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Magazine Chapter 2: Problem 30
Show that \(\lim _{x \rightarrow a}|x|=|a|\)
Short Answer
Expert verified
The limit is proven using the epsilon-delta definition.
Step by step solution
01
Understand the Limit Definition
We are given the function \(|x|\) and need to prove that \(\lim_{x \to a}|x| = |a|\) for some number \(a\). This means as \(x\) approaches \(a\), \(|x|\) should approach \(|a|\).
02
Apply the Epsilon-Delta Definition of a Limit
According to the epsilon-delta definition of a limit, for every \(\varepsilon > 0\), there exists a \(\delta > 0\) such that if \(0 < |x-a| < \delta\), then \(| |x| - |a| | < \varepsilon\). This is what we need to show.
03
Calculate the Absolute Value Difference
The expression to consider is \(| |x| - |a| |\). For any real numbers \(x\) and \(a\), the relation \(| |x| - |a| | \leq |x-a|\) holds. This will help us find a \(\delta\) for a given \(\varepsilon\).
04
Choose an Appropriate Delta
Given \(\varepsilon > 0\), choose \(\delta = \varepsilon\). This ensures that if \(0 < |x-a| < \delta\), then \(| |x| - |a| | \leq |x-a| < \delta = \varepsilon\), which satisfies the epsilon-delta condition.
05
Conclusion
Since we have shown that for every \(\varepsilon > 0\), we can find a \(\delta = \varepsilon\) such that \(0 < |x-a| < \delta\) implies \(| |x| - |a| | < \varepsilon\), the limit \( \lim_{x \to a}|x| = |a|\) is proven.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Epsilon-Delta Definition
The epsilon-delta definition sets the foundation for understanding limits in calculus. This definition provides a precise way to show how a function behaves as the input approaches a particular point. Here's how it works:
- First, select any positive number \(\varepsilon\), which indicates how close the function's value needs to be to the limit.
- We then need to find a corresponding \(\delta > 0\) so that the function's output stays within \(\varepsilon\) of the limit whenever the input \(x\) is within \(\delta\) of \(a\), but not equal to \(a\).
- This translates to: if \(0 < |x-a| < \delta\), then \(|f(x) - L| < \varepsilon\).
Absolute Value
Absolute value is a fundamental concept in mathematics, especially when dealing with real numbers and limits. The absolute value of a number, \(|x|\), is its distance from zero on the number line, regardless of direction. It always yields a non-negative result.
- For a real number \(x\), \(|x| = x\) if \(x \geq 0\).
- For a real number \(x\), \(|x| = -x\) if \(x < 0\).
- Absolute value helps manage expressions that involve potentially negative terms, making these easier to manipulate, particularly when considering limits or inequalities.
Limit Proofs
Limit proofs, especially involving the epsilon-delta definition, are an essential part of calculus. They rigorously demonstrate how a function behaves as its input variable nears a specific point. Here鈥檚 how the limit proof process typically works:
- Start by identifying \(\varepsilon\), the proximity target for the function value to the limit.
- Determine a suitable \(\delta\) to control how close \(x\) should be to the target point \(a\) to keep the function鈥檚 value within \(\varepsilon\) of the limit.
- Use inequalities to relate the difference \(|f(x) - L|\) to \(|x-a|\), and show that one can always keep \(|f(x) - L| < \varepsilon\) with the right choice of \(\delta\).
- The final step is to demonstrate this relationship explicitly, as in the exercise where choosing \(\delta = \varepsilon\) satisfies all conditions, proving \(\lim_{x \to a}|x| = |a|\).
