Chapter 34: Problem 7
Ist \(\mathrm{g}\) auf \([0,1]\) definiert und beschr?nkt, so ist \(x \mapsto x g(x)\) in 0 stetig.
Short Answer
Expert verified
The function \( x \mapsto xg(x) \) is continuous at \( x = 0 \).
Step by step solution
01
Understanding the Problem
We need to prove that if the function \( g \) is defined and bounded on the interval \([0,1]\), then the function \( x \mapsto xg(x) \) is continuous at \( x = 0 \). This means we want to show that \( \lim_{x \to 0} xg(x) = xg(0) = 0 \).
02
Bounding the Function
Since \( g \) is bounded on \([0, 1]\), there exists a constant \( M > 0 \) such that \(|g(x)| \leq M\) for all \( x \in [0, 1] \). This boundedness will be crucial in handling the limit as \( x \to 0 \).
03
Estimating the Expression
For any \( x \in [0, 1] \), we have: \(|xg(x)| \leq |x| \cdot |g(x)| \leq |x| \cdot M\). This inequality helps in controlling the behavior of \( xg(x) \) as \( x \) approaches 0.
04
Taking the Limit
Since \(|xg(x)| \leq M|x|\) and \( \lim_{x \to 0} M|x| = 0 \), by the Squeeze Theorem, we have \( \lim_{x \to 0} |xg(x)| = 0 \). This implies \( \lim_{x \to 0} xg(x) = 0 \).
05
Conclusion of Continuity
Based on the limit evaluated, \( xg(x) \) is continuous at \( x = 0 \) because \( \lim_{x \to 0} xg(x) = 0 = xg(0) \). Thus, the function \( x \mapsto xg(x) \) is continuous at this point.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Squeeze Theorem
The Squeeze Theorem is a handy tool for evaluating limits and is particularly useful when dealing with functions that might behave unpredictably. The theorem states:
In our exercise, the Squeeze Theorem helps show that \( \lim_{x \to 0} xg(x) = 0 \) by comparing \( |xg(x)| \) with \(|x|M\), which approaches zero as \( x \to 0 \). This establishes the continuity at \( 0 \).
- If you have three functions, say \( f(x) \), \( g(x) \), and \( h(x) \), and for all \( x \) in a certain interval except possibly at a point \( c \), \( f(x) \leq g(x) \leq h(x) \).
- And if \( \lim_{x \to c} f(x) = L = \lim_{x \to c} h(x) \), then \( \lim_{x \to c} g(x) = L \) as well.
In our exercise, the Squeeze Theorem helps show that \( \lim_{x \to 0} xg(x) = 0 \) by comparing \( |xg(x)| \) with \(|x|M\), which approaches zero as \( x \to 0 \). This establishes the continuity at \( 0 \).
Bounded Function
A bounded function is a function that does not "go to infinity" either positively or negatively, within a certain interval. In simpler terms, there is a limit to how large or how small the function values can get. For example, a function defined on
This characteristic is crucial because it allows the expression \( xg(x) \) to also be bounded, ensuring that \( \lim_{x \to 0} xg(x) \) can be more easily evaluated using the Squeeze Theorem. By knowing \( g(x) \) is bounded, it aids us in establishing that \( |xg(x)| \) will always be less than or equal to \( |x|M \), effectively controlling its behavior as \( x \) approaches zero.
- an interval \([a, b]\), is bounded if there exists a number \( M \) such that for every \( x \) in the interval, the function satisfies \(|g(x)| \leq M\).
This characteristic is crucial because it allows the expression \( xg(x) \) to also be bounded, ensuring that \( \lim_{x \to 0} xg(x) \) can be more easily evaluated using the Squeeze Theorem. By knowing \( g(x) \) is bounded, it aids us in establishing that \( |xg(x)| \) will always be less than or equal to \( |x|M \), effectively controlling its behavior as \( x \) approaches zero.
Limit of a Function
The concept of a limit deals with how a function behaves as the input approaches a certain value. It helps us understand behavior at points where the function might not be exactly defined, or reveal insights about the trend of values within a small area.
Mathematically, the limit \( \lim_{x \to c} f(x) = L \) illustrates that as \( x \) gets closer to \( c \), the output \( f(x) \) gets closer to \( L \).
In this scenario, the exercise seeks to determine that \( \lim_{x \to 0} xg(x) = 0 \), which speaks to the function \( xg(x) \) approaching a value of 0 precisely as \( x \) zeroes in on the point zero. This limit's evaluation is integral to confirming that \( x \mapsto xg(x) \) remains continuous at \( x = 0 \).
Mathematically, the limit \( \lim_{x \to c} f(x) = L \) illustrates that as \( x \) gets closer to \( c \), the output \( f(x) \) gets closer to \( L \).
- In other words, for every small positive number \( \epsilon \), there exists a \( \delta \) such that whenever \( 0 < |x-c| < \delta \), it follows that \(|f(x) - L| < \epsilon\).
In this scenario, the exercise seeks to determine that \( \lim_{x \to 0} xg(x) = 0 \), which speaks to the function \( xg(x) \) approaching a value of 0 precisely as \( x \) zeroes in on the point zero. This limit's evaluation is integral to confirming that \( x \mapsto xg(x) \) remains continuous at \( x = 0 \).