Problem 70
Question
The function \(f(x)=\left\\{\begin{array}{cll}\frac{x-1}{e^{x-1}}+1 & , & x \neq 1 \\ 0 & , & x=1\end{array}\right.\) (A) is continuous (B) has removable discontinuity (C) has jump discontinuity (D) has infinite discontinuity
Step-by-Step Solution
Verified Answer
(B) The function has removable discontinuity.
1Step 1: Identify and Test Continuity at x = 1
To determine the continuity of the function at a point, we need to evaluate the limit of the function as \( x \) approaches 1 and compare it to \( f(1) \). Let's calculate the limit \( \lim_{x \to 1} f(x) \). For \( x eq 1 \), \( f(x) = \frac{x-1}{e^{x-1}} + 1 \).
2Step 2: Evaluate the Limit as x Approaches 1
Substitute into the expression \( \lim_{x \to 1} \left( \frac{x-1}{e^{x-1}} + 1 \right) \). Use L'Hôpital's Rule, since it results in an indeterminate form \( \frac{0}{0} \), to compute the limit: \[\lim_{x \to 1} \frac{x-1}{e^{x-1}} = \lim_{x \to 1} \frac{1}{e^{x-1}} = 0\] So, the limit is \( 0 + 1 = 1 \).
3Step 3: Check f(1) and Compare
We know that \( f(1) = 0 \). However, from Step 2, the limit as \( x \to 1 \) yields a result of 1. This indicates that \( \lim_{x \to 1} f(x) eq f(1) \), marking a discontinuity.
4Step 4: Determine the Type of Discontinuity
Because \( \lim_{x \to 1} f(x) \) exists but is not equal to \( f(1) \), the discontinuity is removable. In other words, the function can be made continuous by redefining \( f(1) \) to be equal to the limit.
Key Concepts
Removable DiscontinuityFunction ContinuityL'Hôpital's Rule
Removable Discontinuity
A removable discontinuity occurs when there is a point on a graph that, if filled in, would make the function continuous at that point. This usually happens when the limit of a function exists as it approaches a certain point, but the function's value at that point doesn't match the limit.
In the example with the function \( f(x) \), we determined that the limit as \( x \) approaches 1 is 1, but \( f(1) \) is 0. This mismatch indicates a removable discontinuity.
To "remove" this discontinuity, we could redefine \( f(1) \) to be the limit value, which is 1. This adjustment would make the function continuous at \( x = 1 \). Thus, while removable discontinuities indicate a point of otherwise well-behaved functions, they usually suggest the need for small corrections in function definitions.
In the example with the function \( f(x) \), we determined that the limit as \( x \) approaches 1 is 1, but \( f(1) \) is 0. This mismatch indicates a removable discontinuity.
To "remove" this discontinuity, we could redefine \( f(1) \) to be the limit value, which is 1. This adjustment would make the function continuous at \( x = 1 \). Thus, while removable discontinuities indicate a point of otherwise well-behaved functions, they usually suggest the need for small corrections in function definitions.
- They often occur due to human error in defining function values.
- Are correctable by redefining a single point on the function.
- Unlike other discontinuities, don't involve breaks or leaps in the graph.
Function Continuity
Continuity in a function means that there are no breaks, jumps, or gaps in its graph. It's a seamless connection where approaching the function from either side of a point leads to the same value, and that value matches the function's defined value at that point.
A function is continuous at a point \( x = c \) if:
These conditions show that there are no sudden jumps or dips. In our example, the function \( f(x) \) at \( x = 1 \) did not meet the third condition, which led to highlighting the discontinuity there. Achieving continuity often involves carefully checking limits and defining functions properly, especially at problematic points. This ensures smooth graphs and predictable behavior.
A function is continuous at a point \( x = c \) if:
- The function \( f(x) \) is defined at \( x = c \).
- The limit \( \lim_{x \to c} f(x) \) exists.
- \( \lim_{x \to c} f(x) = f(c) \).
These conditions show that there are no sudden jumps or dips. In our example, the function \( f(x) \) at \( x = 1 \) did not meet the third condition, which led to highlighting the discontinuity there. Achieving continuity often involves carefully checking limits and defining functions properly, especially at problematic points. This ensures smooth graphs and predictable behavior.
L'Hôpital's Rule
L'Hôpital's Rule is a handy tool for solving limits that arise in indeterminate forms, like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). When faced with such forms, one can differentiate the numerator and denominator separately to find the limit.
In our problem, evaluating \( f(x) \) as \( x \) approaches 1 leads to an indeterminate form \( \frac{0}{0} \). L'Hôpital's Rule makes it possible to calculate the limit by differentiating the top and bottom:
\[ \lim_{x \to 1} \frac{x-1}{e^{x-1}} = \lim_{x \to 1} \frac{1}{e^{x-1}} = 0 \]
Adding the constant term gives a final limit of 1. This process allows us to determine that while the direct evaluation was indeterminate, the correct limiting value leads us towards understanding the discrepancy at \( x = 1 \). Mastering L'Hôpital's Rule can simplify resolving complex limits and is essential for calculus students.
In our problem, evaluating \( f(x) \) as \( x \) approaches 1 leads to an indeterminate form \( \frac{0}{0} \). L'Hôpital's Rule makes it possible to calculate the limit by differentiating the top and bottom:
\[ \lim_{x \to 1} \frac{x-1}{e^{x-1}} = \lim_{x \to 1} \frac{1}{e^{x-1}} = 0 \]
Adding the constant term gives a final limit of 1. This process allows us to determine that while the direct evaluation was indeterminate, the correct limiting value leads us towards understanding the discrepancy at \( x = 1 \). Mastering L'Hôpital's Rule can simplify resolving complex limits and is essential for calculus students.
- Used for indeterminate forms like \( \frac{0}{0} \).
- Involves differentiating numerator and denominator.
- Essential for finding accurate limits otherwise tough to evaluate.
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