/*! 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 2 Use a calculator to estimate \(\... [FREE SOLUTION] | 91Ó°ÊÓ

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Use a calculator to estimate \(\lim _{x \rightarrow 1^{+}} \frac{|x-1|}{1-x^{2}}\) and \(\lim _{x \rightarrow 1^{-}} \frac{|x-1|}{1-x^{2}} .\)

Short Answer

Expert verified
\(\lim_{x \to 1^+} = -\frac{1}{2}\); \(\lim_{x \to 1^-} = \frac{1}{2}\).

Step by step solution

01

Recognize the Function Components

The expression given is \(\frac{|x-1|}{1-x^{2}}\). Recognize that \( |x-1| \) is the absolute value function, and \( 1-x^{2} \) is a quadratic term, equivalent to \(-(x-1)(x+1)\).
02

Limit from the Right (\( x \to 1^+ \))

For \( x \to 1^+ \), \( x \) is slightly greater than 1, so \( |x-1| = x-1 \). Thus, the expression becomes \( \frac{x-1}{-(x-1)(x+1)} = \frac{1}{-(x+1)} \). As \( x \) approaches 1 from the right, the limit is \( \frac{1}{-2} = -\frac{1}{2} \).
03

Limit from the Left (\( x \to 1^- \))

For \( x \to 1^- \), \( x \) is slightly less than 1, so \( |x-1| = -(x-1) \). The expression becomes \( \frac{-(x-1)}{-(x-1)(x+1)} = \frac{1}{x+1} \). As \( x \) approaches 1 from the left, the limit is \( \frac{1}{2} \).
04

Compare One-Sided Limits

Calculate the limits from both sides to see if they are equal. From the right, \( \lim_{x \to 1^+} \frac{|x-1|}{1-x^{2}} = -\frac{1}{2} \), and from the left, \( \lim_{x \to 1^-} \frac{|x-1|}{1-x^{2}} = \frac{1}{2} \). The limits are different, suggesting the two-sided limit \( \lim_{x \to 1} \frac{|x-1|}{1-x^{2}} \) does not exist.

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Key Concepts

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

One-Sided Limits
When exploring the concept of limits, one-sided limits provide insight into the behavior of functions as they approach a specific point from one direction only. A limit from the right (denoted as \(x \to a^+\)) considers values of \(x\) slightly greater than \(a\), while a limit from the left (\(x \to a^-\)) looks at values slightly smaller.

Calculating both one-sided limits helps determine if a two-sided limit exists. For instance, if \(\lim_{x \to 1^+} f(x) = L_1\) and \(\lim_{x \to 1^-} f(x) = L_2\), the two-sided limit \(\lim_{x \to 1} f(x)\) only exists if \(L_1 = L_2\).

This principle is crucial in the exercise, where the right limit is \(-\frac{1}{2}\) and the left limit is \(\frac{1}{2}\), showing distinct behaviors from both sides, leading to non-existence of the two-sided limit.
Absolute Value Function
The absolute value function, denoted by \(|x|\), is essential in calculus for its non-negative nature. It represents the distance of a number from zero on the number line without regard to direction.

The function is defined as:
  • \(|x| = x\) if \(x \geq 0\)
  • \(|x| = -x\) if \(x < 0\)
This property of flipping a negative sign to positive affects how limits behave, as seen in the given exercise.

For \(x \to 1^+\), \(|x-1| = x-1\) because values are slightly greater than 1. For \(x \to 1^-\), \(|x-1| = -(x-1)\) because values are slightly less than 1. This distinction is crucial in calculating one-sided limits and understanding discontinuities.
Quadratic Expression
A quadratic expression takes the form \(ax^2 + bx + c\). In our exercise, the quadratic term \(1-x^2\) can be expanded to \(-(x-1)(x+1)\). Recognizing its structure is helpful for limit evaluation.

Quadratics have a definitive parabolic shape, and in this case, \(1-x^2\) leads to a downward opening parabola intersecting the x-axis at \(x = 1\) and \(x = -1\). These roots are identified by solving \(1-x^2 = 0\) or factoring as seen.

Understanding these roots assists in identifying critical points of behavior change in functions and aids in the examination of discontinuities.
Discontinuity
Discontinuity refers to points where a function is not continuous. A very useful concept in calculus, discontinuities can occur at points, over intervals, or even be infinite.

In our exercise, the function \(\frac{|x-1|}{1-x^2}\) exhibits a point discontinuity at \(x = 1\). This is evident as the one-sided limits from the left and right do not agree: the right-side limit is \(-\frac{1}{2}\) and the left-side limit is \(\frac{1}{2}\). Such a discontinuity indicates the function's value cannot be continuously defined at that point.

Understanding this helps identify where a function changes its behavior dramatically, which is crucial in both theoretical and applied calculus.

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