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Chapter 2: Problem 75

In Exercises \(74-80\), evaluate the one-sided limits. $$ \lim _{x \rightarrow 3^{-}} \frac{x-3}{|x-3|} $$

Short Answer

Expert verified
The one-sided limit is -1.

Step by step solution

01

Understanding One-Sided Limits

The problem is asking us to evaluate the limit of the given function as the variable \( x \) approaches 3 from the left, denoted \( x \to 3^- \). One-sided limits are useful for determining the behavior of a function as it approaches a specific value from one particular direction only.
02

Simplifying Given Function

The function is defined as \( \frac{x-3}{|x-3|} \). When \( x < 3 \), the expression \( x-3 \) is negative, making \( |x-3| = -(x-3) = 3-x \). Thus, the function becomes \( \frac{x-3}{3-x} \). Simplifying this gives \( -1 \) since \( \frac{x-3}{3-x} = -1 \).
03

Evaluating the Limit Expression

To evaluate the limit as \( x \) approaches 3 from the left, note the simplified expression \(-1\) we derived is constant for all \( x < 3 \). Therefore, the limit of a constant function is just that constant value.
04

Finalizing the Solution

Since the simplified expression for \( x < 3 \) is \(-1\), this means the one-sided limit as \( x \) approaches 3 from the negative side is \(-1\).

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Key Concepts

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

Evaluating Limits
Evaluating limits, especially one-sided limits, is a fundamental concept in calculus. A limit describes the behavior of a function as the input approaches a particular value. The one-sided limit specifically focuses on the function's behavior as the input approaches the limit from only one side 鈥� either from the left (denoted by a superscript minus) or the right (denoted by a superscript plus).

In the given exercise, we are dealing with a one-sided limit approaching from the left:
  • This means we observe how the function behaves as the value of the input gets closer and closer to 3, but never actually reaches it.
  • Since we are considering the limit as \( x \to 3^- \), all values of \( x \) are less than 3.
  • The expression \( \frac{x-3}{|x-3|} \) becomes particularly interesting because of its dependence on the absolute value function.
When evaluating the limit, the key focus is on understanding how the function simplifies when substituting values close to 3 from the left.
Absolute Value Function
The absolute value function is quite pivotal in this problem. Defined generally as \( |a| = a \) if \( a \, \geq \, 0 \) and \( |a| = -a \) if \( a \, < \, 0 \), it essentially measures how far a number is from zero, ignoring any direction (positive or negative).

For the function \( \frac{x-3}{|x-3|} \), understanding the absolute value becomes essential as we determine how \( |x-3| \) behaves based on the value of \( x \):
  • Since we are examining the limit as \( x \to 3^- \), any value of \( x \) we consider will be less than 3.
  • This implies \( x-3 < 0 \), making \( |x-3| = -(x-3) = 3-x \).
This transformation supports simplifying the function to \( \frac{x-3}{3-x} \), which yields \( -1 \). Understanding when and how to break down absolute values is crucial for evaluating such expressions effectively.
Piecewise Functions
Piecewise functions can often arise when evaluating limits involving absolute values. They describe functions defined by different expressions over different intervals.

In the scenario of \( \frac{x-3}{|x-3|} \) approaching from values less than 3:
  • It embodies a piecewise nature since the expression of \( |x-3| \) changes based on whether \( x < 3 \) or \( x > 3 \).
  • When \( x < 3 \), we derived \( |x-3| = 3-x \).
A key skill in evaluating these kinds of limits involves recognizing these intervals and the corresponding expressions that apply. By identifying these partitions, it becomes clearer why the limit was derived as a constant: -1. Recognizing this piecewise construction allows for a simplified view of otherwise complex-looking limits.

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Most popular questions from this chapter

Radiocarbon Dating Two isotopes of carbon, \({ }^{12} \mathrm{C},\) which is stable, and \({ }^{14} \mathrm{C},\) which decays exponentially with a 5700 -year half-life, are found in a known fixed ratio in living matter. After death, carbon is no longer metabolized, and the amount \(m(t)\) of \({ }^{14} \mathrm{C}\) decreases due to radioactive decay. In the analysis of a sample performed \(T\) years after death, the mass of \({ }^{12} \mathrm{C}\), unchanged since death, can be used to determine the mass \(m_{0}\) of \({ }^{14} \mathrm{C}\) that the sample had at the moment of death. The time \(T\) since death can then be calculated from the law of exponential decay and the measurement of \(m(T)\). Use this information for solving Exercises \(95-98\). Prehistoric cave art has recently been found in a number of new locations. At Chauvet, France, the amount of \({ }^{14} \mathrm{C}\) in some charcoal samples was determined to be \(0.0879 \cdot m_{0}\). Ahout how old are the cave drawings?

Prove that, if \(f\) is a continuous function on \([0,1],\) and \(f(0)=f(1)\) then there is a value \(c\) in (0,1) such that \(f(c)=f(c+1 / 2) .\) This is a special case of the Horizontal Chord Theorem. (Hint: Apply the Intermediate Value Theorem to the function \(g\) defined on \([0,1 / 2]\) by \(g(x)=\) \(f(x+1 / 2)-f(x) .)\)

Let \(p\) be a polynomial of odd degree and leading coefficient \(1 .\) It can be shown that \(\lim _{x \rightarrow \infty} p(x)=\infty\) and \(\lim _{x \rightarrow-\infty} p(x)=-\infty .\) Use these facts and the Intermediate Value Theorem to prove that \(p\) has at, least one real root.

In Exercises \(74-80\), evaluate the one-sided limits. $$ \lim _{x \rightarrow 0^{+}} \frac{\sqrt{\sin (x)}}{\sqrt{x}} $$

In Exercises \(67-73,\) use algebraic manipulation (as in Example 5 ) to evaluate the limit. $$ \lim _{x \rightarrow 4}\left(\frac{x-4}{\sqrt{x}-2}\right) $$
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