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Magazine Chapter 2: Problem 40
Prove the limit statements in Exercises 37-50. $$\lim _{x \rightarrow 0} \sqrt{4-x}=2$$
Short Answer
Expert verified
The limit \( \lim_{x \rightarrow 0} \sqrt{4-x} = 2 \) is proved using epsilon-delta definition.
Step by step solution
01
Understand the Limit Definition
The problem asks to show that \( \lim_{x \rightarrow 0} \sqrt{4-x} = 2 \). This means that as \( x \) approaches 0, the function \( \sqrt{4-x} \) gets arbitrarily close to 2.
02
Calculate Function Value at Limit Point
First, substitute \( x = 0 \) into the function \( \sqrt{4-x} \) to check the value of the function at this point. This gives \( \sqrt{4-0} = \sqrt{4} = 2 \). The function equals 2 at \( x = 0 \), which gives us a hint but we need to prove this limit formally.
03
Apply the Epsilon-Delta Definition of a Limit
According to the definition of a limit, for every \( \varepsilon > 0 \), there exists a \( \delta > 0 \) such that if \( 0 < |x - 0| < \delta \), then \( |\sqrt{4-x} - 2| < \varepsilon \). We must find \( \delta \) in terms of \( \varepsilon \).
04
Simplify the Limit Condition
The goal is to show \( |\sqrt{4-x} - 2| < \varepsilon \). Start by manipulating the inequality:\[|\sqrt{4-x} - 2| = |\sqrt{4-x} - \sqrt{4}|.\] Use the identity \( |a - b| = \frac{|a^2 - b^2|}{|a + b|} \) to write:\[|\sqrt{4-x} - 2| = \frac{|(4-x)-4|}{|\sqrt{4-x} + 2|} = \frac{|x|}{|\sqrt{4-x} + 2|}.\]
05
Bound the Expression Below Epsilon
We want \( \frac{|x|}{|\sqrt{4-x} + 2|} < \varepsilon \). Notice that as \( x \to 0 \), \( \sqrt{4-x} \to 2 \), so \( |\sqrt{4-x} + 2| \to 4 \). Therefore, we can say \( \sqrt{4-x} + 2 \geq 2 \) for \( x \) very close to 0. This gives:\[\frac{|x|}{4} < \varepsilon \]which simplifies to \(|x| < 4\varepsilon\).
06
Determine a Suitable Delta
From \( |x| < 4\varepsilon \), choose \( \delta = 4\varepsilon \). Thus, for any \( \varepsilon > 0 \), we have shown there exists a \( \delta = 4\varepsilon \) such that whenever \( 0 < |x| < \delta \), \( |\sqrt{4-x} - 2| < \varepsilon \) holds.
07
Conclusion: Verify the Limit Statement
Because we have shown that for every \( \varepsilon > 0 \), there exists \( \delta > 0 \) (specifically, \( \delta = 4\varepsilon \)), satisfying the condition of the epsilon-delta definition of a limit, we have proven that \( \lim_{x \rightarrow 0} \sqrt{4-x} = 2 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Limit of a function
The concept of a limit is fundamental in calculus, representing how a function behaves as its input approaches a certain value. When we say "the limit of a function," we are interested in the trend of the output value rather than its exact value at a specific point. For example, considering the expression \( \lim_{x \rightarrow a} f(x) = L \), it means that as \( x \) gets closer to \( a \), the function \( f(x) \) approaches \( L \). However, \( f(x) \) doesn't need to equal \( L \) when \( x = a \). The limit simply describes a trend or tendency.Limits are essential tools in understanding behaviors like continuity, the properties of derivatives, and integrals.
Square root function
The square root function, denoted as \( \sqrt{x} \), is a common mathematical function. It represents a value which, when multiplied by itself, gives \( x \). This function plays a key role in mathematics, especially in algebra and calculus.Regarding its properties:
- The square root function is defined only for non-negative numbers, because the square of a real number is never negative.
- Its graph is a curve that starts at the origin and rises gradually.
- When combined with limits and continuity, it requires special attention due to issues at certain points (like zero).
Epsilon-delta proof
The epsilon-delta proof is a rigorous way to confirm the limit of a function at a point. It's a very precise mechanism used primarily in calculus.Here鈥檚 how it works:
- Given a function \( f(x) \) approaching \( L \) as \( x \) approaches \( a \), for any small positive number \( \varepsilon \), we want the function's value to be within \( \varepsilon \) of \( L \) as \( x \) gets close to \( a \).
- We find another small positive number \( \delta \) such that whenever \( 0 < |x - a| < \delta \), \( |f(x) - L| < \varepsilon \) holds true.
Calculating limits
Calculating limits involves determining the value that a function approaches as the input approaches a given point. Several techniques can be employed to solve limit problems, especially when direct substitution doesn't suffice.Steps generally include:
- First, try substituting the approach value directly into the function. If it doesn't produce a meaningful result, other methods are needed.
- Simplifying complex expressions using algebraic manipulation, such as factoring, multiplying by a conjugate, or using identities.
- Applying limit laws or theorems, like L'H么pital's Rule for indeterminate forms.
