Problem 87 - University Calculus: Early Transcendentals

Question

Use limits to find horizontal asymptotes for each function. $$\text { a. } y=x \tan \left(\frac{1}{x}\right)$$ .$$\text { b. } y=\frac{3 x+e^{2 x}}{2 x+e^{3 x}}$$

Step-by-Step Solution

Answer

a. Horizontal asymptote at y = 0. b. Horizontal asymptotes at y = 0 and y = 3/2.

Step 1: Identify Horizontal Asymptote of Function a

To find the horizontal asymptote for the function \( y = x \tan\left(\frac{1}{x}\right) \), evaluate the limit of the function as \( x \to \infty \) and \( x \to -\infty \).

Step 2: Evaluate Limit as x Approaches Infinity for Function a

As \( x \to \infty \), \( \frac{1}{x} \to 0 \). We know that \( \tan(0) = 0 \). Thus, \( y = x \tan\left(\frac{1}{x}\right) \to x \cdot 0 = 0 \). So, \( \lim_{x \to \infty} x \tan\left(\frac{1}{x}\right) = 0 \).

Step 3: Evaluate Limit as x Approaches Negative Infinity for Function a

As \( x \to -\infty \), \( \frac{1}{x} \to 0 \). Similar to the previous step, \( y = x \tan\left(\frac{1}{x}\right) \to x \cdot 0 = 0 \). Hence, \( \lim_{x \to -\infty} x \tan\left(\frac{1}{x}\right) = 0 \).

Step 4: Conclusion for Function a

The horizontal asymptote for \( y = x \tan\left(\frac{1}{x}\right) \) is \( y = 0 \).

Step 4: Identify Horizontal Asymptote of Function b

To find horizontal asymptotes for \( y = \frac{3x + e^{2x}}{2x + e^{3x}} \), evaluate the limit of the function as \( x \to \infty \) and \( x \to -\infty \).

Step 5: Evaluate Limit as x Approaches Infinity for Function b

As \( x \to \infty \), \( e^{2x} \) and \( e^{3x} \) grow significantly faster than \( 3x \) and \( 2x \). Hence, the limit is dominated by \( \frac{e^{2x}}{e^{3x}} = e^{-x} \to 0 \). Thus, \( \lim_{x \to \infty} \frac{3x + e^{2x}}{2x + e^{3x}} = 0 \).

Step 6: Evaluate Limit as x Approaches Negative Infinity for Function b

As \( x \to -\infty \), both terms \( e^{2x} \) and \( e^{3x} \) exponentially approach 0, and the polynomial terms dominate. Therefore, \( \frac{3x}{2x} = \frac{3}{2} \). Thus, \( \lim_{x \to -\infty} \frac{3x + e^{2x}}{2x + e^{3x}} = \frac{3}{2} \).

Step 8: Conclusion for Function b

The function \( y = \frac{3x + e^{2x}}{2x + e^{3x}} \) has horizontal asymptotes at \( y = 0 \) as \( x \to \infty \) and \( y = \frac{3}{2} \) as \( x \to -\infty \).

Key Concepts

Limits

The concept of limits is foundational in calculus and essential when discussing horizontal asymptotes. A limit helps us understand the behavior of a function as it approaches a specific point or infinity. When determining horizontal asymptotes, we look at the limit of a function as it approaches positive or negative infinity.
- When we say, "as \( x \to \infty \)," we refer to the behavior of the function as \( x \) becomes very large.- Conversely, "as \( x \to -\infty \)" means analyzing the function as \( x \) becomes very negative.By finding the limits as \( x \to \infty \) and \( x \to -\infty \), we can identify the horizontal asymptotes of a function. These limits reveal the values that the function approaches, helping to define its end behavior.

Horizontal Asymptotes

Horizontal asymptotes are horizontal lines which represent the value a function approaches as the independent variable (usually \( x \)) goes to positive or negative infinity. Understanding horizontal asymptotes involves analyzing the limits of a function at infinity.When a function has a horizontal asymptote, it means the output gets closer and closer to a specific value without ever actually reaching it, as \( x \) becomes very large (positively or negatively).
In the exercise, for function \( y = x \tan\left(\frac{1}{x}\right) \), the horizontal asymptote is \( y = 0 \), because:- As \( x \to \infty \), the function approaches \( 0 \).- Similarly, as \( x \to -\infty \), the function also approaches \( 0 \).For the function \( y = \frac{3x + e^{2x}}{2x + e^{3x}} \):- As \( x \to \infty \), the horizontal asymptote is \( y = 0 \).- As \( x \to -\infty \), the asymptote is \( y = \frac{3}{2} \). This happens because the polynomial terms dominate when \( x \) is very negative, contrasting with the exponential dominance when \( x \) is very large.

Exponential Functions

Exponential functions are functions where the variable is in the exponent. They often have the form \( f(x) = e^x \), where \( e \) is Euler's number, approximately equal to 2.718. Exponential functions grow or decay quickly and are prevalent in real-world applications like finance, science, and engineering.When analyzing the limits and horizontal asymptotes of exponential functions, it’s important to realize:- As \( x \to \infty \), an exponential function like \( e^{x} \) increases rapidly.- Conversely, as \( x \to -\infty \), \( e^{x} \) approaches zero quickly.In our original exercise, the function \( y = \frac{3x + e^{2x}}{2x + e^{3x}} \) combines exponential and polynomial terms. Understanding the growth rates is crucial:- For \( x \to \infty \), the terms involving \( e^{2x} \) and \( e^{3x} \) become dominant, leading to the asymptote \( y = 0 \).- For \( x \to -\infty \), the exponential terms vanish, allowing the polynomial dominance, which yields the asymptote \( y = \frac{3}{2} \). These properties make exponential functions particularly interesting to study, especially in the context of calculus and asymptotic behavior.