Problem 46 - Calculus: Early Transcendentals

Question

Horizontal asymptotes Determine \(\lim _{x \rightarrow \infty} f(x)\) and \(\lim _{x \rightarrow-\infty} f(x)\) for the following functions. Then give the horizontal asymptotes of \(f(\text {if any})\). $$f(x)=\frac{\sqrt{x^{2}+1}}{2 x+1}$$

Step-by-Step Solution

Answer

In conclusion, the function \(f(x)=\frac{\sqrt{x^{2}+1}}{2 x+1}\) has horizontal asymptotes at \(y=0\) and \(y=1\). The limit as \(x \rightarrow \infty\) is equal to 0, and the limit as \(x \rightarrow -\infty\) is equal to 1.

Step 1: Determine \(\lim_{x \rightarrow \infty} f(x)\)

To find the limit as x approaches infinity, first, we consider the function: $$f(x)=\frac{\sqrt{x^{2}+1}}{2 x+1}$$ Let's perform algebraic manipulation by multiplying the numerator and the denominator by the conjugate of the numerator. $$\lim_{x \rightarrow \infty} f(x)=\lim_{x \rightarrow \infty} \frac{\sqrt{x^{2}+1}}{2 x+1} \cdot \frac{\sqrt{x^{2}+1} - x}{\sqrt{x^{2}+1} - x}$$ Now, perform the multiplication and simplify the expression: $$\lim_{x \rightarrow \infty} \frac{x^2 + 1 - x\sqrt{x^{2}+1}}{(2x + 1)(\sqrt{x^{2}+1} - x)}$$ Divide both the numerator and the denominator by \(x^2\) and simplify: $$\lim_{x \rightarrow \infty} \frac{1 + \frac{1}{x^2} - \frac{\sqrt{x^{2}+1}}{x}}{2 + \frac{1}{x} - \sqrt{1+\frac{1}{x^2}}}$$ Now as x approaches infinity, we can substitute \(\infty\) into the expression to obtain the limit: $$\lim_{x \rightarrow \infty} f(x) = \frac{1 + 0 - 1}{2 - 0} = 0$$

Step 2: Determine \(\lim_{x \rightarrow -\infty} f(x)\)

Similarly, we can find the limit as x approaches negative infinity. We already have the simplified expression: $$\lim_{x \rightarrow -\infty} f(x)=\lim_{x \rightarrow -\infty} \frac{1 + \frac{1}{x^2} - \frac{\sqrt{x^{2}+1}}{x}}{2 + \frac{1}{x} - \sqrt{1+\frac{1}{x^2}}}$$ Now as x approaches negative infinity, we can substitute \(-\infty\) into the expression to obtain the limit: $$\lim_{x \rightarrow -\infty} f(x) = \frac{1 + 0 - \frac{-1}{-1}}{-2 - 0} = 1$$

Step 3: Identify the horizontal asymptotes

We have found the limits of f(x) as x approaches both positive and negative infinity. To identify the horizontal asymptotes, we look at the limits as \(x \rightarrow \infty\) and \(x \rightarrow -\infty\). For \(\lim_{x \rightarrow \infty} f(x) = 0\), the horizontal asymptote is \(y=0\). For \(\lim_{x \rightarrow -\infty} f(x) = 1\), the horizontal asymptote is \(y=1\). Thus, there are horizontal asymptotes at \(y=0\) and \(y=1\).

Key Concepts

Limits

In calculus, limits help us understand the behavior of a function as the input approaches a specific value or even infinity. When dealing with horizontal asymptotes, we're interested in limits as the variable approaches positive or negative infinity.

Consider the function given in the original exercise:

• \(f(x)=\frac{\sqrt{x^{2}+1}}{2 x+1}\)

When solving for horizontal asymptotes, we calculate \(\lim_{x \rightarrow \infty} f(x)\) and \(\lim_{x \rightarrow -\infty} f(x)\). These limits provide us with the horizontal asymptotic values of the function, telling us the y-values the function approaches as \(x\) moves towards positive or negative infinity.

For \(\lim_{x \rightarrow \infty} f(x) = 0\), the function approaches a horizontal line at \(y = 0\). Similarly, for \(\lim_{x \rightarrow -\infty} f(x) = 1\), the function approaches another horizontal line at \(y = 1\). Understanding these limits is crucial for identifying the behavior of functions at their extremes.

Infinity

Infinity is a concept that describes something without any limit, often used to describe a number so large it cannot be quantified. In calculus, we encounter infinity when evaluating the behavior of functions as variables move towards extremely large or extremely small values.

In the exercise, infinity is involved when we look at the limits of \(f(x)\) as \(x\) approaches \(\infty\) or \(-\infty\). Here's why this matters:

• As \(x\) approaches \(\infty\), we examine what happens to \(f(x)\). The function's components grow, and some diminish, showing us long-term tendencies of the function.
• As \(x\) approaches \(-\infty\), we do the same analysis, helping us to see how the negative extremes of \(x\) influence \(f(x)\). This is essential for finding horizontal asymptotes, where the function levels off.

Infinity allows us to view the function's end behavior and determine its horizontal limits, a key aspect of sketching an accurate graph.

Algebraic Manipulation

Algebraic manipulation is a powerful tool in simplifying complex expressions, allowing us to approach limits more easily. In the context of this exercise, we used algebraic manipulation to simplify the function: \(f(x)=\frac{\sqrt{x^{2}+1}}{2 x+1}\).

Here's the step-by-step breakdown:

• First, we multiply by the conjugate of the numerator over itself. This involves using \(\frac{\sqrt{x^{2}+1} - x}{\sqrt{x^{2}+1} - x}\) to simplify the numerator \(\sqrt{x^{2}+1}+x\).
• After multiplication, we divide each term by \(x^2\), streamlining the expression for evaluating limits toward infinity or negative infinity.
• This manipulation reduces complexity, highlighting only significant components as \(x\) grows large or small.

By leveraging these algebraic tricks, we clarify how the function behaves at its extremes, making it easier to identify asymptotic properties.

Calculus

Calculus is the branch of mathematics that deals with rates of change and accumulation. In the context of finding horizontal asymptotes, calculus helps us use limits to describe the behavior of functions at infinite bounds.

To determine horizontal asymptotes:

• We apply limits, a fundamental concept in calculus, to identify the values \(f(x)\) approaches as \(x\) grows infinitely positive or negative.
• Using derivatives, though not required for this problem, often helps find short-term behaviors. Limits guide understanding of long-term trends, essential for asymptotes.
• Calculus principles facilitate the understanding of how functions grow, curve, and ultimately "flatten out" at their asymptotes.

By integrating calculus with algebraic techniques, we solve limits involving infinity, providing insight into the end behavior of functions. This makes calculus indispensable when examining asymptotic trends.