Chapter 2: Problem 49
Let \(g(x)=\frac{x^{2}+x-6}{|x-2|}\) (a) Find \(\begin{array}{ll}{\text { (i) } \lim _{x \rightarrow+} g(x)} & {\text { (ii) } \lim _{x \rightarrow 2^{-}} g(x)} \\ {\text { (b) Does } \lim _{x \rightarrow 2} g(x) \text { exist? }} \\ {\text { (c) Sketch the graph of } g}\end{array}\)
Short Answer
Expert verified
(a)(i) 5, (ii) -5; (b) No, the limit does not exist; (c) The graph has a jump at \(x=2\).
Step by step solution
01
Factorize the Numerator
The function is given as \( g(x) = \frac{x^2 + x - 6}{|x-2|} \). The numerator \( x^2 + x - 6 \) can be factorized as \((x-2)(x+3)\). This helps in simplifying the expression, especially around \(x = 2\).
02
Limits from the Right at x=2
For \( \lim_{x \to 2^+} g(x) \), as \(x\) approaches 2 from the right, \(|x - 2| = x - 2\). Therefore, the function simplifies to \( g(x) = \frac{(x-2)(x+3)}{x-2} = x + 3 \) for \(x > 2\). Thus, \( \lim_{x \to 2^+} g(x) = 2 + 3 = 5 \).
03
Limits from the Left at x=2
For \( \lim_{x \to 2^-} g(x) \), as \(x\) approaches 2 from the left, \(|x - 2| = -(x - 2)\). The function then becomes \( g(x) = \frac{(x-2)(x+3)}{-(x-2)} = -(x + 3) \) for \(x < 2\). Hence, \( \lim_{x \to 2^-} g(x) = -(2 + 3) = -5 \).
04
Determine if the Limit Exists at x=2
The overall limit \( \lim_{x \to 2} g(x) \) exists if and only if the right-hand and left-hand limits are equal. In this case, \( \lim_{x \to 2^+} g(x) = 5 \) and \( \lim_{x \to 2^-} g(x) = -5 \). Since these two limits are not equal, the limit \( \lim_{x \to 2} g(x) \) does not exist.
05
Sketch the Graph
To sketch the graph, note that \( g(x) \) behaves like \( x + 3 \) for \( x > 2 \) and \( -(x + 3) \) for \( x < 2 \). At \( x = 2 \), there is a removable discontinuity or a jump because the single point \(g(2)\) doesn't coincide with these branches. Plot \( x + 3 \) for \( x > 2\) and \( -(x + 3) \) for \( x < 2 \), with a jump at \( x = 2 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Factorization
Understanding factorization is a fundamental step in simplifying mathematical expressions, especially when dealing with complex functions. In the given exercise, we have the function \[g(x) = \frac{x^2 + x - 6}{|x-2|}\].
To simplify this expression, we need to factor the numerator, which is a quadratic expression.
To factor \(x^2 + x - 6\), we look for two numbers that multiply to \(-6\) and add to \(1\). These numbers are \(-2\) and \(3\). Thus, the factorization of the quadratic is \((x-2)(x+3)\).
With this factorization, the expression can be rewritten, helping further evaluation and simplification when analyzing limits. This process is essential as we approach the points of interest around \(x = 2\), where the denominator includes \(|x-2|\).
To simplify this expression, we need to factor the numerator, which is a quadratic expression.
To factor \(x^2 + x - 6\), we look for two numbers that multiply to \(-6\) and add to \(1\). These numbers are \(-2\) and \(3\). Thus, the factorization of the quadratic is \((x-2)(x+3)\).
With this factorization, the expression can be rewritten, helping further evaluation and simplification when analyzing limits. This process is essential as we approach the points of interest around \(x = 2\), where the denominator includes \(|x-2|\).
Piecewise Functions
Piecewise functions consist of different expressions depending on the domain specified. In this exercise, \(g(x)\) is actually a piecewise function due to the presence of \(|x-2|\) in the denominator.
When considering \(x > 2\), we know \(|x-2| = x-2\). This simplification leads the function \(g(x)\) to behave as \(x+3\). However, for \(x < 2\), we have \(|x-2| = -(x-2)\), which makes the function \(-1 imes (x+3)\).
These different expressions work together to create a piecewise function:
When considering \(x > 2\), we know \(|x-2| = x-2\). This simplification leads the function \(g(x)\) to behave as \(x+3\). However, for \(x < 2\), we have \(|x-2| = -(x-2)\), which makes the function \(-1 imes (x+3)\).
These different expressions work together to create a piecewise function:
- For \(x > 2\), function is \(x + 3\)
- For \(x < 2\), function is \(-(x + 3)\)
Discontinuity
Discontinuity occurs when a function is not continuous at a given point. In the context of this exercise, we look at the behavior of \(g(x)\) at \(x = 2\) to determine if any discontinuity exists.
The limit analyses for both left-hand and right-hand approaching \(x = 2\) show different results:
Understanding this discontinuity is vital while analyzing or sketching the function's graph for intuitive learning.
The limit analyses for both left-hand and right-hand approaching \(x = 2\) show different results:
- For \(\lim_{x \to 2^+} g(x) = 5\)
- For \(\lim_{x \to 2^-} g(x) = -5\)
Understanding this discontinuity is vital while analyzing or sketching the function's graph for intuitive learning.
Graph Sketching
Graph sketching involves creating a visual representation of a function over its domain. For \(g(x)\), knowing its piecewise behavior and discontinuity at \(x = 2\) plays a crucial role.
To sketch the graph:
To sketch the graph:
- Start with the portion \(x + 3\) for \(x > 2\). This is a straight line with slope \(1\) intersecting the y-axis at \(3\).
- Next, graph the part \(-(x + 3)\) for \(x < 2\). This is also a line but slopes downwards, creating a mirror image of \(x + 3\).