/*! 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 66 Explain what is wrong with the s... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 66

Explain what is wrong with the statement. $$\lim _{x \rightarrow 1} \frac{x-1}{|x-1|}=1$$

Short Answer

Expert verified
The statement is wrong because the limit doesn't exist; one-sided limits differ as \(x\) approaches 1.

Step by step solution

01

Understanding the Expression

The expression given is \( \lim _{x \rightarrow 1} \frac{x-1}{|x-1|} \). We aim to evaluate this limit as \( x \) approaches 1. The numerator \( x-1 \) indicates the distance from 1, and the denominator \( |x-1| \) is the absolute value of this distance, which deals with the direction depending on whether \( x \) is greater than or less than 1.
02

Consider the Limit from the Right

Let's consider the case where \( x \rightarrow 1^+ \) (from values greater than 1), meaning \( x > 1 \). In this case, \( |x-1| = x-1 \), so our expression becomes \( \lim _{x \rightarrow 1^+} \frac{x-1}{x-1} = \lim _{x \rightarrow 1^+} 1 = 1 \).
03

Consider the Limit from the Left

Now consider the case where \( x \rightarrow 1^- \) (from values less than 1), meaning \( x < 1 \). In this situation, \( |x-1| = -(x-1) \), so our expression becomes \( \lim _{x \rightarrow 1^-} \frac{x-1}{-(x-1)} = \lim _{x \rightarrow 1^-} -1 = -1 \).
04

Conclude About the Limit Existence

Since the limit from the right is 1 and the limit from the left is -1, these one-sided limits are not equal. Therefore, the overall limit \( \lim _{x \rightarrow 1} \frac{x-1}{|x-1|} \) does not exist due to the discrepancy between these two values.

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.

One-sided limits
One-sided limits are an important concept in calculus. They allow us to understand the behavior of a function as it approaches a certain point, but only from one direction. There are two types of one-sided limits:
  • Right-hand limit: This is denoted by approaching a point from the right (larger values). For example, when we say \( x \rightarrow 1^+ \), it means we are approaching 1 from values greater than 1.

  • Left-hand limit: This is denoted by approaching a point from the left (smaller values). For instance, \( x \rightarrow 1^- \) means approaching 1 from values less than 1.

In the given example, we need to separately evaluate the expression as \( x \) approaches 1 from both directions. By comparing these one-sided limits, we can determine whether the overall limit exists.
Absolute value
The absolute value function plays a crucial role in determining the behavior of expressions near certain points. The absolute value of a number \( x \) is written as \( |x| \), and it signifies the distance of \( x \) from zero on a number line, irrespective of direction. Here are some simple facts about absolute values:
  • If \( x \) is positive or zero, \( |x| = x \).

  • If \( x \) is negative, \( |x| = -x \).

In our problem, as \( x \) approaches 1, the use of \(|x-1|\) indicates a shift around the point 1. When \( x < 1 \), \(|x-1| = -(x-1)\), due to the reversal of sign from the absolute value. Conversely, for \( x > 1 \), \(|x-1| = x-1\). Understanding this distinction is key to evaluating the limits properly.
Limit does not exist
A limit does not exist when the behavior of a function at a particular point does not converge to a single value. To verify if a limit exists, we often compare the one-sided limits. For the limit to exist at a point \( x = a \), the left-hand limit and the right-hand limit must be equal. Consider:
  • If \( \lim _{x \rightarrow a^-} f(x) = \lim _{x \rightarrow a^+} f(x) = L \), then \( \lim _{x \rightarrow a} f(x) = L \).

  • If the left-hand limit differs from the right-hand limit, the overall limit at that point does not exist.

In the original exercise, the limit from the right as \( x \rightarrow 1^+ \) was found to be 1, and the limit from the left as \( x \rightarrow 1^- \) was -1. Since these values do not match, \( \lim _{x \rightarrow 1} \frac{x-1}{|x-1|} \) does not exist.

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

Is the statement true or false? Assume that \(\lim _{x \rightarrow 7^{+}} f(x)=2\) and \(\lim _{x \rightarrow 7^{-}} f(x)=-2 .\) Explain. $$\lim _{x \rightarrow 7}(f(x))^{2} \text { exists. }$$

In \(2014,\) the populations of China and India were approximately 1.355 and 1.255 billion people, \(^{45}\) respectively. However, due to central control the annual population growth rate of China was \(0.44 \%\) while the population of India was growing by \(1.25 \%\) each year. If these growth rates remain constant, when will the population of India exceed that of China?

Are the statements true or false? Explain. $$\text { If } \lim _{x \rightarrow 0} \frac{f(x)}{g(x)} \text { exists, then } \lim _{x \rightarrow 0} f(x) \text { exists and } \lim _{x \rightarrow 0} g(x)$$exists

Give an example of: A function \(f(x)\) such that \(\ln (f(x))\) is only defined for \(x<0.\)

If \(h(x)=\ln (x+a),\) where \(a>0,\) what is the effect of increasing \(a\) on the vertical asymptote?
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Calculus

Read Explanation

Pure Maths

Read Explanation

Decision Maths

Read Explanation

Geometry

Read Explanation

Discrete 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.