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Chapter 1: Problem 35

Find the values of \(x\) (if any) at which \(f\) is not continuous, and determine whether each such value is a removable discontinuity. (a) \(f(x)=\frac{|x|}{x}\) (b) \(f(x)=\frac{x^{2}+3 x}{x+3}\) (c) \(f(x)=\frac{x-2}{|x|-2}\)

Short Answer

Expert verified
(a) Jump at x=0, not removable. (b) Removable at x=-3. (c) Removable at x=2, jump at x=-2.

Step by step solution

01

Analyze Discontinuity for f(x)=\frac{|x|}{x}

The function \(f(x) = \frac{|x|}{x}\) is defined for all \(x eq 0\). At \(x = 0\), the function is not defined, so there is a potential point of discontinuity. To determine the type of discontinuity, we examine the limit on both sides of \(x = 0\).For \(x > 0\), \(|x| = x\), so \(f(x) = 1\). For \(x < 0\), \(|x| = -x\), so \(f(x) = -1\). Thus, as \(x\) approaches 0 from the positive side, the limit is 1, while from the negative side, the limit is -1. These limits are not equal, indicating a jump discontinuity at \(x = 0\). Since the limits from either side are not equal, it is not removable.
02

Analyze Discontinuity for f(x)=\frac{x^{2}+3x}{x+3}

The function \(f(x) = \frac{x^{2}+3x}{x+3}\) is initially undefined at \(x = -3\) due to division by zero. However, we can factor the numerator: \(x^{2} + 3x = x(x + 3)\). Thus, the function simplifies to:\[ f(x) = x \text{ for } x eq -3 \]Now, check the limit as \(x\) approaches -3. As \(x \rightarrow -3\), \(f(x) = x\), so the limit is -3. However, since the original function is not defined at \(x = -3\), this represents a removable discontinuity because the simplified form of the function can be defined to have a continuous value at \(x = -3\) by setting \( f(-3) = -3 \).
03

Analyze Discontinuity for f(x)=\frac{x-2}{|x|-2}

The function \(f(x) = \frac{x-2}{|x|-2}\) is undefined where \(|x|-2 = 0\), which occurs at \(x = 2\) and \(x = -2\). We need to evaluate the limits around these points:- At \(x = 2\), \(|x|=x\), thus \(|x|-2 = x-2\). The function simplifies to \(f(x) = 1\) for \(x > 2\), and remains 1 as \(x\) approaches 2 from both sides. Thus, \(x = 2\) is a removable discontinuity as the limits from both sides match.- At \(x = -2\), \(|x|=-x\), thus \(|x|-2 = -x-2\), and the function becomes \(f(x) = \frac{x-2}{-x-2}\). As \(x\) approaches -2 from the positive side and negative side, the terms approach forms like \(\frac{-4}{0^+}\) and \(\frac{-4}{0^-}\) respectively, indicating a jump discontinuity at \(x = -2\). Therefore, it is not removable.

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Key Concepts

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

Removable Discontinuity
A removable discontinuity in a function occurs when a point on the function is undefined or does not match the limit of the function, but adjusting the function at this point can make it continuous. This means that with a small alteration, usually by filling in a single point or redefining it, the graph could be made smooth or continuous.
Let's examine an example. For the function \[f(x) = \frac{x^2 + 3x}{x + 3}\]we notice it is undefined at \(x = -3\) since the denominator becomes zero. However, when we simplify it, \[x(x + 3)/(x + 3) = x \]for all \(x eq -3\), we see that the function's limit as \(x\) approaches \(-3\) is indeed \(-3\). By assigning \(f(-3) = -3\), we eliminate the discontinuity, making it removable. This ability to "patch" or "plug" the function at an isolated point signifies a removable discontinuity.
Generally, to find if a discontinuity is removable, check if the limits from both sides match, and if redefining the function at that point will restore continuity.
Jump Discontinuity
A jump discontinuity is a type of discontinuity wherein the limits from the left and right of a point do not agree. This results in a 'jump' in the graph at that point, causing a break or leap from one value to another. Unlike removable discontinuities, jump discontinuities cannot be easily mended by merely adjusting a single point.
For instance, consider the function \[f(x) = \frac{|x|}{x}\]which is defined for all \(x eq 0\). As \(x\) approaches zero, the function equals \(1\) from the positive side (as \(x > 0\)), but equals \(-1\) from the negative side (as \(x < 0\)). Since these two limits differ, a jump occurs at \(x = 0\), making it a clear example of a jump discontinuity.
To identify a jump discontinuity, look for points where the left-hand and right-hand limits are unequal, indicating an abrupt transition. Unlike removable discontinuity, these cannot be resolved by simply redefining the function at one point.
Limit Analysis
Limit analysis is a key tool to evaluate the behavior of functions as they approach specific points. By analyzing limits, we can understand whether functions have discontinuities and determine their types.
Consider the function \[f(x) = \frac{x-2}{|x|-2}\]which is undefined at \(x = 2\) and \(x = -2\).
  • At \(x = 2\), by simplifying the expression, \[f(x) = 1\]as \(x\) approaches \(2\) from both sides, the output remains constant. Since both side limits are equal, \(x = 2\) is a removable discontinuity that can be eliminated by defining \(f(2) = 1\).
  • Conversely, for \(x = -2\), the expression becomes undefined due to differing approach values, leading to -inf and +inf. Here, the limits do not match, indicating a jump discontinuity.
For thorough limit analysis, observe how the function behaves from both directions at the troublesome points. If side limits coincide, it's a hint for removable discontinuity; if not, it indicates a jump discontinuity.

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