Problem 59 - Calculus: Early Transcendentals

Question

A function \( f \) is a ratio of quadratic functions and has a vertical asymptote \( x = 4 \) and just one \( x \)-intercept, \( x = 1 \). It is known that \( f \) has a removable discontinuity at \( x = -1 \) and \( \displaystyle \lim_{x \to -1} f(x) = 2 \). Evaluate (a) \( f (0) \) (b) \( \displaystyle \lim_{x \to \infty} f(x) \)

Step-by-Step Solution

Answer

(a) \( f(0) = \frac{5}{8} \); (b) \( \lim_{x \to \infty} f(x) = \frac{5}{2} \).

Step 1: Understanding the Function Structure

Since the function \( f(x) \) is a ratio of quadratic functions, let \( f(x) = \frac{P(x)}{Q(x)} \), where both \( P(x) \) and \( Q(x) \) are polynomials of degree 2. Given the vertical asymptote at \( x = 4 \), \( Q(x) \) must have a factor of \( (x-4) \). The removable discontinuity at \( x = -1 \) implies both \( P(x) \) and \( Q(x) \) have a factor of \( (x+1) \). The \( x \)-intercept at \( x = 1 \) means \( P(x) \) has a factor of \( (x-1) \). Therefore, \( P(x) = a(x-1)(x+1) \) and \( Q(x) = b(x-4)(x+1) \).

Step 2: Constructing the Complete Function

We need to use the given limit condition \( \displaystyle \lim_{x \to -1} f(x) = 2 \) to find the constants \( a \) and \( b \). Substitute \( P(x) \) and \( Q(x) \) into \( f(x) = \frac{a(x-1)(x+1)}{b(x-4)(x+1)} \). Cancel the \( (x+1) \) from the numerator and denominator to get \( f(x) = \frac{a(x-1)}{b(x-4)} \). Evaluate \( \lim_{x \to -1} \frac{a(x-1)}{b(x-4)} = 2 \) using substitution, which gives \( \frac{-2a}{-5b} = 2 \), resulting in \( \frac{2a}{5b} = 2 \). Simplify to find \( a: b = 5:2 \). Assume they are 5 and 2 for simplicity.

Step 3: Evaluating \( f(0) \)

Now, substitute \( a = 5 \) and \( b = 2 \) into the function to get \( f(x) = \frac{5(x-1)}{2(x-4)} \). Evaluate \( f(0) \):\[ f(0) = \frac{5(0-1)}{2(0-4)} = \frac{-5}{-8} = \frac{5}{8} \].

Step 4: Finding \( \lim_{x \to \infty} f(x) \)

To find \( \lim_{x \to \infty} f(x) \), use the fact that for rational functions where the degrees of numerator and denominator are the same, the horizontal asymptote is the ratio of leading coefficients. Here, both \( P(x) \) and \( Q(x) \) have a leading coefficient of 5 and 2, respectively. Hence, the limit is:\[ \lim_{x \to \infty} f(x) = \frac{5}{2} \].

Key Concepts

Vertical Asymptote

A vertical asymptote in a rational function is essentially a line where the function grows without bound, approaching infinity or negative infinity. It occurs where the denominator of the function equals zero, but the numerator does not, causing the function to "shoot up" to infinity or "drop down" to negative infinity. In the function given in the exercise, the vertical asymptote is at

• \( x = 4 \)

This means that as \( x \) gets closer and closer to 4, the value of the function grows very large either positively or negatively.

This is because the factor \( (x-4) \) in the denominator causes it to approach zero, thereby making the quotient very large. The numerator doesn't equal zero at this point, which is what makes the function undefined and results in the asymptotic behavior.

Removable Discontinuity

A removable discontinuity in a rational function occurs when both the numerator and the denominator share a common factor that can be canceled. This creates a hole in the graph rather than a vertical line. In the context of the problem, both \(

• P(x)\) and \(Q(x)\)

... have a common factor of \( (x+1) \). Because we can divide \((x+1)\) out of both the numerator and the denominator, the discontinuity "disappears" except at the exact point of \(x = -1\).

Despite being undefined at this specific point, the rest of the function behaves normally. The given information that \( \lim_{x \to -1} f(x) = 2 \) tells us the value the function "wants" to be at this point. Therefore, the discontinuity is considered "removable."

Limits

Limits are a fundamental concept in calculus that describe the behavior of a function as it approaches a specific point or infinity. They help us understand and define points where the function might be undefined. In the provided exercise, two limits are considered:

• \(\lim_{x \to -1} f(x) = 2\)
• \(\lim_{x \to \infty} f(x) = \frac{5}{2}\)

The first limit tells us that as \(x\) approaches -1, the function tries to get close to the value 2, even though it can't actually reach a defined value there due to the removable discontinuity.

The second limit describes the behavior of the function as \(x\) moves far to the left or right, approaching infinity. In this scenario, since both the numerator and denominator are quadratic, the function approaches the ratio of their leading coefficients. Thus, the function stabilizes towards the value \(\frac{5}{2}\), which is the horizontal asymptote.