Question

Finding a \(\delta\) for a Given \(\varepsilon\) In Exercises \(33-36\) , find the limit \(L\) . Then find \(\delta>0\) such that \(|f(x)-L|<0.01\) whenever \(0<|x-c|<\delta .\) $$ \lim _{x \rightarrow 4}\left(x^{2}+6\right) $$

Short answer

The limit 'L' is 22 and the \(\delta\) value such that \(|f(x) - L| < 0.01\) exists and is less than 0.01.

Step-by-step solution

Find the Limit

The limit of a function \(f(x)\) as \(x\) approaches a value \(c\) can be found by simply substituting the value of \(c\) into the function. Hence, the limit 'L' as \(x\) approaches 4, for the function \(x^2 + 6\) will be found by substituting \(x=4\) into \(x^2 + 6\), which will give us \(L=4^2 + 6 = 22\).

Find the Value of f(x)

We substitute the values of \(x\) which satisfy \(|x - 4| < 0.01\) into the function \(x^2 + 6\). Considering the two ends, for \(x = 4.01\) and \(x = 3.99\), we will get \(16.0801\) and \(15.9201\) respectively. This gives us the range where \(f(x)\) lies.

Calculate Delta

Given that \(|f(x) - L| < 0.01\), we need to solve for \(\delta\). The value of \(f(x) - L\) lies within the range \([16.0801 - 22, 15.9201 - 22]\), which simplifies to \([-5.9199, -6.0799]\). Taking the absolute value, we get \([5.9199, 6.0799]\). Since we need the \(\delta\) value such that the \(|f(x) - L| < 0.01\), the value of \(\delta\) must be less than 0.01. Hence, we are within the required \(\delta\) for the provided \(\varepsilon=0.01\).

Key concepts

Epsilon-Delta Definition

The epsilon-delta (\(\varepsilon\)-\(\delta\)) definition is a cornerstone concept of calculus, used to define the limit of a function. It gives a formal framework to understand what it means for a function to approach a certain value. Imagine you are standing on a number line. Your challenge is to move close to a specific mark, say 22 in this example, without actually stopping at it.
### How It WorksThis is where \(\varepsilon\) and \(\delta\) come into play:

• \(\varepsilon\) (epsilon) represents how close the output, \(f(x)\), is to the limit \(L\). So, |\(f(x) - L\)| < \(\varepsilon\) means the output should be within \(\varepsilon\) distance from \(L\).
• \(\delta\) (delta) refers to how close \(x\) must be to \(c\), the value approaching. |\(x - c\)| < \(\delta\) ensures \(x\)'s proximity to \(c\).

The goal is to show, for any chosen \(\varepsilon > 0\), there exists a \(\delta > 0\) ensuring that if \(x\) is within \(\delta\) of \(c\), then \(f(x)\) is within \(\varepsilon\) of \(L\). Here, it's finding a \(\delta\) when \(\varepsilon = 0.01\) for \(L = 22\). This gives us a precise way to talk about approaching behavior in functions.

Finding Delta Given Epsilon

In this step, we are tasked with finding the appropriate \(\delta\) after specifying our \(\varepsilon\). It’s like fine-tuning your position on a number line to ensure outputs stay within the defined \(\varepsilon\) boundary.
### Process Explanation

• Identify the limit \(L\) first. For \(x^2 + 6\), \(L = 22\) when \(x = 4\).
• Confirm your \(\varepsilon\) value. In this exercise, \(\varepsilon\) is 0.01.
• Solve |\(f(x) - L\)| < \(\varepsilon\) to find \(x\) values and thus \(\delta\). This involves substituting \(|x - 4| < \delta\) into the equation and ensuring that outputs still fall within 0.01 of \(L\).

Here, we checked around \(x = 4\), finding \(|16.0801 - 22|\) and \(|15.9201 - 22|\). Simplifying these gives the range of acceptable \(f(x)\), assisting us in deducing \(\delta = 0.01\), validating \(\varepsilon = 0.01\). This careful balancing ensures \(f(x)\) remains predictably close to \(L\) given any movement towards \(x = 4\).

Polynomial Functions

Polynomial functions, like the one in our exercise, are algebraic expressions composed of variables and constants using addition, subtraction, multiplication, and non-negative integer exponents. The function in question is \(x^2 + 6\). Recognizing and understanding them is key for solving various limit problems.
### Key Characteristics:

• Smooth and Continuous: Their graphs are lines or curves without breaks. This property makes them relatively predictable and easy to work with.
• Power Exponents: The degree of the polynomial (highest power of \(x\)) helps determine end behavior. Here, \(x^2\) signifies a quadratic shape.
• Standard Form: Typically arranged with terms in descending powers of \(x\). The function \(x^2 + 6\) is simplified, showing all terms explicitly.

Polynomial problems often require substituting values or working familiar operations like factoring. Their stability and consistency in formily handling make them a frequent subject for limit investigations in calculus.

Limits in Calculus

Limits are foundational in calculus, providing a way to understand the behavior of functions as variables approach certain values. In our example, we find the limit \(\lim_{x \to 4} (x^2 + 6)\), which simplifies to 22.
### Understanding Limits:

• Approaching Behavior: Not necessarily about reaching a point, but about understanding closeness—how \(f(x)\) behaves near \(c\).
• Existence of Limits: A limit exists if \(f(x)\), as \(x\) approaches \(c\), converges to a finite number, \(L\).
• Consistency: A limit can be seen from both sides of \(c\), and both approaches should result in the same \(L\).

Limits help us delve deeper into calculus concepts like derivatives and integrals. They carve the path from simple algebraic manipulations to dynamic systems analysis, shaping our understanding of continuous change in mathematics.