Problem 75 - Thomas Calculus in SI Units

Question

Which one is correct, and which one is wrong? Give reasons for your answers. a. \(\lim _{x \rightarrow 3} \frac{x-3}{x^{2}-3}=\lim _{x \rightarrow 3} \frac{1}{2 x}=\frac{1}{6}\) b. \(\lim _{x \rightarrow 3} \frac{x-3}{x^{2}-3}=\frac{0}{6}=0\)

Step-by-Step Solution

Answer

Statement a is correct; b is incorrect. a resolves the indeterminate form correctly using L'Hôpital's Rule.

Step 1: Understand the Limit Expression

We need to evaluate the limit expression \( \lim _{x \rightarrow 3} \frac{x-3}{x^{2}-3}\). As \(x\) approaches 3, both the numerator \((x-3)\) and the denominator \((x^2-3)\) approach 0, resulting in an indeterminate form \(\frac{0}{0}\).

Step 2: Simplify the Expression

To resolve the indeterminate form, we simplify the expression. Notice that \(x^2 - 3\) can be factored using a difference of squares: \(x^2 - 3 = (x - \sqrt{3})(x + \sqrt{3})\). However, factoring directly doesn't help simplify with \(x-3\) readily; instead, we consider an identity transformation or a suitable factorization if applicable.

Step 3: Use L'Hospital's Rule

Given the indeterminate form \(\frac{0}{0}\), we apply L'Hôpital's Rule, which states that for a limit \( \lim _{x \rightarrow c} \frac{f(x)}{g(x)}\) resulting in \(\frac{0}{0}\), we can instead evaluate \( \lim _{x \rightarrow c} \frac{f'(x)}{g'(x)}\). Compute the derivatives: \(f'(x) = 1\) for the numerator, and \(g'(x) = 2x\) for the denominator. Thus, the limit simplifies to \( \lim _{x \rightarrow 3} \frac{1}{2x} = \frac{1}{6} \).

Step 4: Verify Both Statements

Statement a correctly applies L'Hôpital's Rule to reach \(\frac{1}{6}\). Statement b incorrectly evaluates the limit as \(\frac{0}{6}=0\), without addressing the indeterminate form using proper techniques. Statement a is correct, and statement b is incorrect.

Key Concepts

Indeterminate Forms

In calculus, an indeterminate form is a mathematical expression that does not lead to a specific limit or value immediately. The most common example is when we have a limit taking the form \(\frac{0}{0}\). This is because both the numerator and the denominator approach zero, making it unclear what the limit of the expression should actually be. Various types of indeterminate forms exist, such as:

• \(\frac{0}{0}\)
• \(\frac{\infty}{\infty}\)
• \(\infty - \infty\)
• Products like \(0 \cdot \infty\)
• Powers like \(0^0\) or \(\infty^0\)

Dealing with indeterminate forms usually involves applying specific techniques like L'Hôpital's Rule, algebraic manipulation, or series expansion to resolve them. Once resolved, these forms can be analyzed, allowing for a proper limit or value to be found.

Limit Evaluation

Limit evaluation is a fundamental concept in calculus used to determine the behavior of a function as the input approaches a particular value. In the given exercise, you are tasked with evaluating the limit \(\lim _{x \rightarrow 3} \frac{x-3}{x^{2}-3}\). As \(x\) approaches 3, the direct substitution into the expression leads to the indeterminate form \(\frac{0}{0}\).To evaluate this limit, we can:

• Simplify the function algebraically, if possible.
• Use L'Hôpital's Rule when the direct evaluation leads to an indeterminate form.
• Consider any pre-existing limits or properties of the function that might eliminate the need for complex calculations.

In the given example, applying L'Hôpital's Rule allows us to find that the correct limit is \(\frac{1}{6}\). This simplifies the complex behavior of the function into a straightforward evaluation.

Differential Calculus

Differential calculus is primarily concerned with the concept of the derivative, which represents the rate of change of a function with respect to a variable. In the context of L'Hôpital's Rule, differential calculus becomes invaluable. When an indeterminate form \(\frac{0}{0}\) arises, L'Hôpital's Rule utilizes the derivatives of the numerator and the denominator. You calculate \(f'(x)\) and \(g'(x)\) to transform the indeterminate expression into one that can be easily evaluated. For example, in the limit \(\lim _{x \rightarrow 3} \frac{x-3}{x^2-3}\), the derivatives are:

• Numerator: \(f'(x) = 1\)
• Denominator: \(g'(x) = 2x\)

By substituting these derivatives back into the limit, the expression simplifies to \(\lim _{x \rightarrow 3} \frac{1}{2x}\), which is easily evaluated as \(\frac{1}{6}\). Thus, differential calculus serves as a powerful tool in resolving challenging limit problems.