Chapter 10: Problem 38
Use series to evaluate the limits. \begin{equation} \lim _{x \rightarrow 2} \frac{x^{2}-4}{\ln (x-1)} \end{equation}
Short Answer
Expert verified
The limit evaluates to 4.
Step by step solution
01
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.
02
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 \( (x-2) - \frac{(x-2)^2}{2} + \cdots \).
03
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.
04
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: \( 2 + 2 = 4 \).
05
Conclude with Final Answer
Now, complete the limit evaluation by substituting back the resolved expression. Thus, the evaluated limit is \( 4 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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:
- \( x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots \)
- \( (x-2) - \frac{(x-2)^2}{2} + \frac{(x-2)^3}{3} - \cdots \)
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:
- \( \lim _{x \rightarrow 2} \frac{x^{2}-4}{\ln (x-1)} \)
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:
- \( \lim _{x \rightarrow 2} \frac{x^{2}-4}{\ln (x-1)} \)
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:
- \( \lim_{x \to a} \frac{f'(x)}{g'(x)} \)