Problem 49 - Calculus with Concepts in Calculus

Question

Let $$ f(x)=\frac{\sqrt{2+4 x^{2}}}{x} $$ Show that the graph of \(f\) has two horizontal asymptotes, and determine them.

Step-by-Step Solution

Answer

The function has two horizontal asymptotes at y=2.

Step 1: Understand Horizontal Asymptotes

A horizontal asymptote occurs when the function approaches a certain value as \(x\) tends to \(\infty\) or \(-\infty\). To find this, we need to evaluate the limits of \(f(x)\) as \(x\) approaches \(\infty\) and \(-\infty\).

Step 2: Simplify the Function for Asymptotic Behavior

To analyze the asymptotic behavior of \(f(x) = \frac{\sqrt{2+4x^2}}{x}\), we first focus on the expression inside the square root. \(\sqrt{2 + 4x^2}\) can be rewritten as \(\sqrt{4x^2(\frac{1}{x^2} + \frac{1}{2x^2})}\), which simplifies to \(2x\sqrt{1 + \frac{1}{2x^2}}\). This allows us to approximate \(f(x)\) as \(\frac{2x\sqrt{1 + \frac{1}{2x^2}}}{x}\) = \(2\sqrt{1 + \frac{1}{2x^2}}\).

Step 3: Determine the Horizontal Asymptote for x -> ∞

As \(x\to\infty\), \(\frac{1}{2x^2} \to 0\), so \(\sqrt{1 + \frac{1}{2x^2}} \to \sqrt{1} = 1\). Therefore, \(f(x) \to 2\cdot 1 = 2\), indicating that \(y=2\) is a horizontal asymptote.

Step 4: Determine the Horizontal Asymptote for x -> -∞

Similarly, as \(x \to -\infty\), \(\frac{1}{2x^2} \to 0\) again, and \(f(x)\) simplifies in a similar manner as for \(x \to \infty\). Hence, \(f(x) \to 2\), implying that \(y=2\) is also the horizontal asymptote as \(x\to -\infty\).

Key Concepts

Limits at Infinity

When we talk about limits at infinity, we mean examining what happens to a function's value as the input, or "x," grows extremely large or extremely negative. For the function \( f(x) = \frac{\sqrt{2+4x^2}}{x} \), understanding its behavior as \( x \to \infty \) and \( x \to -\infty \) is crucial to finding the horizontal asymptotes.
To calculate these, we evaluate the limit of \( f(x) \) as \( x \) approaches infinity in both positive and negative directions. These evaluations help us determine if the function approaches a constant y-value at the extremes of its domain.
In this context, limits at infinity are our main tool in confirming the presence of horizontal asymptotes.

Asymptotic Behavior

The concept of asymptotic behavior is about understanding how functions behave as they extend toward infinity or negative infinity. For many functions, like our given \( f(x) \), they tend to hover around or reach a horizontal asymptote.
Horizontal asymptotes at \( y = 2 \) for both \( x \to \infty \) and \( x \to -\infty \) signify that as \( x \) becomes very large or very small, the value of \( f(x) \) approaches 2. This means that the function's graph gets closer to the line \( y = 2 \) but doesn't cross or touch it, especially at very large or very small values of \( x \).
It’s important to realize that while curves can have multiple vertical asymptotes, they usually only have at most two horizontal ones, as is the case here.

Simplification of Functions

Simplifying a function is key to spotting the asymptotic behavior and determining horizontal asymptotes. By simplifying \( f(x) = \frac{\sqrt{2+4x^2}}{x} \), you can make complex expressions easier to manage, which is instrumental in predicting the end behaviors.
In this exercise, rewriting \( \sqrt{2+4x^2} \) as \( 2x\sqrt{1+\frac{1}{2x^2}} \) allows us to see that for very large \( x \) values, \( \frac{1}{2x^2} \) becomes negligible, leading to the approximation \( 2\sqrt{1+\frac{1}{2x^2}} \approx 2 \).
Simplification not only aids in computation but also clarifies the function’s potential limits and asymptotic tendencies, particularly for functions expressed in terms of radicals and other complex structures.