Problem 36 - Calculus for Biology and Medicine

Question

Let $$ f(x)=-\frac{2}{x^{2}-1}, \quad x \neq-1,1 $$ (a) Show that $$ \lim _{x \rightarrow+\infty} f(x)=0 $$ and $$ \lim _{x \rightarrow-\infty} f(x)=0 $$ That is, show that \(y=0\) is a horizontal asymptote of \(f(x)\). (b) Show that $$ \lim _{x \rightarrow-1^{-}} f(x)=-\infty $$ and $$ \lim _{x \rightarrow-1^{+}} f(x)=+\infty $$ and that $$ \lim _{x \rightarrow 1^{-}} f(x)=+\infty $$ and $$ \lim _{x \rightarrow 1^{+}} f(x)=-\infty $$ That is, show that \(x=-1\) and \(x=1\) are vertical asymptotes of \(f(x)\)

Step-by-Step Solution

Answer

The function has a horizontal asymptote at \( y = 0 \) and vertical asymptotes at \( x = -1 \) and \( x = 1 \).

Step 1: Show the Horizontal Asymptote for Positive Infinity

To find \( \lim_{x \to +\infty} f(x) \), substitute \( f(x) = -\frac{2}{x^2 - 1} \). As \( x \to +\infty \), \( x^2 \) becomes very large, so the expression \( x^2 - 1 \approx x^2 \). Thus, \( -\frac{2}{x^2 - 1} \approx -\frac{2}{x^2} \), which approaches 0. Therefore, \( \lim_{x \to +\infty} f(x) = 0 \).

Step 2: Show the Horizontal Asymptote for Negative Infinity

Similarly, for \( \lim_{x \to -\infty} f(x) \), again substitute \( f(x) = -\frac{2}{x^2 - 1} \). As \( x \to -\infty \), \( x^2 \) is still very large, making \( x^2 - 1 \approx x^2 \). Therefore, \( -\frac{2}{x^2 - 1} \approx -\frac{2}{x^2} \), which goes to 0. Accordingly, \( \lim_{x \to -\infty} f(x) = 0 \).

Step 3: Identify Vertical Asymptote at x = -1 from the Left

For \( \lim_{x \to -1^{-}} f(x) \), as \( x \) approaches \( -1 \) from the left, \( x^2 - 1 \to 0^+ \), causing the denominator to become very small positive. Therefore, \( f(x) = -\frac{2}{x^2 - 1} \to -\infty \). Thus, \( \lim_{x \to -1^{-}} f(x) = -\infty \).

Step 4: Identify Vertical Asymptote at x = -1 from the Right

For \( \lim_{x \to -1^{+}} f(x) \), as \( x \) approaches \( -1 \) from the right, \( x^2 - 1 \to 0^- \). The denominator becomes very small negative, and hence, \( f(x) = -\frac{2}{x^2 - 1} \to +\infty \). Thus, \( \lim_{x \to -1^{+}} f(x) = +\infty \).

Step 5: Identify Vertical Asymptote at x = 1 from the Left

For \( \lim_{x \to 1^{-}} f(x) \), as \( x \) approaches \( 1 \) from the left, \( x^2 - 1 \to 0^- \). The denominator becomes very small negative, hence \( f(x) = -\frac{2}{x^2 - 1} \to +\infty \). Thus, \( \lim_{x \to 1^{-}} f(x) = +\infty \).

Step 6: Identify Vertical Asymptote at x = 1 from the Right

For \( \lim_{x \to 1^{+}} f(x) \), as \( x \) approaches \( 1 \) from the right, \( x^2 - 1 \to 0^+ \). The denominator becomes very small positive, hence \( f(x) = -\frac{2}{x^2 - 1} \to -\infty \). Thus, \( \lim_{x \to 1^{+}} f(x) = -\infty \).

Key Concepts

Vertical Asymptote

Vertical asymptotes occur at the values of \( x \) where a rational function has division by zero, leading to the function approaching infinity or negative infinity. These are the points where the function is undefined due to the denominator being equal to zero.

• For the function \( f(x) = -\frac{2}{x^2 - 1} \), the denominator \( x^2 - 1 = 0 \) when \( x = \pm 1 \).
• Thus, we have potential vertical asymptotes at \( x = -1 \) and \( x = 1 \).

Approaching \( x = -1 \) or \( x = 1 \) from different sides gives us different limit behaviors. As you approach these x-values:

• If the denominator approaches zero from the positive side (tiny positive values), the function heads towards \(+\infty\).
• If it approaches from the negative side (tiny negative values), the function heads towards \(-\infty\).

This analysis confirms vertical asymptotes at both \( x = -1 \) and \( x = 1 \).

Limits

Limits help us understand the behavior of functions as input values approach a particular point, and it's fundamental in describing both vertical and horizontal asymptotes.

• The concept of limits is powerful because it tells us what happens even when a function is not explicitly defined at certain points, such as where we have division by zero in rational functions.
• For examining horizontal asymptotes, we look at the limits as \( x \) approaches \( \pm\infty \). These limits show the end behavior of the function.
• With vertical asymptotes, the limits help describe how the function behaves near certain x-values where it might shoot up or down indefinitely.

Calculating limits involves understanding how the numerator and denominator change in size relative to each other, which can be simplified using polynomial long division or recognizing growth patterns.
By calculating limits around the troublesome points (where denominator is nearly zero), we understand the nature of asymptotes in rational functions and how these functions behave infinitely close to those points.

Rational Functions

Rational functions are ratios of polynomials. They have applications ranging from physics to economics due to their simple yet versatile form involving division.

• A typical rational function is of the form \( \frac{P(x)}{Q(x)} \), where \( P(x) \) and \( Q(x) \) are polynomial functions.
• The vertical asymptotes are found at points where \( Q(x) = 0 \) provided \( P(x) \) is not also zero, as this would cause undefined values.
• Horizontal asymptotes describe the function's behavior as \( x \) approaches large positive or negative values, often determined by the degrees of \( P(x) \) and \( Q(x) \).

For instance, if the degree of \( Q(x) \) (denominator) is greater than the degree of \( P(x) \) (numerator), the horizontal asymptote is \( y = 0 \).

• This happens in \( f(x) = -\frac{2}{x^2 - 1} \), establishing a horizontal asymptote at \( y = 0 \).

Understanding these properties allows us to predict function behavior across domains and detect potential issues such as asymptotes or discontinuities, making rational functions essential in mathematical analysis.