}\right)^{x} \)

As \( x \to \infty \), \( \frac{1}{x} \to 0 \) and \( 2+\frac{1}{x} \to 2 \). Hence \( f(x) \to 2^x \), which means it increases without bound. Thus, there are no horizontal asymptotes.For \( x \to -\infty \), \( 2+\frac{1}{x} \to 2 \), but the behavior differs as the exponent is negative, making it negligible, indicating no horizontal asymptote as well.
04

Evaluate \( f(x) \) for \( f(x)=\frac{\sin x}{x} \)

As \( x \to \pm \infty \), \( \sin(x) \) oscillates between -1 and 1, making \( \frac{\sin x}{x} \to 0 \). Therefore, the horizontal asymptote is \( y = 0 \).
05

Graphical Confirmation

Graph each function over a large domain to visually confirm the horizontal asymptotes:- For \( f(x)=\frac{x^{2}-1}{5x^{2}+1} \): Confirm \( y = \frac{1}{5} \).- For \( f(x)=\left(2+\frac{1}{x}\right)^{x} \): Observe no horizontal asymptotes as it grows undetermined.- For \( f(x)=\frac{\sin x}{x} \): Confirm \( y = 0 \).
06

Use CAS for Verification (Optional)

Use CAS (Computer Algebra System) to analytically determine the limits at \( \pm \infty \) for additional confirmation:- \( \lim_{{x \to \pm \infty}} \frac{x^{2}-1}{5x^{2}+1} = \frac{1}{5} \)- \( \lim_{{x \to \pm \infty}} \left(2+\frac{1}{x}\right)^{x} \) shows unbounded behavior.- \( \lim_{{x \to \pm \infty}} \frac{\sin x}{x} = 0 \)

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Key Concepts

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

Behavior at Infinity
When we analyze the behavior at infinity of a given function, we are essentially investigating what happens to the function values as the input, or the variable \( x \), becomes extremely large or extremely small. This type of analysis is crucial for understanding the horizontal asymptotes of a function.

Horizontal asymptotes provide insight into the long-term behavior of the function. When a function has a horizontal asymptote at \( y = L \), it means that the function values approach \( L \) as \( x \) approaches infinity (either positive or negative). To determine this behavior, it's helpful to substitute increasingly large values of \( x \) and observe the resulting function values.

For example, consider the function \( f(x) = \frac{x^2 - 1}{5x^2 + 1} \). As \( x \) becomes very large, the terms \(-1\) and \(+1\) in the numerator and denominator become negligible, and \( f(x) \) simplifies to approximately \( \frac{x^2}{5x^2} = \frac{1}{5} \). Therefore, the function approaches the horizontal asymptote \( y = \frac{1}{5} \) as \( x \to \pm\infty \).
Another example is \( f(x) = \frac{\sin x}{x} \), where \( \sin x \) oscillates between \(-1\) and \(+1\) regardless of \( x \), but \( x \) keeps increasing or decreasing, leading \( \frac{\sin x}{x} \to 0 \) as \( x \to \pm\infty \). Therefore, the horizontal asymptote is \( y = 0 \).
Limiting Behavior of Functions
The limiting behavior of a function describes what happens to the function's output as the input approaches a particular value, notably infinity. Investigating limits is a fundamental concept in calculus, particularly with regard to asymptotic behavior.

To grasp limiting behavior, consider again the function \( f(x) = \frac{x^2 - 1}{5x^2 + 1} \). As discussed, for large values of \( x \), the \(-1\) and \(+1\) terms become irrelevant, leading us to conclude that \( \lim_{{x \to \pm\infty}} \frac{x^2 - 1}{5x^2 + 1} = \frac{1}{5} \). This means that no matter how large or small \( x \) gets, the function value will be close to \( \frac{1}{5} \).
Another example is \( f(x) = \left( 2 + \frac{1}{x} \right)^x \), which poses an intriguing behavior as \( x \to \infty \) because the \( \frac{1}{x} \) term diminishes, causing the function value to increase without limit. Thus, there is no horizontal asymptote, as the function doesn’t stabilize toward a particular value.
Calculating limits such as these helps in understanding if the function stabilizes—approaches a finite, constant value—as \( x \) heads towards positive or negative infinity, or if the function keeps growing or shrinking indefinitely.
Graphical Analysis of Functions
Graphical analysis is a valuable method to visually confirm the presence and nature of horizontal asymptotes derived from analytical calculations. By plotting the function over a vast interval, you can observe the trends and patterns in the graph.

Take for instance \( f(x) = \frac{x^2 - 1}{5x^2 + 1} \). Graphically, as you plot \( f(x) \), especially over large values of \( x \), the graph tends to hover around \( y = \frac{1}{5} \), visually confirming the horizontal asymptote deduced mathematically.
In contrast, plotting \( f(x) = \left(2 + \frac{1}{x}\right)^x \) reveals a graph that continues to rise without settling into a horizontal asymptote. This visual cue helps affirm that the function lacks a horizontal asymptote and supports the understanding that \( f(x) \) increases indefinitely beyond any finite bound.
Lastly, for \( f(x) = \frac{\sin x}{x} \), visualizing the graph depicts how \( f(x) \) approaches \( y = 0 \) as \( x \) becomes very large in the positive or negative direction. Even with the oscillations of \( \sin x \), \( f(x) \) moves closer to the x-axis, confirming \( y = 0 \) as the horizontal asymptote. Graphical analysis thus helps in solidifying both analytical predictions and enhances understanding through visual confirmation.

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