/*! 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 48 Let $$ g(t)=\left\\{\begin{a... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 48

Let $$ g(t)=\left\\{\begin{array}{ll}{\frac{2+3 t}{5 t^{2}+6},} & {t<1,000,000} \\\ {\frac{\sqrt{36 t^{2}-100}}{5-t},} & {t>1,000,000}\end{array}\right. $$ Find $$ \begin{array}{ll}{\text { (a) } \lim _{t \rightarrow-\infty} g(t)} & {\text { (b) } \lim _{t \rightarrow+\infty} g(t)}\end{array} $$

Short Answer

Expert verified
(a) 0; (b) -6.

Step by step solution

01

Understanding the function

We are given a piecewise function \( g(t) \), which is defined differently for \( t < 1,000,000 \) and \( t > 1,000,000 \). The two expressions are: \( g(t) = \frac{2 + 3t}{5t^2 + 6} \) for \( t < 1,000,000 \), and \( g(t) = \frac{\sqrt{36t^2 - 100}}{5-t} \) for \( t > 1,000,000 \). We need to find the limits as \( t \to -\infty \) and \( t \to +\infty \).
02

Find \( \lim_{t \to -\infty} g(t) \)

For \( t < 1,000,000 \), the function is \( g(t) = \frac{2 + 3t}{5t^2 + 6} \). As \( t \to -\infty \), the higher degree terms dominate, so we approximate the limit using: \[ g(t) \approx \frac{3t}{5t^2} = \frac{3}{5t} \]. As \( t \to -\infty \), \( \frac{3}{5t} \to 0 \). Hence, \( \lim_{t \to -\infty} g(t) = 0 \).
03

Analyze the second part of the function

For \( t > 1,000,000 \), the function is \( g(t) = \frac{\sqrt{36t^2 - 100}}{5-t} \). As \( t \to +\infty \), we need to simplify this expression to find the limit.
04

Simplify the expression for large \( t \)

For large \( t \), \( \sqrt{36t^2 - 100} \approx \sqrt{36t^2} = 6t \). The expression simplifies to \( g(t) \approx \frac{6t}{5-t} \). Further simplify this as \( t \to +\infty \); the highest power of \( t \) in both numerator and denominator is \( t \), which simplifies to \( \frac{6t}{-t} = -6 \).
05

Find \( \lim_{t \to +\infty} g(t) \)

Therefore, as \( t \to +\infty \), the expression reduces to \( -6 \). Thus, \( \lim_{t \to +\infty} g(t) = -6 \).

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 91影视!

Key Concepts

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

Limits
The concept of limits is foundational in calculus, used to understand the behavior of functions as the input approaches certain values. It helps in determining the value that a function approaches as the variable within it reaches a particular point, which can be finite or infinite. For example, when we examine the limit of a function as the variable approaches infinity, we are trying to ascertain how the function behaves in the extremes.
  • Approaching a Value: Limits describe the trend of a function as it gets very close to a particular_value.
  • Infinite Limits: These occur when a function approaches infinity or negative infinity as the variable itself approaches a particular value.
  • Applications: Limits are used to define derivatives, integrals, and continuity, forming the backbone of calculus.
In our specific exercise, for the function defined differently for very large and very small values of t, limits help determine the function's behavior towards the extremes, such as when t becomes significantly positive or negative.
Piecewise Functions
Piecewise functions are defined by different expressions based on the value of the input variable. They can appear complex at first, but the key is to break them down into manageable parts and understand how each piece behaves.
  • Definition: A piecewise function is expressed with different formulas based on different intervals of the independent_variable.
  • Relevance: They model real-world situations where a rule or relationship changes based on different scenarios.
  • Analyzing: To analyze a piecewise function, consider each 'piece' separately and evaluate its behavior over its specific interval.
In the provided exercise, the function g(t) is defined differently for t values less or greater than 1,000,000. This means that for any evaluation, we need to consider which part of the function applies to the value of t we are dealing with. When t is less than 1,000,000, we use one formula, and for values greater, we switch to another.
Infinity
In calculus, infinity (\(\infty\)) is a concept to describe something without bounds. It is not a number but a way to represent a limitless approach in calculus, particularly useful in understanding behavior at the extremes of a function.
  • Not a Number: Infinity is a concept, not a calculable number.
  • Used in Limits: It is used extensively in limits to describe behavior of functions as they progress toward endlessly large or small values.
  • Symbol: The symbol for infinity is \(\infty\), and can be positive or negative depending on the direction of approach.
In the exercise, as t approaches negative or positive infinity, we needed to calculate how the function g(t) behaves. Understanding this helps predict long-term trends and behaviors of various mathematical models and function behaviors.

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

Suppose that \(M\) is a positive number and that for all real numbers \(x\), a function \(f\) satisfies $$ -M \leq f(x) \leq M $$ Then $$ \lim _{x \rightarrow 0} x f(x)=0 \quad \text { and } \quad \lim _{x \rightarrow+\infty} \frac{f(x)}{x}=0 $$

Find the limits. $$ \lim _{x \rightarrow+\infty} \sin ^{-1}\left(\frac{x}{1-2 x}\right) $$

Find the limits. $$ \lim _{x \rightarrow 0^{+}} \frac{\sin x}{5 \sqrt{x}} $$

Sketch the graphs of the curves \(y=1-x^{2}, y=\cos x,\) and \(y=f(x),\) where \(f\) is a function that satisfies the inequalities $$ 1-x^{2} \leq f(x) \leq \cos x $$ for all \(x\) in the interval \((-\pi / 2, \pi / 2)\). What can you say about the limit of \(f(x)\) as \(x \rightarrow 0 ?\) Explain.

(a) Use the Intermediate-Value Theorem to show that the equation \(x=\cos x\) has at least one solution in the interval \([0, \pi / 2] .\) (b) Show graphically that there is exactly one solution in the interval. (c) Approximate the solution to three decimal places.
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Logic and Functions

Read Explanation

Mechanics Maths

Read Explanation

Statistics

Read Explanation

Applied Mathematics

Read Explanation
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.