Problem 284 - Calculus Volume 1

Question

Find the horizontal and vertical asymptotes. \(\quad f(x)=\frac{1}{x}-\sqrt{x}\)

Step-by-Step Solution

Answer

Vertical asymptote at \(x = 0\); no horizontal asymptote.

Step 1: Identify discontinuities

The function \(f(x) = \frac{1}{x} - \sqrt{x}\) consists of two terms: \(\frac{1}{x}\) and \(-\sqrt{x}\). Let's identify where each term becomes problematic. The term \(\frac{1}{x}\) is undefined at \(x = 0\) and \(-\sqrt{x}\) is defined for \(x \geq 0\). Thus, \(f(x)\) is undefined at \(x < 0\) and it is essential to investigate further where potential asymptotes could exist.

Step 2: Find vertical asymptote

A vertical asymptote occurs where \(f(x)\) is undefined and near-infinite in value. The term \(\frac{1}{x}\) leads to a vertical asymptote at \(x = 0\). Since this is the only point where the function is undefined within its domain (\(x \geq 0\)), a vertical asymptote is found at \(x = 0\).

Step 3: Analyze horizontal behavior

To determine horizontal asymptotes, we evaluate the limits of \(f(x) = \frac{1}{x} - \sqrt{x}\) as \(x\) approaches infinity. We check \(\lim_{x \to \infty} \frac{1}{x}\) and \(\lim_{x \to \infty} \sqrt{x}\) separately. As \(x\) tends to infinity, \(\frac{1}{x} \to 0\) and \(\sqrt{x} \to \infty\), making \(f(x)\) approach negative infinity. Since \(f(x)\) does not approach a finite number, there is no horizontal asymptote as \(x \to \infty\).

Step 4: Confirm the asymptotes

From the calculations above, we know that the function has a vertical asymptote at \(x = 0\) and no horizontal asymptote exists since \(f(x)\) does not converge to a finite value as \(x \to \infty\).

Key Concepts

Vertical Asymptote

A vertical asymptote occurs at points where a function becomes undefined and its values tend to infinity. In simpler terms, it's where the graph shoots up towards infinity or dives down to negative infinity.

In the function \( f(x) = \frac{1}{x} - \sqrt{x} \), the term \( \frac{1}{x} \) creates a vertical asymptote at \( x = 0 \). Why? Because dividing by zero is undefined. As \( x \) gets closer and closer to 0 from the positive side, \( \frac{1}{x} \) increases without bound towards infinity.

Simply put, you can't plug in \( x = 0 \) into the function because you'd be dividing by zero, creating this vertical boundary in the function's graph. This is why the vertical asymptote is at \( x = 0 \).

Horizontal Asymptote

Horizontal asymptotes indicate the behavior of a function as \( x \) approaches infinity or negative infinity. Essentially, it tells us what value \( f(x) \) is approaching as \( x \) gets very large or very small.

In our function \( f(x) = \frac{1}{x} - \sqrt{x} \), let's explore the limits of its components as \( x \) tends to infinity:

• The \( \frac{1}{x} \) term tends towards 0 because as you divide 1 by a larger number, the result gets closer and closer to zero.
• The \( -\sqrt{x} \) term, however, continues to increase negatively without bound, moving towards negative infinity.

Thus, \( f(x) \) itself tends towards negative infinity as \( x \to \infty \). This means that \( f(x) \) does not approach a specific finite number, leading to the conclusion that there is no horizontal asymptote.

Limits

Limits help us understand the behavior of a function as the input values get closer to a certain point, or become very large. They are essential in identifying both vertical and horizontal asymptotes.

Consider the function \( f(x) = \frac{1}{x} - \sqrt{x} \):

• For vertical asymptotes, we look at what happens when \( x \) is close to any point where the function is undefined, such as \( x = 0 \). Here, \( \frac{1}{x} \) becomes very large as it approaches this point, defining the vertical asymptote.
• For horizontal asymptotes, we examine the limit as \( x \to \infty \). We found that \( \frac{1}{x} \to 0 \) and \( -\sqrt{x} \to -\infty \), so overall \( f(x) \to -\infty \).

By understanding limits, we can see whether a function approaches a specific value or tends to infinity, aiding in the identification of asymptotic behavior.