Problem 38 - Thomas Calculus

Question

Use series to evaluate the limits. \begin{equation} \lim _{x \rightarrow 2} \frac{x^{2}-4}{\ln (x-1)} \end{equation}

Step-by-Step Solution

Answer

The limit evaluates to 4.

Step 1: Recognize the Indeterminate Form

The given limit \( \lim _{x \rightarrow 2} \frac{x^{2}-4}{\ln (x-1)} \) is initially in the indeterminate form \( \frac{0}{0} \). As \( x \rightarrow 2 \), \( x^2 - 4 \rightarrow 0 \) and \( \ln(x-1) \rightarrow \ln(1) = 0 \). This signals the need for further manipulation, such as using Taylor expansion.

Step 2: Apply Taylor Series Expansion

For \( \ln(1+x) \) around 0, the Taylor series is \( x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots \). Rewriting \( \ln(x-1) = \ln(1 + (x-2)) \), the series expansion gives \( (x-2) - \frac{(x-2)^2}{2} + \cdots \).

Step 3: Simplify the Expression

We need to simplify the expression with this series expansion: \( \frac{x^2-4}{\ln(x-1)} = \frac{(x-2)(x+2)}{(x-2) - \frac{(x-2)^2}{2} + \cdots} \). Since \( x \to 2 \), focus on the leading terms. The major term here contributes \( \frac{x+2}{1} \) because the series expansion shows \( (x-2) \) is the leading behavior in the numerator and denominator.

Step 4: Calculate the Limit

Since the series expansion indicates \( \ln(x-1) \approx (x-2) \) for \( x \to 2 \), the other terms become negligible. The expression simplifies to a form that's clearer to handle: \( \frac{(x-2)(x+2)}{(x-2)} \). Cancel the \( (x-2) \) terms, resulting in \( x+2 \). Evaluating at \( x = 2 \) gives: \( 2 + 2 = 4 \).

Step 5: Conclude with Final Answer

Now, complete the limit evaluation by substituting back the resolved expression. Thus, the evaluated limit is \( 4 \).

Key Concepts

Taylor Series

The Taylor series is a powerful mathematical tool used for approximating functions. It expresses a function as an infinite sum of terms calculated from the values of its derivatives at a single point. For the function \( \ln(1+x) \), the Taylor series expansion around zero (Maclaurin series) is given by:

• \( x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots \)

This series is very useful in calculus, especially for approximating functions near the value where \( x \) is small. In the original problem, \( \ln(x-1) \) is expressed as \( \ln(1 + (x-2)) \). This allows us to apply the Taylor series where the variable is \((x-2)\), resulting in:

• \( (x-2) - \frac{(x-2)^2}{2} + \frac{(x-2)^3}{3} - \cdots \)

Using Taylor series, the behavior of \( \ln(x-1) \) is approximated effectively for limit evaluation.

Limit Evaluation

Limit evaluation is the process of finding the value that a function approaches as the input approaches a certain value. In calculus, limits are essential for defining derivatives and integrals. In the problem:

• \( \lim _{x \rightarrow 2} \frac{x^{2}-4}{\ln (x-1)} \)

we aim to find what the expression approaches as \( x \) nears 2. Recognizing it as an indeterminate form \( \frac{0}{0} \) requires further analysis, because direct substitution results in undefined expressions. Using the Taylor series expands \( \ln(x-1) \) and simplifies limit evaluation by focusing on leading behaviors. The simplification results in a more tractable form, making it easier to evaluate the outcomes.

Indeterminate Forms

Indeterminate forms occur in calculus when evaluating a limit leads to expressions like \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), or similar forms that are not directly resolvable. In these scenarios, the true limiting behavior isn't obvious, requiring further methods to solve.When faced with:

• \( \lim _{x \rightarrow 2} \frac{x^{2}-4}{\ln (x-1)} \)

it initially presents \( \frac{0}{0} \) because both numerator and denominator approach zero as \( x \to 2 \). Indeterminate forms signal that there is no straightforward plug-and-play solution, thus prompting mathematical tools like Taylor series or algebraic manipulation to redefine or adjust the terms involved for clearer evaluation. Tackling this indeterminate form supports simplifying otherwise obscure outcomes.

L'Hopital's Rule

L'Hopital's rule is another method commonly used to evaluate limits involving indeterminate forms. It states that if \( \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{0}{0} \) or \( \frac{\infty}{\infty} \), then the limit can be found as:

• \( \lim_{x \to a} \frac{f'(x)}{g'(x)} \)

This means taking the derivative of both the numerator and the denominator, and then re-evaluating the limit. However, it is essential to first confirm the function is continuously differentiable near the point in question.Although not employed in the initial solution of the exercise, understanding L’Hopital's rule can provide alternative paths to solving limit problems. It's a powerful technique allowing direct evaluation of certain hard-to-solve limits by focusing on derivative behavior.