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Find \(\lim _{x \rightarrow a^{-}} f(x), \lim _{x \rightarrow a^{+}} f(x),\) and \(\lim _{x \rightarrow a} f(x)\) at the indicated value for the indicated function. Do not use a computer or graphing calculator. $$ a=2, f(x)=\frac{x-2}{|x-2|} $$

Short Answer

Expert verified
\(\lim_{x \to 2^-} = -1, \lim_{x \to 2^+} = 1, \lim_{x \to 2}\) does not exist.

Step by step solution

01

Identify the Function Behavior as x Approaches from the Left

For the left-hand limit, we consider values of \(x\) that are slightly less than \(2\) since \(x \to 2^-\). In this case, \(|x - 2| = 2 - x\). So the function becomes \(f(x) = \frac{x-2}{2-x}\). This simplifies to \(-1\) because \(x - 2\) and \(2 - x\) differ only by a sign. Thus, \(\lim_{x \to 2^-} \frac{x-2}{|x-2|} = -1\).
02

Identify the Function Behavior as x Approaches from the Right

For the right-hand limit, we consider values of \(x\) that are slightly greater than \(2\) since \(x \to 2^+\). Here, \(|x - 2| = x - 2\). So \(f(x) = \frac{x-2}{x-2} = 1\) for \(x > 2\). Thus, \(\lim_{x \to 2^+} \frac{x-2}{|x-2|} = 1\).
03

Compare the One-Sided Limits to Determine the Two-Sided Limit

The two-sided limit \(\lim_{x \to 2} f(x)\) exists if and only if the left-hand limit equals the right-hand limit. From Step 1, we found \(\lim_{x \to 2^-} f(x) = -1\) and from Step 2, \(\lim_{x \to 2^+} f(x) = 1\). Since these are not equal, the limit \(\lim_{x \to 2} f(x)\) 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
One-sided limits help us understand the behavior of functions as they get closer to a specific point from either side. Unlike a regular limit, which looks at both sides simultaneously, one-sided limits focus on just one direction:
- **Left-Hand Limit (LHL)**: This is denoted as \(\lim_{x \to a^-} f(x)\). It examines what happens to the value of the function as \(x\) approaches \(a\) from the left, meaning with values slightly smaller than \(a\).
- **Right-Hand Limit (RHL)**: This is symbolized by \(\lim_{x \to a^+} f(x)\). It checks the function's behavior as \(x\) comes close to \(a\) from the right, with values just larger than \(a\).
In our exercise, we calculated both of these limits for \(a=2\):
- As \(x\) approaches from the left (\
Discontinuity in Functions
A function is said to be discontinuous at a point if it jumps, breaks, or has a hole at that point.
In terms of limits, discontinuity typically occurs when the two-sided limit doesn't exist or doesn't equal the function's value at that point.
**Types of Discontinuities:**
  • Jump Discontinuity: Happens when the left-hand limit and right-hand limit exist but are not equal. This is exactly what we found in our exercise, with a jump from -1 to 1 at \(x=2\).

  • Removable Discontinuity: This occurs when the limit exists but it's not equal to the function's value at that point.

  • Infinite Discontinuity: Involves a limit that approaches infinity, creating a vertical asymptote.

These types help in categorizing how the function behaves around points of interest, particularly in piecewise functions where different rules apply for different intervals.
Piecewise Functions
Piecewise functions are those defined by different expressions for different intervals of the domain. Instead of a single formula for all values, piecewise functions have multiple 'pieces' with specific rules.
**Why Piecewise Functions Matter:**
  • They capture real-world situations: Many scenarios don't follow a single rule. For instance, tax brackets or shipping costs may follow different formulas for different income levels or weights.

  • Useful for discontinuous functions: Functions with jumps or holes can be accurately described using multiple equations.

Our given function \(f(x) = \frac{x-2}{|x-2|}\) can be seen as piecewise because it acts differently when \(x\) is less than or greater than 2.
For \(x < 2\), it simplifies to \(-1\), and for \(x > 2\), it turns into \(1\). This separation into distinct behaviors at \(x=2\) highlights the utility of piecewise definitions.

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Most popular questions from this chapter

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On your computer or graphing calculator, graph \(y=\) \(f(x)=\cos x\) in radian mode, using a window with dimensions [-6.14,6.14] by [-1,1] to familiarize yourself with this function. As you see, this function moves back and forth between -1 and \(1 .\) We wish to estimate \(f^{\prime}(\pi / 2)\) where \(\pi / 2 \approx 1.57\). For this purpose, graph using a window with dimensions [1.07,2.07] by \([-0.5,0.5] .\) From the graph, estimate \(f^{\prime}(1.57)\).

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