Problem 23
Question
Prove the statement using the \( \varepsilon \), \( \delta \) definition of a limit. \( \displaystyle \lim_{x \to a} x = a \)
Step-by-Step Solution
Verified Answer
The limit \( \lim_{x \to a} x = a \) holds because choosing \( \delta = \varepsilon \) satisfies the conditions for any \( \varepsilon > 0 \).
1Step 1: Understanding the Limit Definition
The limit statement \( \lim_{{x \to a}} x = a \) means that as \( x \) approaches \( a \), the value of \( x \) gets arbitrarily close to \( a \). To prove this using the \( \varepsilon \) and \( \delta \) definition of a limit, for every \( \varepsilon > 0 \), we need to find a \( \delta > 0 \) such that if \( 0 < |x - a| < \delta \), then \( |x - a| < \varepsilon \).
2Step 2: Expression of the Condition \(|x - a| < \varepsilon\)
From the limit definition, our task translates to ensuring that \(|x - a|\) (distance between \(x\) and \(a\)) remains less than \(\varepsilon\). Our goal is to prove \(|x - a| < \varepsilon\) given \(0 < |x - a| < \delta\).
3Step 3: Choosing \( \delta \)
To satisfy \( |x - a| < \varepsilon \), we can straightforwardly choose \( \delta = \varepsilon \). This choice works because as long as \( |x - a| < \varepsilon \), both conditions \( |x - a| < \varepsilon \) and \( |x-a| < \delta \) are satisfied by definition since we set \( \delta = \varepsilon \).
4Step 4: Conclusion of Proof
With \( \delta = \varepsilon \), the conditions \( 0 < |x - a| < \delta \) naturally imply \( |x - a| < \varepsilon \). Hence, the \( \varepsilon \), \( \delta \) definition for the limit \( \lim_{{x \to a}} x = a \) holds true for any given \( \varepsilon > 0 \). This completes the proof.
Key Concepts
Limit of a FunctionProof TechniquesContinuity
Limit of a Function
The concept of a limit is central to calculus and mathematical analysis. The limit of a function describes how a function behaves as the input (usually denoted as \( x \)) approaches a particular value (denoted \( a \)). In simple terms, you can think of a limit as the value that a function approaches, but does not necessarily reach, as the input gets closer and closer to the specified number.
When we write \( \lim_{{x \to a}} f(x) = L \), it means as \( x \) gets closer to \( a \), the value of the function \( f(x) \) approaches \( L \). The limit doesn't always have to be the function's value at \( a \).
When we write \( \lim_{{x \to a}} f(x) = L \), it means as \( x \) gets closer to \( a \), the value of the function \( f(x) \) approaches \( L \). The limit doesn't always have to be the function's value at \( a \).
- We focus on behavior very close to \( a \), not necessarily exactly at \( a \).
- If you can make \( f(x) \) as close as you want to \( L \) just by choosing \( x \) sufficiently near \( a \), then \( L \) is the limit of \( f \) as \( x \) approaches \( a \).
Proof Techniques
Proofs come in many forms and serve as a way to rigorously demonstrate the truth of mathematical statements. One common method is the epsilon-delta proof, which is particularly used for proving limits in calculus.
In epsilon-delta proofs, we need to show that for every \( \varepsilon > 0 \) (no matter how small), there exists a \( \delta > 0 \) such that if **\( 0 < |x - a| < \delta \)**, then **\( |f(x) - L| < \varepsilon \)**. This statement ensures that \( f(x) \) can get arbitrarily close to \( L \) as \( x \) approaches \( a \).
In epsilon-delta proofs, we need to show that for every \( \varepsilon > 0 \) (no matter how small), there exists a \( \delta > 0 \) such that if **\( 0 < |x - a| < \delta \)**, then **\( |f(x) - L| < \varepsilon \)**. This statement ensures that \( f(x) \) can get arbitrarily close to \( L \) as \( x \) approaches \( a \).
- Identify your \( \varepsilon \): the level of closeness to the limit.
- Determine your \( \delta \): the acceptable range for the input \( x \).
- Show that if \( x \) is within \( \delta \) of \( a \), then \( f(x) \) is within \( \varepsilon \) of \( L \).
Continuity
Continuity is a property of functions that describes smooth and unbroken behavior over an interval. Intuitively, if you can draw a function without lifting your pencil from the paper, it is continuous. In mathematical terms, a function \( f(x) \) is continuous at a point \( a \) if three conditions are satisfied:
Understanding continuity is important as it underpins various operations in calculus, like differentiation and integration. Continuous functions guarantee smoother transitions and are fundamental in defining problem setups in calculus applications.
- The function \( f(x) \) is defined at \( a \).
- The limit \( \lim_{{x \to a}} f(x) \) exists.
- The limit is equal to the function's value at \( a \), that is, \( \lim_{{x \to a}} f(x) = f(a) \).
Understanding continuity is important as it underpins various operations in calculus, like differentiation and integration. Continuous functions guarantee smoother transitions and are fundamental in defining problem setups in calculus applications.
Other exercises in this chapter
Problem 23
Find the limit or show that it does not exist. \( \displaystyle \lim_{x \to \infty}\frac{\sqrt{1 + 4x^6}}{2 - x^3} \)
View solution Problem 23
How would you "remove the discontinuity" of \( f \)? In other words, how would you define \( f(2) \) in order to make \( f \) continuous at 2? \( f(x) = \dfrac
View solution Problem 23
Evaluate the limit, if it exists. \( \displaystyle \lim_{x \to 3}\frac{\frac{1}{x}-\frac{1}{3}}{x - 3} \)
View solution Problem 24
Find the derivative of the function using the definition of derivative. State the domain of the function and the domain of its derivative. \( f(x) = 4 + 8x - 5x
View solution