Problem 42
Question
Determine whether the function is continuous at the given point \(c .\) If the function is not continuous, determine whether the discontinuity is removable or non-removable. $$ f(x)=\frac{x^{2}-100}{x-10} ; c=10 $$
Step-by-Step Solution
Verified Answer
The function is not continuous at \( x = 10 \) and has a removable discontinuity.
1Step 1: Identify Continuity Requirement
A function is continuous at a point \( c \) if the following three conditions are satisfied: 1. \( f(c) \) is defined. 2. \( \lim_{x \to c} f(x) \) exists. 3. \( \lim_{x \to c} f(x) = f(c) \). We'll analyze each in this context.
2Step 2: Check if \( f(10) \) is Defined
Substitute \( c = 10 \) into \( f(x) \): \[ f(10) = \frac{10^{2} - 100}{10 - 10} = \frac{0}{0} \]. Since this results in division by zero, \( f(10) \) is undefined.
3Step 3: Examine the Limit \( \lim_{x \to 10} f(x) \)
The expression \( f(x) = \frac{x^2 - 100}{x - 10} \) can be simplified: \[ x^2 - 100 = (x + 10)(x - 10) \]. Thus, \[ f(x) = \frac{(x + 10)(x - 10)}{x - 10} = x + 10 \text{ for } x eq 10 \]. The limit becomes: \[ \lim_{x \to 10} f(x) = \lim_{x \to 10} (x + 10) = 20 \].
4Step 4: Determine the Type of Discontinuity
Since \( \lim_{x \to 10} f(x) = 20 \) but \( f(10) \) is undefined, the function is not continuous at \( x = 10 \). However, the limit exists, implying a removable discontinuity, which can be "fixed" by defining \( f(10) = 20 \).
Key Concepts
Understanding LimitsRemovable Discontinuity ClarifiedExploring Undefined Functions
Understanding Limits
The concept of limits is fundamental in calculus and is crucial when analyzing function behavior at specific points. A limit estimates the value a function approaches as the input gets closer to a specified point. In our exercise, we examine the function \( f(x) = \frac{x^2 - 100}{x - 10} \) as \( x \) approaches 10.
First, notice that directly substituting \( x = 10 \) into the function results in division by zero. This doesn't mean the limit doesn't exist. Instead, it suggests the need for further analysis to simplify the function.
By factoring the polynomial in the numerator as \( (x + 10)(x - 10) \), we simplify the expression to \( x + 10 \) for \( x eq 10 \). Here, the limit calculation becomes straightforward: \( \lim_{x \to 10} (x + 10) = 20 \), estimating the behavior near \( x = 10 \) without substitution issues.
Limits help identify potential continuity in functions, contributing to understanding the behavior and characteristics at precise points.
First, notice that directly substituting \( x = 10 \) into the function results in division by zero. This doesn't mean the limit doesn't exist. Instead, it suggests the need for further analysis to simplify the function.
By factoring the polynomial in the numerator as \( (x + 10)(x - 10) \), we simplify the expression to \( x + 10 \) for \( x eq 10 \). Here, the limit calculation becomes straightforward: \( \lim_{x \to 10} (x + 10) = 20 \), estimating the behavior near \( x = 10 \) without substitution issues.
Limits help identify potential continuity in functions, contributing to understanding the behavior and characteristics at precise points.
Removable Discontinuity Clarified
A removable discontinuity occurs in a function when the limit exists at a point, but the actual function value does not. This type of discontinuity is 'removable' because it can be "fixed" by defining the function at that point. In our exercise, the function \( f(x) = \frac{x^2 - 100}{x - 10} \) is non-continuous at \( x = 10 \), primarily due to \( f(10) \) being undefined.
However, since we've found that \( \lim_{x \to 10} f(x) = 20 \), there is a clear path to continuity by setting \( f(10) = 20 \). By doing so, we are "filling the hole" in the function, making it continuous across that interval.
It's key to remember that removable discontinuities are unique in their ability to be corrected by redefining function values, unlike more severe discontinuities where gaps or infinite values cannot be resolved.
However, since we've found that \( \lim_{x \to 10} f(x) = 20 \), there is a clear path to continuity by setting \( f(10) = 20 \). By doing so, we are "filling the hole" in the function, making it continuous across that interval.
It's key to remember that removable discontinuities are unique in their ability to be corrected by redefining function values, unlike more severe discontinuities where gaps or infinite values cannot be resolved.
Exploring Undefined Functions
An undefined function value occurs typically due to division by zero, as demonstrated in our exercise, where \( f(10) \) results in \( \frac{0}{0} \). Functions become undefined at points where operations violate mathematical rules, such as dividing by zero, leading to those spots lacking a proper function value.
To address this, it often involves algebraic manipulation, such as factoring polynomials, to explore the function's behavior around the problematic point. Once transformed, we can analyze limits and potentially define the function appropriately for continuity.
It's crucial to differentiate between undefined due to removable discontinuity, where limits exist and corrections are possible, and non-removable scenarios where such simplifications and definitions are not feasible. This understanding shapes how mathematicians treat and interpret these voids in function outputs.
To address this, it often involves algebraic manipulation, such as factoring polynomials, to explore the function's behavior around the problematic point. Once transformed, we can analyze limits and potentially define the function appropriately for continuity.
It's crucial to differentiate between undefined due to removable discontinuity, where limits exist and corrections are possible, and non-removable scenarios where such simplifications and definitions are not feasible. This understanding shapes how mathematicians treat and interpret these voids in function outputs.
Other exercises in this chapter
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