Problem 277 - Calculus Volume 1

Question

For the following exercises, find the horizontal and vertical asymptotes. $$ f(x)=\frac{x \sin (x)}{x^{2}-1} $$

Step-by-Step Solution

Answer

Vertical asymptotes at \( x = 1 \) and \( x = -1 \); horizontal asymptote at \( y = 0 \).

Step 1: Understanding Asymptotes

Horizontal asymptotes are found by considering the behavior of the function as \( x \to \pm \infty \). Vertical asymptotes occur where the function is undefined, usually where the denominator is zero.

Step 2: Determine Vertical Asymptotes

To find vertical asymptotes, set the denominator equal to zero and solve for \( x \): \( x^2 - 1 = 0 \). This gives \( x = \pm 1 \). Therefore, there are potential vertical asymptotes at \( x = 1 \) and \( x = -1 \). We must check if the numerator, \( x \sin(x) \), is not zero at these points.

Step 3: Check the Numerator at Vertical Asymptote Points

Evaluate the numerator at \( x = 1 \) and \( x = -1 \): \( x \sin(x) \). Both \( 1 \sin(1) \) and \( -1 \sin(-1) \) are non-zero. Therefore, vertical asymptotes exist at \( x = 1 \) and \( x = -1 \).

Step 4: Determine Horizontal Asymptotes

Examine \( \lim_{x \to \infty} \frac{x \sin(x)}{x^2 - 1} \). As \( x \to \infty \), \( \sin(x) \) oscillates between -1 and 1, while \( x^2 - 1 \to \infty \). Therefore, \( \lim_{x \to \infty} \frac{x \sin(x)}{x^2 - 1} = 0 \). Similarly, for \( x \to -\infty \), \( \lim_{x \to -\infty} \frac{x \sin(x)}{x^2 - 1} = 0 \). Thus, there is a horizontal asymptote at \( y = 0 \).

Key Concepts

Horizontal Asymptotes

When we talk about horizontal asymptotes, we are looking at the behavior of a function as it moves towards the extreme ends of the x-axis, both positive and negative infinity. Imagine you are stretching the x-axis infinitely in both directions. At some point, you want to know how the function behaves. Does it level off or approach a certain line? This is where horizontal asymptotes come into play. A horizontal asymptote is a horizontal line that the graph of a function approaches as the absolute value of x grows larger. Mathematically, we find horizontal asymptotes by taking the limit of the function as x approaches infinity, both positive and negative:

• If \( \lim_{x \to \infty} f(x) = L \), the horizontal asymptote is \( y = L \).
• If \( \lim_{x \to -\infty} f(x) = L \), again, the horizontal asymptote is \( y = L \).

For the function \( f(x) = \frac{x \sin(x)}{x^2 - 1} \), as x approaches either positive or negative infinity, the term \( \sin(x) \) continues to oscillate between -1 and 1. Meanwhile, the denominator, \( x^2 - 1 \), grows very large positive. This causes the value of the fraction to squeeze closer to zero. Hence, the horizontal asymptote for this function is \( y = 0 \). This tells us that as x extends towards infinity or negative infinity, the function's output values get very close to zero.

Vertical Asymptotes

Vertical asymptotes are quite different from horizontal ones. They occur where a function tends to infinity as it gets closer to a certain x-value, often where the function is not defined. Usually, this involves zeros in the denominator of a rational function, which cause the function to "break" or "blow up." To find vertical asymptotes, set the denominator equal to zero and solve for x. This gives the potential points where the function may become undefined. For the function \( f(x) = \frac{x \sin(x)}{x^2 - 1} \):

• Set \( x^2 - 1 = 0 \).
• Solve to find \( x = \pm 1 \).

These are the x-values where the denominator becomes zero. Next, check the numerator at these points. Evaluate \( x \sin(x) \) at \( x = 1 \) and \( x = -1 \), ensuring it's non-zero:

• \( 1 \sin(1) \) and \( -1 \sin(-1) \) do not equal zero.

Thus, the function cannot be simplified at \( x = 1 \) or \( x = -1 \), making these true vertical asymptotes. As the x-values get closer to 1 or -1, the function shoots off towards infinity, solidifying the presence of vertical asymptotes at these points.

Limits and Infinity

Understanding limits and infinity is crucial when exploring asymptotic behavior. Limits help us capture the behavior of a function as it approaches a particular x-value or as x stretches out towards infinity. The idea of a limit captures both the essence and behavior of a function near certain points or as it moves out to distant values. When you see notation like \( \lim_{x \to a} f(x) \), it means we observe what value \( f(x) \) nears as x approaches a particular number \( a \). Similarly, \( \lim_{x \to \infty} f(x) \) investigates what \( f(x) \) approaches as x grows large positively or negatively.
In the context of the given function \( f(x) = \frac{x \sin(x)}{x^2 - 1} \), we applied limits to determine how the function behaves at infinity, hence establishing the horizontal asymptote. We considered:

• \( \lim_{x \to \infty} \frac{x \sin(x)}{x^2 - 1} = 0 \)
• \( \lim_{x \to -\infty} \frac{x \sin(x)}{x^2 - 1} = 0 \)

Both limits indicate that the function values approach zero as the x-values spread wide, regardless of the function's oscillating numerator term.