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Chapter 13: Problem 30

Find the limit, if it exists. If the limit does not exist, explain why. $$\text { 0. } \lim _{x \rightarrow-4^{-}} \frac{|x+4|}{x+4}$$

Short Answer

Expert verified
The limit is -1.

Step by step solution

01

Understand the Limit Definition

We need to find the limit as \(x\) approaches \(-4\) from the left of the function \(\frac{|x+4|}{x+4}\). This means we evaluate the expression when \(x\) is slightly less than \(-4\).
02

Analyze the Absolute Value

The expression \(|x+4|\) depends on the value of \(x+4\). For values of \(x\) close to \(-4\) from the left (i.e., \(x < -4\)), \(x+4\) will be a negative number. Thus, \(|x+4|\) is equal to \(-(x+4)\).
03

Simplify the Function

Substitute \(|x+4|\) with \(-(x+4)\). The function becomes \(\frac{-(x+4)}{x+4}\). When simplifying this, cancel \(x+4\) from the numerator and the denominator (assuming not zero), so it simplifies to \(-1\).
04

State the Result of the Limiting Process

As \(x\) approaches \(-4\) from the left, the simplified expression \(-1\) is constant. Hence, the limit of \(\frac{|x+4|}{x+4}\) as \(x\) approaches \(-4^{{-}}\) is \(-1\).

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Key Concepts

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

Understanding the Absolute Value Function
The absolute value function is a fundamental mathematical concept that deals with the magnitude of a real number, without considering its sign. In essence, it tells us how far a number is from zero.
  • The notation for the absolute value of a number is denoted by vertical bars, such as \(|x|\).
  • Mathematically, it's defined as \(|x| = x\) if \(x \geq 0\); and \(|x| = -x\) if \(x < 0\).
This definition is pivotal when working with expressions like \(|x+4|\), particularly because we must determine the sign of \(x+4\) to evaluate the absolute value correctly.
For values of \(x\) less than \(-4\), we have \(x+4 < 0\), hence \(|x+4| = -(x+4)\). Applying this understanding is crucial for correctly evaluating limits involving absolute value functions.
Exploring the Left-Hand Limit
Calculating the left-hand limit involves evaluating a function as the input approaches a specific value from the left side. For our exercise, we are primarily concerned with how the function \(\frac{|x+4|}{x+4}\) behaves as \(x\) approaches \(-4\) but specifically from the left, denoted as \(x \to -4^{-}\).
  • When the expression \(x+4\) is negative, approaching from the left naturally results in values of \(x\) that remain less than \(-4\).
  • Subsequently, we specify \(|x+4| = -(x+4)\) for \(x < -4\).
This leads to the simplification \(\frac{-(x+4)}{x+4} = -1\). Therefore, the limit as \(x \to -4^{-}\) exists and equals \(-1\). The left-hand limit helps confirm that the function stabilizes at this value from one side, crucial for understanding continuity and limits.
Delving into Continuous Functions
A function is continuous if its graph has no breaks, holes, or jumps. For a function \(f(x)\) to be continuous at \(x = a\), three conditions must be satisfied:
  • The function \(f(a)\) must be defined.
  • The limit of \(f(x)\) as \(x\) approaches \(a\) must exist.
  • This limit must equal \(f(a)\).
Considering our function \(\frac{|x+4|}{x+4}\) as \(x \to -4^{-}\), we observe its behavior at a specific borderline such as \(-4\). Given that on the left side of \(-4\), the function consistently approaches \(-1\), it suggests continuity from that direction. However, to ensure continuity overall, one must also evaluate the right-hand limit at \(-4\).
While our evaluation only discusses the left hand, continuous functions need bilateral agreement at points like \(-4\) to genuinely reflect continuous behavior.

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Most popular questions from this chapter

(a) Estimate the area under the graph of \(f(x)=25-x^{2}\) from \(x=0\) to \(x=5\) using five approximating rectangles and right endpoints. Sketch the graph and the rectangles. Is your estimate an underestimate or an overestimate? (b) Repeat part (a) using left endpoints.

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Find the limit, if it exists. If the limit does not exist, explain why. $$\lim _{x \rightarrow 15} \frac{2 x^{2}-3 x}{|2 x-3|}$$
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