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Magazine Chapter 2: Problem 39
Prove the limit statements $$\lim _{x \rightarrow 9} \sqrt{x-5}=2$$
Short Answer
Expert verified
The limit \( \lim_{x \rightarrow 9} \sqrt{x-5} = 2 \) is proven using the epsilon-delta definition.
Step by step solution
01
Understand the Limit Statement
We are given the limit \( \lim_{x \rightarrow 9} \sqrt{x-5} = 2 \). This means as \( x \) approaches 9, the value of \( \sqrt{x-5} \) should approach 2.
02
Substitute the Limit Point
Substitute \( x = 9 \) into the function \( \sqrt{x-5} \) to verify the limit point. This gives \( \sqrt{9-5} = \sqrt{4} = 2 \). Hence, the function value at \( x = 9 \) is indeed 2, which is consistent with the limit statement.
03
Use Epsilon-Delta Definition
We use the epsilon-delta definition of a limit to prove this statement. According to the definition, for every \( \varepsilon > 0 \), there exists a \( \delta > 0 \) such that if \( 0 < |x - 9| < \delta \), then \( |\sqrt{x-5} - 2| < \varepsilon \).
04
Determine \( \delta \)
Starting from \( |\sqrt{x-5} - 2| < \varepsilon \), express \( \sqrt{x-5} - 2 \) as \( \sqrt{x-5} = 2 + \varepsilon \). Square both sides to get \( x-5 = (2 + \varepsilon)^2 \). So, \( x = 9 + 4\varepsilon + \varepsilon^2 \). Simplify this to express \( |x-9| < \delta \), and solve for \( \delta \): \( \delta = 4\varepsilon + \varepsilon^2 \).
05
Verify the Supporting Conditions
To ensure \( \delta \) is valid, assume the chosen \( \varepsilon \) is small enough to control \( \sqrt{x-5} \sim 2 \), meaning variations are minor. Since \( \delta = 4\varepsilon + \varepsilon^2 \), small \( \varepsilon \) yields a sufficiently small \( \delta \) ensuring \( |\sqrt{x-5} - 2| < \varepsilon \).
06
Conclusion
We have shown using the epsilon-delta definition of a limit that \( \lim_{x \rightarrow 9} \sqrt{x-5} = 2 \). Thus, for any small \( \varepsilon \), we can find a \( \delta \) which satisfies the condition, confirming the limit.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Limit of a Function
A limit describes how a function behaves as its input (usually denoted as \( x \)) approaches a certain value. Evaluating limits is a foundational concept in calculus, as it helps us understand functions that can be complex.
For instance, consider the limit notation \( \lim_{x \rightarrow 9} \sqrt{x-5} = 2 \). This statement tells us that as \( x \) gets closer to 9, the value of \( \sqrt{x-5} \) will get closer to 2.
Several key ideas are helpful when understanding limits:
For instance, consider the limit notation \( \lim_{x \rightarrow 9} \sqrt{x-5} = 2 \). This statement tells us that as \( x \) gets closer to 9, the value of \( \sqrt{x-5} \) will get closer to 2.
Several key ideas are helpful when understanding limits:
- Approaching, not reaching: Limits express behavior as we approach a point, but do not need the function to reach a specific value at that point.
- Function behavior: Limits can describe increasingly large values (infinite limits) or finite values that the function approaches, regardless of whether the function is defined at the approach point.
- Practical application: Limits help define concepts like derivatives and integrals and help in solving real-world problems such as motion and growth rates.
Square Root Function
The square root function, denoted \( \sqrt{x} \), is a common mathematical function that returns the non-negative root of a number. In simple terms, it is the value that, when multiplied by itself, equals \( x \).
For our given problem, we examine \( \sqrt{x-5} \). The subtraction inside the root makes the expression a shifted version of the basic square root function, adjusted to account for input values minus 5.
Key characteristics of the square root function include:
For our given problem, we examine \( \sqrt{x-5} \). The subtraction inside the root makes the expression a shifted version of the basic square root function, adjusted to account for input values minus 5.
Key characteristics of the square root function include:
- Domain: The domain of \( \sqrt{x} \) is \( x \geq 0 \), meaning it only accepts non-negative numbers. However, an expression like \( \sqrt{x-5} \) shifts this domain to \( x \geq 5 \).
- Range: The range of a square root function is non-negative real numbers (\([0, \infty)\)).
- Graph: The graph of \( \sqrt{x} \) is a curve starting from the origin, gradually flattening as \( x \) increases. Shifting this graph as \( \sqrt{x-5} \) starts at \( x = 5 \).
Proving Limits
Proving limits often involves the epsilon-delta definition, an analytical approach to confirming that a function's value approaches a specific limit. This rigorous process helps ensure precision in calculus.
In our limit proof for \( \lim_{x \rightarrow 9} \sqrt{x-5} = 2 \), we directly apply the epsilon-delta definition:
In our limit proof for \( \lim_{x \rightarrow 9} \sqrt{x-5} = 2 \), we directly apply the epsilon-delta definition:
- Epsilon-delta definition: For a limit \( L \) of function \( f(x) \) as \( x \) approaches \( c \), we want to show: For any \( \varepsilon > 0 \), there exists a \( \delta > 0 \) such that if \( 0 < |x - c| < \delta \), then \( |f(x) - L| < \varepsilon \).
- Choosing \( \delta \): The challenge is to express \( \delta \) in terms of \( \varepsilon \). Solve for \( \delta \) such that the inequality \( |\sqrt{x-5} - 2| < \varepsilon \) holds, ensuring \( x \) remains close to 9.
- Assumptions: Assure that \( \varepsilon \) chosen is small enough, prompting a similarly small \( \delta \), sustaining \( \sqrt{x-5} \approx 2 \) within our acceptable margin.
