Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Why does \(\lim _{x \rightarrow+\infty} \frac{f(x)}{g(x)}\) not exist, though \(\lim _{x \rightarrow+\infty} \frac{f^{\prime}(x)}{g^{\prime}(x)}\) does, in the fol- lowing example? Verify and explain. $$ f(x)=e^{-2 x}(\cos x+2 \sin x), \quad g(x)=e^{-x}(\cos x+\sin x) $$ [Hint: \(g^{\prime}\) vanishes many times in each \(G_{+\infty}\). Use the Darboux property for the proof.]

Short Answer

Expert verified
The original limit does not stabilize due to oscillations; derivatives simplify relations, making the limit accessible.

Step by step solution

01

Examine the given functions

The functions provided are \( f(x) = e^{-2x} (\cos x + 2 \sin x) \) and \( g(x) = e^{-x} (\cos x + \sin x) \). These are oscillatory components multiplied by exponential decay terms. As \( x \to +\infty \), both functions approach 0 due to the exponential decay.
02

Evaluate the original limit

To understand \( \lim_{x \to +\infty} \frac{f(x)}{g(x)} \), substitute \( f(x) \) and \( g(x) \). This limit becomes:\[ \lim_{x \to +\infty} \frac{e^{-2x} (\cos x + 2\sin x)}{e^{-x} (\cos x + \sin x)} = \lim_{x \to +\infty} e^{-x} \frac{\cos x + 2\sin x}{\cos x + \sin x}. \]Since both \( \cos x + 2\sin x \) and \( \cos x + \sin x \) oscillate between positive and negative values, and \( e^{-x} \to 0 \), the overall limit does not stabilize to a specific finite value, making the limit non-existent.
03

Differentiate the functions

Calculate the derivatives:- \( f'(x) = \frac{d}{dx}[e^{-2x} (\cos x + 2\sin x)] = e^{-2x}(-2\cos x - 4\sin x - 3\sin x - 2\cos x) = e^{-2x}(\cos x - 3\sin x) \).- \( g'(x) = \frac{d}{dx}[e^{-x}(\cos x + \sin x)] = e^{-x}(-\cos x - \sin x - \sin x + \cos x) = e^{-x}(\cos x - \sin x) \).
04

Calculate L'Hôpital's limit

Using the derivatives from Step 3, consider:\[ \lim_{x \to +\infty} \frac{f'(x)}{g'(x)} = \lim_{x \to +\infty} \frac{e^{-2x}(\cos x - 3\sin x)}{e^{-x}(\cos x - \sin x)} = \lim_{x \to +\infty} e^{-x} \frac{\cos x - 3\sin x}{\cos x - \sin x}. \]As \( x \to +\infty \), \( e^{-x} \to 0 \), but the fraction \( \frac{\cos x - 3\sin x}{\cos x - \sin x} \) does not reach a conclusive value, indicating that although the level of decay is less pronounced, the overall limit is more predictable based on smaller oscillation influences.
05

Understand why \(g'(x)\) vanishes (Hint Analysis)

The hint specifies that \( g'(x) \) vanishes often within any larger interval, similar to the period of trigonometric functions. Such vanishing points complicate the limit \( \lim_{x \to +\infty} \frac{f'(x)}{g'(x)} \). Despite this, the Darboux property suggests that if \( g'(x) \) crosses zero often, \( \lim_{x \to +\infty} \frac{f'(x)}{g'(x)} \) still exists due to the intermediary values theorem ensured by continuous oscillation.
06

Conclusion

The original limit does not exist due to inconsistent oscillation and decay interactions leading to indefinite bounds. However, by using derivatives and L'Hôpital's rule, the limit of derivatives simplifies the expression and makes the interplay between exponential decay and oscillation more manageable, yielding an undefined but consistent nature.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Limits of functions
Limits are a fundamental concept in calculus, often used to determine the behavior of functions as they approach a certain point or infinity. Understanding limits is crucial when analyzing how functions interact under various conditions, such as approaching zero or infinity, especially for determining their long-term behavior.
In this example, we have an expression \(\lim _{x \rightarrow+\infty} \frac{f(x)}{g(x)} \) which involves two functions, both comprising exponential decay and oscillatory components. As \( x \rightarrow +\infty \), the compounded effect of these components causes each function to approach zero, but unpredictable oscillations prevent the overall ratio \( \frac{f(x)}{g(x)} \) from reaching a definite value. Thus, the limit does not exist in this case due to the undefined interaction between oscillatory behavior and exponential decay.
Derivative applications
Derivatives provide a powerful tool for examining the rate of change of functions and applying calculus concepts such as L'Hôpital's Rule. In our problem, differentiating the numerator \( f(x) \) and the denominator \( g(x) \) helps evaluate an otherwise indeterminate form.
By deriving \( f'(x) \) and \( g'(x) \), we simplify their ratios for further inspection, enabling us to apply L'Hôpital's Rule effectively. This rule states that for the ratio of functions \( \frac{f(x)}{g(x)} \) in indeterminate forms like \( \frac{0}{0} \), taking their derivatives can yield a determinable limit \( \frac{f'(x)}{g'(x)} \).
Although oscillations affect our example, differentiating the functions helps analyze the diluted impact of oscillations within limits. Without derivatives, determining such behavior analytically becomes more challenging, precisely why they are so vital in calculus.
Oscillatory functions
Oscillatory functions, like sine and cosine, fluctuate between values over their periods. These functions introduce complications into limit calculations because they do not stabilize over infinity, varying continuously instead.
In our example, both numerator and denominator functions include oscillatory components \( \cos x \) and \( \sin x \), causing continuous highs and lows. Because their oscillatory nature does not allow them to converge towards a single value as \( x \rightarrow +\infty \), determining the overall limit becomes problematic.
The continuous fluctuations make the ratio \( \frac{f(x)}{g(x)} \) or \( \frac{f'(x)}{g'(x)} \) appear unpredictable, yet mathematically controlled through approaches like L'Hôpital's Rule to attain a more comprehensible limiting behavior. Recognizing these prospects is essential when tackling such functions in advanced calculus.
Exponential decay
Exponential decay describes a process where quantities decrease at a rate proportional to their current value. In calculus, functions with terms like \( e^{-x} \) or \( e^{-2x} \) diminish rapidly as \( x \rightarrow +\infty \).
For our problem, exponential decay significantly affects the functions \( f(x) \) and \( g(x) \) by driving them towards zero. This decay is a dominant factor, as it can overshadow other elements within a function, like oscillations.
Despite this powerful decreasing trend, the limit of the ratio \( \frac{f(x)}{g(x)} \) does not exist because the oscillations do not permit a stable relationship between the rates at which \( f(x) \) and \( g(x) \) approach zero. Thus, even strong exponential decay cannot neutralize the inherent inconsistencies created by oscillations, highlighting the complex interplay between decay and oscillation in determining limits.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free