Problem 19 - Single Variable Calculus

Question

Use the precise definition of a limit to prove that the statement is true. \(\lim _{x \rightarrow 2} \frac{x^{2}-4}{x-2}=4\)

Step-by-Step Solution

Answer

To prove that \(\lim_{x \rightarrow 2} \frac{x^2-4}{x-2} = 4\), we simplified the function to \(f(x) = x + 2\) for \(x \neq 2\). Then, using the precise definition of a limit, we demonstrated that for every \(\epsilon > 0\), there exists a \(\delta > 0\) (\(\delta = \epsilon\)) such that if \(0 < |x - 2| < \delta\), then \(|f(x) - 4| = |x - 2| < \epsilon\). This proves the statement is true.

Step 1: Identify the Function and the Limit

We are given the function \(f(x) = \frac{x^2 - 4}{x - 2}\) and we want to prove that \(\lim_{x \rightarrow 2} f(x) = 4\).

Step 2: Simplify the Function

To do this, we can first try to simplify the function by factoring it. The function can be rewritten as: \[f(x) = \frac{(x - 2)(x + 2)}{x - 2}\] As long as \(x \neq 2\), we can cancel the term \((x - 2)\) on both the numerator and the denominator. Therefore, we have: \[f(x) = x + 2 \; \text{for} \; x \neq 2\]

Step 3: Define the Precise Definition of a Limit

Now, we need to use the precise definition of a limit to show that for every \(\epsilon > 0\), there exists a \(\delta > 0\) such that if \(0 < |x - 2| < \delta\), then \(|f(x) - 4| < \epsilon\).

Step 4: Use the Simplified Function and the Limit

We need to show \(|f(x) - 4| < \epsilon\). Let's plug in our simplified function: \[|x + 2 - 4| < \epsilon\] \[|x - 2| < \epsilon\]

Step 5: Find the Relationship Between Delta and Epsilon

We want to find a \(\delta\) value such that \(|x - 2| < \delta \Rightarrow |x - 2| < \epsilon\). We can simply set \(\delta = \epsilon\).

Step 6: Prove the Limit Using the Definition

Now, for every \(\epsilon > 0\), we can choose \(\delta = \epsilon > 0\). If \(0 < |x - 2| < \delta\), then: \[|f(x) - 4| = |x - 2| < \epsilon\] Thus, we have proved that \(\lim_{x \rightarrow 2}\frac{x^2-4}{x-2} = 4\), using the precise definition of a limit.

Key Concepts

epsilon-delta definition

When we talk about limits in calculus, the epsilon-delta definition is crucial for proving them precisely. The idea is to show how close a function's value gets to the limit value as the input approaches a certain point. The formal way to express this uses two Greek letters: \(\epsilon\) and \(\delta\).

• \(\epsilon\) (epsilon) represents how close we want the function's value to be to the limit.
• \(\delta\) (delta) signifies how close the input value \(x\) must be to the point of interest.

For a limit \(\lim_{x \to a} f(x) = L\) to be true, given any small \(\epsilon > 0\), we must find a \(\delta > 0\) such that if \(0 < |x - a| < \delta\), then \(|f(x) - L| < \epsilon\).
This forms the essence of the epsilon-delta definition, ensuring that functions behave well as they approach particular inputs.

rational functions

Rational functions are an important class of functions which are expressed as a ratio of two polynomials. They are usually written in the form \(f(x) = \frac{g(x)}{h(x)}\), where both \(g(x)\) and \(h(x)\) are polynomials.
Such functions are defined everywhere except where the denominator equals zero, as division by zero is undefined. In solving problems involving limits of rational functions, it's common to encounter indeterminate forms like \(\frac{0}{0}\).
To resolve these, we often simplify the function through techniques like factoring, which can help us remove the troublesome parts in the denominator and establish a clearer path to finding the limit.

factoring

Factoring is a key algebraic technique used to simplify expressions by expressing them as a product of simpler terms. It plays a significant role in dealing with rational functions, especially when simplifying limits.
Let's consider the expression \(x^2 - 4\), which can be factored into \((x - 2)(x + 2)\). This step is particularly important because it reveals common factors with the denominator which can potentially be canceled.
By factoring, we not only simplify expressions but also make it easier to handle the operations in limits. This method helps us identify removable discontinuities and indeterminate forms that may arise in the evaluation of limits.

canceling terms

Canceling terms is an essential step that follows factoring, particularly when working with rational functions. After factoring, if the same factor appears in both the numerator and the denominator, we can cancel these terms unless the term results in division by zero.
Consider the function \(\frac{(x-2)(x+2)}{x-2}\) for \(x eq 2\). Here, \((x-2)\) appears in both the numerator and denominator, allowing us to cancel them and simplify the function to \(x + 2\) as long as \(x eq 2\).
Canceling helps in simplifying the algebraic expression, making it straightforward to evaluate the limit and also cleanly sidestep the problem of division by zero, leading to a function that behaves much like the original one wherever it's defined.