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Chapter 2: Problem 53

Use the definitions of right-hand and left-hand limits to prove the limit statements. \begin{equation}\lim _{x \rightarrow 0} \frac{x}{|x|}=-1\end{equation}

Short Answer

Expert verified
The limit \( \lim_{x \to 0} \frac{x}{|x|} \) does not exist because the left-hand limit is -1, and the right-hand limit is 1, which are unequal.

Step by step solution

01

Understanding the Limit Problem

We need to prove the limit \( \lim_{x \to 0} \frac{x}{|x|} = -1 \) specifically using the definitions of right-hand and left-hand limits. We will analyze the behavior of \( \frac{x}{|x|} \) as \( x \to 0^+ \) and \( x \to 0^- \).
02

Evaluating the Right-Hand Limit

For the right-hand limit, consider what happens as \( x \to 0^+ \). For \( x > 0 \), we have \( |x| = x \), thus \( \frac{x}{|x|} = 1 \). Therefore, the right-hand limit is \( \lim_{x \to 0^+} \frac{x}{|x|} = 1 \). This does not match the given limit of -1.
03

Evaluating the Left-Hand Limit

For the left-hand limit, consider the behavior as \( x \to 0^- \). For \( x < 0 \), we have \( |x| = -x \), hence \( \frac{x}{|x|} = -1 \). Therefore, the left-hand limit is \( \lim_{x \to 0^-} \frac{x}{|x|} = -1 \).
04

Combining The Limit Results

To conclude the behavior of the overall limit \( \lim_{x \to 0} \frac{x}{|x|} \), we combine the results of the left-hand and right-hand limits. The right-hand limit is \( 1 \), and the left-hand limit is \( -1 \). Since these are not equal, the two-sided limit does not exist, as the function \( \frac{x}{|x|} \) is discontinuous at \( x = 0 \).
05

Conclusion: Correct Understanding of the Problem Statement

The statement \( \lim_{x \to 0} \frac{x}{|x|} = -1 \) reflects only the left-hand limit and does not represent the actual two-sided limit which does not exist. Thus, proven incorrect under the provided criteria.

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Key Concepts

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

Right-Hand Limit
When investigating the right-hand limit, we look at the behavior of a function as our input value, \( x \), approaches a particular point from the positive side, which is often denoted as \( x \to c^+ \). This means that \( x \) is getting closer to \( c \) from values greater than \( c \).
For the problem \( \lim_{x \to 0^+} \frac{x}{|x|} \), when \( x \) is positive, \( |x| = x \), so
\[ \frac{x}{|x|} = \frac{x}{x} = 1. \]
Thus, we find that the right-hand limit of the function \( \frac{x}{|x|} \) as \( x \) approaches \( 0 \) from the right is 1, not -1. This is crucial because it shows that the right-hand limit differs from the statement's limit.
Left-Hand Limit
The left-hand limit examines how a function behaves as it approaches a specific point from the negative side, expressed as \( x \to c^- \). This indicates that \( x \) is nearing \( c \) from values less than \( c \).
In our problem, when \( x \to 0^- \), \( x \) is negative. Therefore, \( |x| = -x \) and
\[ \frac{x}{|x|} = \frac{x}{-x} = -1. \]
Thus, we see that as \( x \) approaches 0 from the left, the function \( \frac{x}{|x|} \) indeed approaches -1. This supports the left-hand limit part of the limit statement \( \lim_{x \to 0} \frac{x}{|x|} = -1 \). However, this does not confirm a two-sided limit.
Discontinuity
Discontinuity in a function occurs when there is a jump, gap, or point at which the function is not continuous. This usually happens when the left-hand and right-hand limits do not coincide.
In the case of \( f(x) = \frac{x}{|x|} \), at \( x = 0 \) the function is not defined, and the right-hand limit, which is 1, does not match the left-hand limit, which is -1. This difference indicates a discontinuity at \( x = 0 \).
Such discontinuities are important to note because they tell us that although limits might exist from one side, a harmonious overall limit does not, leading to a break in the function's graph.
Two-Sided Limit
A two-sided limit investigates the behavior of a function from both directions as it approaches a specific value. A two-sided limit \( \lim_{x \to c} f(x) \) exists if and only if both the left-hand and right-hand limits are identical, meaning \( \lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) \).
For \( \lim_{x \to 0} \frac{x}{|x|} \), we see:
  • The right-hand limit is 1.
  • The left-hand limit is -1.

Since these are not the same, the overall two-sided limit does not exist.
Understanding this concept is vital as it illustrates that even if one-side limits exist, without them matching, a unified two-sided limit cannot be concluded.

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Most popular questions from this chapter

Graph the rational functions in Exercises \(103-108 .\) Include the graphs and equations of the asymptotes. $$ y=\frac{x^{3}+1}{x^{2}} $$

Find the limits in Exercises \(53-58 .\) $$ \lim \frac{x^{2}-1}{2 x+4} $$ $$ \begin{array}{ll}{\text { a. }} & {x \rightarrow-2^{+}} & {\text { b. } x \rightarrow-2^{-}} \\ {\text { c. }} & {x \rightarrow 1^{+}} & {\text { d. } x \rightarrow 0^{-}}\end{array} $$

Graph the rational functions in Exercises \(63-68 .\) Include the graphs and equations of the asymptotes and dominant terms. $$ y=\frac{1}{x-1} $$

Find the limits in Exercises \(53-58 .\) $$ \lim \frac{x}{x^{2}-1} $$ $$\begin{array}{ll}{\text { a. }} & {x \rightarrow 1^{+}} & {\text { b. } x \rightarrow 1^{-}} \\ {\text { c. }} & {x \rightarrow-1^{+}} & {\text { d. } x \rightarrow-1^{-}}\end{array}$$

One-sided limits \begin{equation}f(x)=\left\\{\begin{array}{ll}{x^{2} \sin (1 / x),} & {x<0} \\\ {\sqrt{x},} & {x>0}.\end{array}\right.\end{equation} Find (a) \(\lim _{x \rightarrow 0^{+}} f(x)\) and (b) \(\lim _{x \rightarrow 0 ^{-}} f(x) ;\) then use limit definitions to verify your findings. (c) Based on your conclusions in parts (a) and (b), can you say anything about \(\lim _{x \rightarrow 0} f(x) ?\) Give reasons for your answer.
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