Problem 39 - Calculus with Concepts in Calculus

Question

Find all vertical and horizontal asymptotes of the graph of \(f\). You may wish to use a graphics calculator to assist you. $$ f(x)=\frac{\ln (1-x)}{\ln (1+x)} $$

Step-by-Step Solution

Answer

The vertical asymptotes are at \( x = -1, 0, 1 \). No horizontal asymptotes.

Step 1: Identify potential vertical asymptotes

To find vertical asymptotes, we need to see where the function is undefined. The function \( f(x) = \frac{\ln (1-x)}{\ln (1+x)} \) is undefined where its denominator is zero. So, set \( \ln (1+x) = 0 \). This implies \( 1+x = e^0 = 1 \), or \( x = 0 \). So, \( x=0 \) is a potential vertical asymptote. Additionally, the function will also be undefined where the arguments of the logarithms become zero or negative, meaning \( 1-x \leq 0 \) or \( x\geq 1 \), and \( 1+x \leq 0 \) or \( x\leq -1 \). Therefore, \( x=1 \) and \( x=-1 \) are also points to consider.

Step 2: Analyze vertical asymptotes further

We confirm the vertical asymptote at \( x = 1 \) by checking the behavior of the function when \( x \to 1^- \). In this case, \( \ln(1-x) \to \ln(0) = -\infty \), making the function tend to \(-\infty\) if \( \ln(1+x) eq 0 \). Similarly, check around \( x = -1^+ \): \( \ln(1+x) \to \ln(0) = -\infty \), and also the function becomes undefined around \( x = -1 \). Conclusively, vertical asymptotes are at \( x = -1 \), \( x = 0 \), and \( x = 1 \).

Step 3: Find horizontal asymptotes

Horizontal asymptotes occur at \( \lim_{x \to \pm \infty} f(x) \). Check \( \lim_{x \to \infty} \frac{\ln (1-x)}{\ln (1+x)} \). As \( x \to \infty \), \( \ln(1-x) \) and \( \ln(1+x) \) both tend to \( \infty \). The quotient's limit will depend on their rate of approach. Using L'Hôpital’s Rule may help to compute this limit more accurately; however, the equality of their logarithmic forms suggests it tends toward a finite value if properly bounded. In this case, evaluate \( \lim_{x \to \infty} \frac{\ln(1-x)}{-x} \approx \lim_{x \to \infty} \frac{\ln(1+x)}{x} \) leading to 1, suggesting no horizontal asymptote but rather that \( f(x) \to c eq 0 \).

Step 4: Conclusion

The function \( f(x) = \frac{\ln (1-x)}{\ln (1+x)} \) has vertical asymptotes at \( x = -1, 0, \text{and} 1 \). It does not have a horizontal asymptote as it doesn't approach a horizontal line as \( x \to \pm \infty \).

Key Concepts

Vertical Asymptotes

Vertical asymptotes are critical characteristics of a function where the graph tends to infinity. For the function \( f(x) = \frac{\ln(1-x)}{\ln(1+x)} \), we determine vertical asymptotes by identifying values of \( x \) that make the denominator zero or lead to the logarithms being undefined.

These occur when the denominator \( \ln(1+x) = 0 \), implying \( x = 0 \). Another situation happens when the arguments of the logarithms are zero or negative, leading to potential division by zero or a negative logarithmic value, which are invalid in real-valued functions.

• \( \ln(1-x) \) is undefined for \( x \geq 1 \) thus indicating a concern at \( x = 1 \).
• \( \ln(1+x) \) is undefined for \( x \leq -1 \) which flags an asymptote at \( x = -1 \).

Thus, the vertical asymptotes in this specific function are at \( x = -1, 0, \text{and} 1 \). These are points where the graph will "shoot" up or down dramatically as displayed on its graph.

Horizontal Asymptotes

Horizontal asymptotes are horizontal lines that a graph approaches as \( x \) becomes increasingly large in the positive or negative direction. For the function \( f(x) = \frac{\ln(1-x)}{\ln(1+x)} \), finding horizontal asymptotes involves limit analysis towards infinity.

We calculate \( \lim_{x \to \pm \infty} f(x) \). Both \( \ln(1-x) \) and \( \ln(1+x) \) trend towards infinity at \( x \to \infty\), but their growth rates are equivalent due to their similar logarithmic base, implying a potential factor of simplification between positive infinity of the numerator and the denominator.

• If the function truly reaches a constant value, it forms an asymptote, perhaps a slanted one, rather than purely horizontal.
• In this case, however, \( f(x) \to c \, eq \, 0 \), showing no horizontal asymptote is present. The function does respect a non-zero constant but does not level to a strict horizontal line.

Therefore, the calculation indicates that \( f(x) \) approaches a consistent, constant behavior but not a true horizontal asymptote.

Limit Analysis

Limit analysis helps us understand the behavior of a function as \( x \) approaches critical values or as \( x \) goes to infinity. In assessing \( f(x) = \frac{\ln(1-x)}{\ln(1+x)} \), this involves both vertical and horizontal assessments.

• Vertical asymptotes arise when approaching values result in the denominator equating to zero, causing the function to move infinitely high or low.
• When assessing horizontal tendencies, look at \( \lim_{x \to \pm \infty} \), to evaluate end behavior, seeking constant real limits.

Limit analysis is crucial for functions involving indeterminate or undefined behaviors and is often simplified by techniques like L'Hôpital’s Rule, which handle cases like \( \frac{\infty}{\infty} \) or \( \frac{0}{0} \).

By focusing on limit analysis, one can better perceive where vertical asymptotes form \((x = -1, 0, 1)\), and why no horizontal asymptotes exist even as the function's behavior approaches stability without strictly flattening on a line.