Question

Finding \(\delta\) algebraically Let \(f(x)=x^{2}-2 x+3\) a. For \(\varepsilon=0.25,\) find the largest value of \(\delta>0\) satisfying the statement $$ |f(x)-2|<\varepsilon \quad \text { whenever } \quad 0<|x-1|<\delta $$ b. Verify that \(\lim _{\vec{r} \rightarrow 1} f(x)=2\) as follows. For any \(\varepsilon>0\), find the largest value of \(\delta>0\) satisfying the statement $$ |f(x)-2|<\varepsilon \text { whenever } 0<|x-1|<\delta $$

Short answer

Answer: The largest value is δ = √ε.

Step-by-step solution

a. Finding the value of δ for ε = 0.25

1. First, we need to rewrite the inequality \(|f(x) - 2| < \varepsilon\) : $$ |x^{2} - 2x + 3 - 2| < 0.25 \\ |x^{2} - 2x + 1| < 0.25 \\ |(x - 1)^{2}| < 0.25 $$ 2. Now that we have the inequality, we can find the largest value of \(\delta\) such that \(0 <|x - 1|< \delta\) satisfying: $$ |(x - 1)^{2}| < 0.25 $$ 3. Note that for \(0<|x - 1|<\delta\), we have: $$ 0 < (x - 1)^{2} < 0.25 $$ 4. Next, we solve for x: $$ -0.25 < x - 1 < 0.25 \\ 0.75 < x < 1.25 $$ 5. We see that the largest value of \(\delta\) that satisfies the inequality is the maximum distance from x to 1 in the interval [0.75, 1.25]. The largest distance occurs at x = 0.75 or x = 1.25: $$ \delta = \max\{ |0.75 - 1|, |1.25 - 1| \} = 0.25 $$ So the largest value of \(\delta\) for \(\varepsilon=0.25\) is \(\delta=0.25\).

b. Finding the value of δ for any ε > 0

1. We need to rewrite the inequality \(|f(x) - 2| < \varepsilon\) again: $$ |x^{2} - 2x + 3 - 2| < \varepsilon \\ |x^{2} - 2x + 1| < \varepsilon \\ |(x - 1)^{2}| < \varepsilon $$ 2. For \(0<|x - 1|<\delta\), we have the inequality: $$ 0 < (x - 1)^{2} < \varepsilon $$ 3. Now we consider the interval \(-\sqrt{\varepsilon} + 1 < x < \sqrt{\varepsilon} + 1\): $$ -\sqrt{\varepsilon} + 1 < x < \sqrt{\varepsilon} + 1 $$ 4. By comparing the inequality to the condition \(0<|x - 1|<\delta\), we figure out the maximum distance from x to 1: $$ \delta = \max\{ |\text{-} \sqrt{\varepsilon} + 1 - 1|, |\sqrt{\varepsilon} + 1 - 1| \} = \sqrt{\varepsilon} $$ Therefore, the largest value of \(\delta\) that satisfies the statement for any \(\varepsilon > 0\) is \(\delta = \sqrt{\varepsilon}\).

Key concepts

Understanding the Delta-Epsilon Proof

The delta-epsilon proof is a formal method used to define the concept of a limit in calculus. This proof plays a crucial role in understanding the behavior of functions as they approach a certain value. In essence, the delta-epsilon proof is used to show that the limit of a function at a particular point is a specific value.

In this proof, \(\text{ε} (epsilon)\) represents an arbitrarily small positive number that defines how close the function's value should be to the limit value. Likewise, \(\text{δ} (delta)\) is a positive number that corresponds to a particular distance from the point at which we are taking the limit. The goal is to show that for every \(\text{ε}\), there exists a \(\text{δ}\) such that whenever \(0 < |x - c| < \text{δ}\), the inequality \( |f(x) - L| < \text{ε}\) holds true, where \(c\) is the point we're approaching and \(L\) is the proposed limit value.

In the given exercise, the function \( f(x) = x^2 - 2x + 3 \text{ has a proposed limit of 2 as} x \text{ approaches 1}\). Through an algebraic process, we find a valid \(\text{δ}\) that proves the limit statement within the delta-epsilon framework for a specific \(\text{ε}\) and generalize it for all positive \(\text{ε}\) values.

Grasping the Limit of a Function

The limit of a function is a fundamental concept in calculus that describes the behavior of a function as its argument approaches a certain point. It's important to distinguish that the limit does not necessarily have to equal the function’s value at that point; it’s about the trend as it gets arbitrary close.

To understand the concept intuitively, imagine a function's graph: as we zoom in on a point where we're taking the limit, the values of the function should get closer and closer to a specific number, representing the limit. This conception helps define continuity and the behavior of functions at points where they may not be well-defined.

In the provided exercise, we are examining the function \( f(x) \text{ as} x \text{ approaches 1}\), and we're interested in knowing its limit. The steps discussed illustrate the precise method by which we can confirm that the given function approaches the value of 2 as \(x\) approaches 1. This thorough analysis assures us of the behavior of \( f(x) \text{ in the vicinity of the point 1}\).

The Process of Algebraic Limit Computation

Algebraic computations are often used to find the limits of functions. This method involves manipulating the function algebraically to discern the limit. Algebraic limit computation may involve factoring, simplifying, or performing other algebraic operations to make the limit clear.

In the previously mentioned exercise, the algebraic steps involve expanding and simplifying the expression to isolate \( (x - 1)^2 \text{ and evaluating the inequalities to find} \text{δ}\).

Key Points in Algebraic Limit Computation

• Understand the original function and the point at which the limit is being taken.
• Rewrite the function in a way that highlights the trend as the variable approaches the limit point.
• Compare the resulting expression to the epsilon condition to find the appropriate delta.

Performing these steps allows us to apply the delta-epsilon proof and establish the validity of our limit computation. When done correctly, this process not only finds the value of the limit but also reinforces the reasoning behind the result.