Problem 24
Question
Prove the statement using the \(\varepsilon, \delta\) definition of limit. \(\lim _{x \rightarrow a} c=c\)
Step-by-Step Solution
Verified Answer
The limit is proven as constant functions always have the limit equal to the constant.
1Step 1: Understanding the Limit Definition
The statement \[\lim_{x \rightarrow a} c = c\] claims that as \(x\) approaches \(a\), the constant function \(f(x) = c\) approaches \(c\). We need to prove this using the \(\varepsilon, \delta\) definition of limit.
2Step 2: The \\(\varepsilon, \delta\\)-Definition
According to the \(\varepsilon, \delta\) definition of limit, for every \(\varepsilon > 0\), there exists a \(\delta > 0\) such that if \(0 < |x - a| < \delta\), then \(|f(x) - L| < \varepsilon\). Here, \(f(x) = c\) and \(L = c\).
3Step 3: Substituting in the Definition
Substitute \(f(x) = c\) and \(L = c\) into the limit definition. We obtain \[|c - c| < \varepsilon.\] This simplifies to \[0 < \varepsilon.\]
4Step 4: Choosing \\(\delta\\)
Since \(0 < \varepsilon\) is always true for any \(\varepsilon > 0\), \[|x - a| < \delta\] has no effect on the inequality \(|c - c| < \varepsilon\). Thus, we can choose any positive \(\delta\) satisfying \(0 < \delta\). Typically, \(\delta = 1\) suffices.
5Step 5: Proving the Limit Statement
Since any \(\delta > 0\) satisfies the condition derived from the limit definition \[|c - c| < \varepsilon\], the statement is proven for any \(\varepsilon > 0\). Hence, \[\lim_{x \rightarrow a} c = c.\]
Key Concepts
Constant Function LimitLimit of a FunctionEpsilon-Delta ProofFormal Definition of Limit
Constant Function Limit
When dealing with a constant function, the output value of the function remains the same regardless of the input. This can be represented as \(f(x) = c\), where \(c\) is a constant. So, whenever you plug in any \(x\) value, the function simply returns \(c\).
In terms of limits, the idea is pretty straightforward. If \(f(x) = c\) for all \(x\), the limit as \(x\) approaches any number \(a\) is also \(c\). This is because the function doesn’t change its output at all. Therefore, the limit \(\lim_{x \rightarrow a} f(x)\) is \(c\), since \(f(x)\) does not depend on \(x\).
This exercise asks for a proof that the limit of a constant function is the constant itself using an epsilon-delta approach. Although constants might seem too simple at first glance, they provide a great way to become comfortable with more complex limit proofs.
In terms of limits, the idea is pretty straightforward. If \(f(x) = c\) for all \(x\), the limit as \(x\) approaches any number \(a\) is also \(c\). This is because the function doesn’t change its output at all. Therefore, the limit \(\lim_{x \rightarrow a} f(x)\) is \(c\), since \(f(x)\) does not depend on \(x\).
This exercise asks for a proof that the limit of a constant function is the constant itself using an epsilon-delta approach. Although constants might seem too simple at first glance, they provide a great way to become comfortable with more complex limit proofs.
Limit of a Function
The concept of a limit is foundational in calculus and involves understanding how a function behaves as its inputs approach a particular point. The limit \(\lim_{x \rightarrow a} f(x) = L\) means that as \(x\) gets arbitrarily close to \(a\), \(f(x)\) approaches the value \(L\).
This doesn’t necessarily mean \(f(x)\) will ever be exactly \(L\), only that it will get closer and closer to it. Limits help smoothly transition from algebra to calculus by enabling calculations about points that aren't easily evaluated through direct substitution, like points of discontinuity or points where the function isn't defined.
In this simple exercise where \(f(x) = c\), the solution demonstrates that the limit equals the function's constant value \(c\), plainly illustrating one way limits don’t always arise from change in \(x\).
