/*! 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 83 Sometimes a change of variable c... [FREE SOLUTION] | 91Ó°ÊÓ

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Sometimes a change of variable can change an unfamiliar expression into one whose limit we know how to find. For example, $$ \begin{aligned} \lim _{x \rightarrow \infty} \sin \frac{1}{x} &=\lim _{\theta \rightarrow 0^{+}} \sin \theta \\ &=0 \end{aligned} $$ This suggests a creative way to "see" limits at infinity. Describe the procedure and use it to picture and determine limits in Exercises \(79-84 .\) $$ \lim _{x \rightarrow \pm \infty}\left(3+\frac{2}{x}\right)\left(\cos \frac{1}{x}\right) $$

Short Answer

Expert verified
The limit is 3.

Step by step solution

01

Identify the limit expression

The expression we need to find the limit of as \( x \) approaches \( \pm \infty \) is \( \left(3+\frac{2}{x}\right)\left(\cos \frac{1}{x}\right) \).
02

Apply change of variable

Substitute \( \theta = \frac{1}{x} \). As \( x \rightarrow \pm \infty \), \( \theta \rightarrow 0 \). This allows us to rewrite the expression in terms of \( \theta \): \( \left(3+2\theta\right)\left(\cos \theta\right) \).
03

Analyze the limits of the components

Evaluate the limits separately: - Since \( \theta \rightarrow 0 \), \( 3+2\theta \rightarrow 3 \).- \( \cos \theta \rightarrow 1 \) as \( \theta \rightarrow 0 \).
04

Determine the limit of the entire expression

Multiply the limits of the components: \( \lim_{\theta \rightarrow 0} \left(3+2\theta\right)\left(\cos \theta\right) = 3 \times 1 = 3 \).

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Key Concepts

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

Change of Variable
When it comes to solving limit problems, particularly those involving limits at infinity, a useful technique is the change of variable. Changing variables can transform complex expressions into simpler ones that we already understand. For the exercise we explored, the original expression was \( \left(3+\frac{2}{x}\right)\left(\cos \frac{1}{x}\right) \).

To simplify this, we introduced a new variable \( \theta \) such that \( \theta = \frac{1}{x} \). As \( x \to \pm \infty \), \( \theta \to 0 \). This change of variable helped translate our complex expression into a simpler form: \( \left(3+2\theta\right)\left(\cos \theta\right) \).

With this substitution, we can easily analyze the behavior of the function as \( x \to \pm \infty \) by instead examining how it behaves as \( \theta \to 0 \). This approach thus provides a fresh perspective making it easier to calculate limits, especially when initial expressions seem intractable.
  • Simplifies complex problems.
  • Transforms the approach to finding limits.
  • Makes difficult functions more manageable.
Limits at Infinity
Understanding limits at infinity involves determining the behavior of a function as the input grows without bound, either positively or negatively. In mathematical expressions like \( \lim_{x \to \pm \infty} \), we seek to find what value, if any, the function approaches as \( x \) becomes extremely large or extremely small in the negative direction.

To solve such problems, breaking down the components of the expression can often make it clearer. In our example, using the change of variable simplified the expression to \( \left(3+2\theta\right)\left(\cos \theta\right) \). As \( \theta \to 0 \), we see that \( 3+2\theta \) converges to 3, while \( \cos \theta \) tends to 1. The product of these two limits, therefore, will be 3.

This illustrates a key principle: by analyzing the simpler limits of individual components, we often find that the overall limit at infinity reveals itself easily. This can demystify challenging problems by showing that at their limits, complex expressions can become surprisingly straightforward.
  • Essential for understanding function behavior.
  • Analyzes component limits separately.
  • Simplifies complex expressions by understanding their infinite bounds.
Trigonometric Functions
Trigonometric functions like sine and cosine are periodic and bounded, meaning they oscillate within a particular range of values. For instance, the cosine function swings between -1 and 1. In limit problems, particularly at infinity, handling these oscillations requires understanding their behaviors at critical points.

In our scenario, due to the change of variable where \( \theta = \frac{1}{x} \), as \( x \to \pm \infty \), \( \theta \to 0 \) which brings \( \cos \theta \to \cos 0 = 1 \). This implies that the function stabilizes to its value at zero, aiding in simplifying the evaluation of the limit.

The periodic nature of cosine is crucial because even as \( x \) gets exceedingly large, cosine remains well-behaved, offering a predictable result. When analyzing limits that include trigonometric terms, always consider the values that these functions approach as their inputs converge to singular points like zero.
  • Periodic and bounded behavior define trigonometric functions.
  • Cosine stabilizes the limit evaluation by approaching consistent values.
  • Valuable in simplifying the determination of limits at infinity.

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Most popular questions from this chapter

Graph the rational functions in Exercises \(27-38 .\) Include the graphs and equations of the asymptotes and dominant terms. $$ y=\frac{2 x}{x+1} $$

Stretching a rubber band Is it true that if you stretch a rubber band by moving one end to the right and the other to the left, some point of the band will end up in its original position? Give reasons for your answer.

A function discontinuous at every point a. Use the fact that every nonempty interval of real numbers contains both rational and irrational numbers to show that the function $$ f(x)=\left\\{\begin{array}{ll}{1,} & {\text { if } x \text { is rational }} \\\ {0,} & {\text { if } x \text { is rrational }} \\ {0,} & {\text { if } x \text { is irrational }}\end{array}\right. $$ is discontinuous at every point. b. Is \(f\) right-continuous or left-continuous at any point?

Graphing Secant and Tangent Lines Use a CAS to perform the following steps for the functions in Exercises \(45-48 .\) a. Plot \(y=f(x)\) over the interval \(\left(x_{0}-1 / 2\right) \leq x \leq\left(x_{0}+3\right)\) b. Holding \(x_{0}\) fixed, the difference quotient $$ q(h)=\frac{f\left(x_{0}+h\right)-f\left(x_{0}\right)}{h} $$ at \(x_{0}\) becomes a function of the step size \(h .\) Enter this function into your CAS workspace. c. Find the limit of \(q\) as \(h \rightarrow 0\) d. Define the secant lines \(y=f\left(x_{0}\right)+q \cdot\left(x-x_{0}\right)\) for \(h=3,2\) and \(1 .\) Graph them together with \(f\) and the tangent line over the interval in part (a). $$ f(x)=x+\frac{5}{x}, \quad x_{0}=1 $$

In Exercises \(5-10\) , find an equation for the tangent to the curve at the given point. Then sketch the curve and tangent together. $$ y=2 \sqrt{x}, \quad(1,2) $$

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