/*! 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 1 Evaluate each limit (or state th... [FREE SOLUTION] | 91Ó°ÊÓ

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Evaluate each limit (or state that it does not exist). $$ \lim _{x \rightarrow \infty} \frac{1}{x^{2}} $$

Short Answer

Expert verified
The limit is 0.

Step by step solution

01

Determine the Type of Limit

This is a limit where the variable \(x\) is approaching infinity. Consider the form of the expression \( \frac{1}{x^2} \). As \( x \to \infty \), evaluate how each component behaves.
02

Analyze the Expression

For very large values of \(x\), the expression \(x^2\) becomes very large (tending towards infinity). The expression \(\frac{1}{x^2}\), therefore, becomes very small as the denominator becomes very large.
03

Evaluate the Limit

Since \(\frac{1}{x^2}\) tends to 0 as \(x\) grows without bound, conclude that the limit is \(0\). Therefore: \[\lim_{x \to \infty} \frac{1}{x^2} = 0\]

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Key Concepts

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

Infinity in Calculus
Infinity in calculus is a concept that expresses the idea of something growing without bounds. In calculus, infinity is often encountered when evaluating limits as variables approach extremely large or small values. When we talk about limits approaching infinity, we are interested in understanding how a function behaves as the input or independent variable becomes indefinitely large. In our original exercise, the variable \( x \) approaches infinity.
  • Infinity can be positive (\( \infty \)) or negative (\( -\infty \)).
  • It is not a number, but rather a concept indicating boundlessness.
When a function approaches infinity, it either shoots up without bound or down (for negative infinity). Understanding these concepts helps us comprehend the behavior of functions as they grow larger. This is essential for evaluating limits involving infinity.
Evaluating Limits
Evaluating limits is the process of finding the value that a function approaches as the input gets closer to a specified point, often infinite. In calculus, limits provide a way of understanding the behavior of functions as variables get very large or very small. In the original exercise, we evaluated the limit \( \lim_{x \to \infty} \frac{1}{x^2} \).
  • Consider the behavior of \( x^2 \) as \( x \) grows; it becomes extremely large.
  • The fraction \( \frac{1}{x^2} \) decreases because as the denominator increases, the overall value becomes smaller.
  • Ultimately, as \( x \to \infty \), \( \frac{1}{x^2} \) approaches \( 0 \).
By applying these principles, we determined that the limit is \( 0 \). This process involves identifying how each component of the expression behaves, especially when dealing with powers and fractions.
Behavior of Functions at Infinity
The behavior of functions at infinity explains how functions behave as the variable approaches very large values. For the function \( \frac{1}{x^2} \), as \( x \) becomes extremely large, its behavior is such that the value of the function diminishes.
  • As \( x \to \infty \), \( x^2 \) becomes infinitely large.
  • The fraction \( \frac{1}{x^2} \) thus shrinks towards \( 0 \) because the denominator overwhelms the numerator.
This understanding of function behavior at infinity is crucial for recognizing limits and predicting what happens to the output of functions at extreme input values. It's the basis for understanding horizontal asymptotes and how they reflect the long-term behavior of functions on a graph.

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Most popular questions from this chapter

In the reservoir model, the heart is viewed as a balloon that swells as it fills with blood (during a period called the systole), and then at time \(t_{0}\) it shuts a valve and contracts to force the blood out (the diastole). Let \(p(t)\) represent the pressure in the heart at time \(t\) a. During the diastole, which lasts from \(t_{0}\) to time \(T, p(t)\) satisfies the differential equation $$\frac{d p}{d t}=-\frac{K}{R} p$$ Find the general solution \(p(t)\) of this differential equation. ( \(K\) and \(R\) are positive constants determined, respectively, by the strength of the heart and the resistance of the arteries. The differential equation states that as the heart contracts, the pressure decreases \((d p / d t\) is negative) in proportion to itself.) b. Find the particular solution that satisfies the condition \(p\left(t_{0}\right)=p_{0} . \quad\left(p_{0}\right.\) is a constant representing the pressure at the transition time \(t_{0}\).) c. During the systole, which lasts from time 0 to time \(t_{0}\), the pressure \(p(t)\) satisfies the differential equation $$\frac{d p}{d t}=K I_{0}-\frac{K}{R} p$$ Find the general solution of this differential equation. \(\left(I_{0}\right.\) is a positive constant representing the constant rate of blood flow into the heart while it is expanding.) [Hint: Use the same \(u\) -substitution technique that was used in Example 7.] d. Find the particular solution that satisfies the condition \(p\left(t_{0}\right)=p_{0}\). e. In parts (b) and (d) you found the formulas for the pressure \(p(t)\) during the diastole \(\left(t_{0} \leq t \leq T\right)\) and the systole \(\left(0 \leq t \leq t_{0}\right)\). Since the heart behaves in a cyclic fashion, these functions must satisfy \(p(T)=p(0)\). Equate the solutions at these times (use the correct formula for each time) to derive the important relationship $$ R=\frac{p_{0}}{I_{0}} \frac{1-e^{-K T / R}}{1-e^{-K t_{0} / R}} $$

Solve each differential equation and initial condition and verify that your answer satisfies both the differential equation and the initial condition. $$ \left\\{\begin{array}{l} y^{\prime}=a x^{2} y \\ y(0)=2 \end{array}\right. $$

An annuity is a fund into which one makes equal payments at regular intervals. If the fund earns interest at rate \(r\) compounded continuously, and deposits are made continuously at the rate of \(d\) dollars per year (a "continuous annuity"), then the value \(y(t)\) of the fund after \(t\) years satisfies the differential equation \(y^{\prime}=d+r y\). (Do you see why?) Solve the differential equation above for the continuous annuity \(y(t)\) with deposit rate \(d=\$ 1000\) and continuous interest rate \(r=0.05\), subject to the initial condition \(y(0)=0 \quad\) (zero initial value).

An annuity is a fund into which one makes equal payments at regular intervals. If the fund earns interest at rate \(r\) compounded continuously, and deposits are made continuously at the rate of \(d\) dollars per year (a "continuous annuity"), then the value \(y(t)\) of the fund after \(t\) years satisfies the differential equation \(y^{\prime}=d+r y\). (Do you see why?) Solve the differential equation above for the continuous annuity \(y(t)\), where \(d\) and \(r\) are unknown constants, subject to the initial condition \(y(0)=0\) (zero initial value).

Your company has developed a new product, and your marketing department has predicted how it will sell. Let \(y(t)\) be the (monthly) sales of the product after \(t\) months. a. Write a differential equation that says that the rate of growth of the sales will be four times the one-half power of the sales. b. Write an initial condition that says that at time \(t=0\) sales were 10,000 . c. Solve this differential equation and initial value. d. Use your solution to predict the sales at time \(t=12\) months.

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