This doesn’t necessarily mean \(f(x)\) will ever be exactly \(L\), only that it will get closer and closer to it. Limits help smoothly transition from algebra to calculus by enabling calculations about points that aren't easily evaluated through direct substitution, like points of discontinuity or points where the function isn't defined.
In this simple exercise where \(f(x) = c\), the solution demonstrates that the limit equals the function's constant value \(c\), plainly illustrating one way limits don’t always arise from change in \(x\).
Epsilon-Delta Proof
To rigorously prove a limit using the epsilon-delta definition, we need to demonstrate a precise relationship between the closeness of \(x\) to \(a\) and the closeness of \(f(x)\) to \(L\). The epsilon-delta definition states for every \(\varepsilon > 0\), there must exist a \(\delta > 0\) such that whenever \(0 < |x - a| < \delta\), it follows that \(|f(x) - L| < \varepsilon\).
The epsilon \(\varepsilon\) represents how close \(f(x)\) must be to \(L\), while \(\delta\) shows how close \(x\) must be to \(a\).
In case of our exercise, since \(f(x) = c\), \(|f(x) - L|\) becomes \(|c - c| = 0\), which is less than any positive \(\varepsilon\), always true. Therefore, any positive \(\delta\) will suffice, usually chosen as \(\delta = 1\) for simplicity. This eliminates the influence of \(x\) entirely here, reflecting the constancy of the constant function.
The epsilon \(\varepsilon\) represents how close \(f(x)\) must be to \(L\), while \(\delta\) shows how close \(x\) must be to \(a\).
In case of our exercise, since \(f(x) = c\), \(|f(x) - L|\) becomes \(|c - c| = 0\), which is less than any positive \(\varepsilon\), always true. Therefore, any positive \(\delta\) will suffice, usually chosen as \(\delta = 1\) for simplicity. This eliminates the influence of \(x\) entirely here, reflecting the constancy of the constant function.
Formal Definition of Limit
The formal definition of a limit is built around the precise language of the epsilon-delta approach, solidifying the intuitive concept of limits found in early calculus. This definition strengthens the foundational logic required to analyze the behavior of functions closely. It is written as:
This definition allows mathematicians not only to state, but also rigorously prove properties about functions as they approach a given point.
In showing \(\lim_{x \rightarrow a} c = c\), our task is elegantly simple due to the constant nature of the function \(f(x) = c\). By choosing any arbitrary \(\varepsilon\) and finding a corresponding \(\delta\) if even necessary, we show that \(|f(x) - L|\) is indeed less than \(\varepsilon\), using the unchanged difference of zero to elegantly meet conditions of the definition.
- For every \(\varepsilon > 0\), there exists \(\delta > 0\) such that,
- whenever \(0 < |x - a| < \delta\), then \(|f(x) - L| < \varepsilon\).
This definition allows mathematicians not only to state, but also rigorously prove properties about functions as they approach a given point.
In showing \(\lim_{x \rightarrow a} c = c\), our task is elegantly simple due to the constant nature of the function \(f(x) = c\). By choosing any arbitrary \(\varepsilon\) and finding a corresponding \(\delta\) if even necessary, we show that \(|f(x) - L|\) is indeed less than \(\varepsilon\), using the unchanged difference of zero to elegantly meet conditions of the definition.
Other exercises in this chapter
Problem 23
\(21-28\) Explain, using Theorems \(4,5,7,\) and \(9,\) why the function is continuous at every number in its domain. State the domain. \(R(x)=x^{2}+\sqrt{2 x-1
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. \(f(x)=x+\sqrt{x}\)
View solution Problem 24
(a) If $$G(x)=4 x^{2}-x^{3},\( find$$G^{\prime}(a)\) and use it to find equations of the tangent lines to the curve $$y=4 x^{2}-x^{3}$$ at the points \((2,8)\)
View solution Problem 24
Evaluate the limit, if it exists. $$\lim _{x \rightarrow-1} \frac{x^{2}+2 x+1}{x^{4}-1}$$
View solution