Problem 28
Question
Prove the statement using the \( \varepsilon \), \( \delta \) definition of a limit. \( \displaystyle \lim_{x \to -6^+} \sqrt[5]{6 + x} = 0 \)
Step-by-Step Solution
Verified Answer
The limit is proven by showing \\( 0 < \sqrt[5]{6 + x} < \varepsilon \\) for \\( \delta = \varepsilon^5 \\).
1Step 1: Identify Limit Definition
The \( \varepsilon \, \delta \) definition of a limit states that for every \( \varepsilon > 0 \), there exists a \( \delta > 0 \) such that if \( 0 < x + 6 < \delta \, \) then \(|\sqrt[5]{6 + x} - 0| < \varepsilon\). Our goal is to show this holds true for \( x \) approaching \(-6^+\).
2Step 2: Relate Function to Epsilon
Since \( \lim_{x \to -6^+} \sqrt[5]{6 + x} = 0 \), we want \( |\sqrt[5]{6 + x}| < \varepsilon \). This simplifies to \( \sqrt[5]{6 + x} < \varepsilon \) since \( \sqrt[5]{6 + x} \) is positive for \( x \to -6^+ \).
3Step 3: Express Delta in Terms of Epsilon
To satisfy \( |\sqrt[5]{6 + x}| < \varepsilon \, \) we take \( \sqrt[5]{6 + x} < \varepsilon \. \) This implies \( 6 + x < \varepsilon^5 \. \) Solving for \( x \) gives \( x < \varepsilon^5 - 6 \. \) Choose \( \delta = \varepsilon^5\).
4Step 4: Validate the Choice of Delta
Assume \( 0 < x + 6 < \delta = \varepsilon^5 \. \) Then \( 0 < \sqrt[5]{6 + x} < \varepsilon \. \) This ensures that \( |\sqrt[5]{6 + x} - 0| < \varepsilon \, \) thus fulfilling the definition of the limit.
Key Concepts
Limit of a FunctionReal AnalysisCalculus Proofs
Limit of a Function
When dealing with the limit of a function, we are interested in understanding how a function behaves as the input approaches a particular value. In this case, the focus is on the limit of the function \( f(x) = \sqrt[5]{6 + x} \) as \( x \) approaches \( -6^+ \):
- A limit can be intuitively thought of as the value a function "approaches" as the input gets arbitrarily close to a given point.
- For the specific function \( \sqrt[5]{6 + x} \), as \( x \) approaches \(-6^+\), the function approaches \( 0 \).
Real Analysis
Real analysis provides the rigorous foundation for studying limits, continuity, and other fundamental concepts in calculus. It employs precise definitions and logical reasoning to delve into the behavior of real-valued functions. The epsilon-delta definition is a cornerstone concept here:
- The \( \varepsilon \), \( \delta \) method is essential for establishing the existence of a limit, offering a formal way to express the approach of a function towards a specific value.
- It relies on creating a "neighborhood" around the point of interest, ensuring the function value stays within a prescribed vicinity, called \( \varepsilon \), by controlling \( \delta \), the distance in the domain.
Calculus Proofs
Calculus proofs often require the use of structured reasoning and logical arguments to establish mathematical truths. With the epsilon-delta definition of limits, the proof process involves:
- Clearly stating the conditions of the problem, such as the input value \( x \) approaching \(-6^+\) in our case.
- Determining transformations of the conditions into manageable mathematical expressions, like expressing \( \delta \) in terms of \( \varepsilon \).
- Validating that the transformations hold under the defined constraints, confirming that the function's output meets the given criteria.
Other exercises in this chapter
Problem 28
If \( g(x) = x^4 - 2 \), find \( g'(1) \) and use it to find an equation of the tangent line to the curve \( y = x^4 - 2 \) at the point \( (1, -1) \).
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Find the limit or show that it does not exist. \( \displaystyle \lim_{x \to -\infty} \left(\sqrt{4x^2 + 3x} + 2x \right) \)
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Evaluate the limit, if it exists. \( \displaystyle \lim_{x \to 2}\frac{x^2 - 4x + 4}{x^4 - 3x^2 - 4} \)
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Use a table of values to estimate the value of the limit. If you have a graphing device, use it to confirm your result graphically. \( \displaystyle \lim_{x \to
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