Problem 87 - Calculus Early Transcendentals

Question

Let \(f(x)=\frac{2 e^{x}+10 e^{-x}}{e^{x}+e^{-x}} .\) Analyze \(\lim _{x \rightarrow 0} f(x), \lim _{x \rightarrow-\infty} f(x),\) and \(\lim _{x \rightarrow \infty} f(x) .\) Then give the horizontal and vertical asymptotes of \(f .\) Plot \(f\) to verify your results.

Step-by-Step Solution

Answer

Answer: The limits of the function as \(x\) approaches 0, \(-\infty\), and \(\infty\) are 6, 10, and 2, respectively. The horizontal asymptotes are the lines \(y = 10\) and \(y = 2\).

Step 1: Analyze the limit as \(x \rightarrow 0\)

To find the limit of \(f(x)\) as \(x\) approaches 0, we can substitute \(x\) with 0 and check if the function has a meaning in this case. \(\lim _{x \rightarrow 0} f(x) = \frac{2 e^{0}+10 e^{-0}}{e^{0}+e^{-0}} = \frac{2+10}{1+1} = \frac{12}{2}=6\). So, the limit of \(f(x)\) as \(x\) approaches 0 is 6.

Step 2: Analyze the limit as \(x \rightarrow -\infty\)

To find the limit of \(f(x)\) as \(x\) approaches \(-\infty\), we need to observe the behavior of the function when \(x\) approaches \(-\infty\). As \(x \rightarrow -\infty\), the term \(e^{-x}\) grows very large, and the other terms, \(e^x\), become very small compared to it. Thus, \(\lim _{x \rightarrow -\infty} f(x) = \frac{2 e^{x}+10 e^{-x}}{e^{x}+e^{-x}} \approx \frac{10 e^{-x}}{e^{-x}} = 10\). So, the limit of \(f(x)\) as \(x\) approaches \(-\infty\) is 10.

Step 3: Analyze the limit as \(x \rightarrow \infty\)

To find the limit of \(f(x)\) as \(x\) approaches \(\infty\), we need to observe the behavior of the function when \(x\) approaches \(\infty\). As \(x \rightarrow \infty\), the term \(e^{x}\) grows very large, and the other terms, \(e^{-x}\), become very small compared to it. Thus, \(\lim _{x \rightarrow \infty} f(x) = \frac{2 e^{x}+10 e^{-x}}{e^{x}+e^{-x}} \approx \frac{2 e^{x}}{e^{x}} = 2\). So, the limit of \(f(x)\) as \(x\) approaches \(\infty\) is 2.

Step 4: Determine horizontal and vertical asymptotes

Horizontal asymptotes are the lines parallel to the x-axis that the function approaches as \(x \rightarrow -\infty\) or \(x \rightarrow \infty\). From steps 2 and 3, we already found that the limits as \(x\) approaches \(-\infty\) and \(\infty\) are 10 and 2, respectively. Therefore, the horizontal asymptotes are given by the lines \(y = 10\) and \(y = 2\). There are no vertical asymptotes for this function because the denominator of \(f(x)\), \(e^{x}+e^{-x}\), is never equal to zero.

Step 5: Plot the function to verify the results

After plotting the function, you will see that the function approaches 6 when \(x\) approaches 0, which confirms our result from step 1. The function approaches the line \(y = 10\) as \(x\) approaches \(-\infty\), and approaches the line \(y = 2\) as \(x\) approaches \(\infty\), which confirms our results from steps 2 and 3. Finally, as there are no vertical asymptotes in the graph, our results from step 4 are also verified.

Key Concepts

Horizontal Asymptotes

Horizontal asymptotes are the imaginary lines that a graph of a function gets closer to as it moves infinitely far in either the positive or negative direction along the x-axis. In simpler terms, horizontal asymptotes show us where a function "flattens out." These lines take the form of \(y = c\), where \(c\) is a constant value that the function approaches. In our exercise, after analyzing the limits as \(x\) tends to \(-\infty\) and \(+\infty\), we discovered the horizontal asymptotes for the function \(f(x) = \frac{2 e^{x} + 10 e^{-x}}{e^{x} + e^{-x}}\).

• As \(x \rightarrow -\infty\), the function approaches the line \(y = 10\).
• As \(x \rightarrow \infty\), it approaches the line \(y = 2\).

These results tell us that no matter how large or small \(x\) gets, the function will get closer and closer to these two lines.

Vertical Asymptotes

Vertical asymptotes appear in a function when there is a division by zero and the function can become infinitely large. Unlike horizontal asymptotes that show behavior along the x-axis, vertical asymptotes are like walls the function can't cross on the y-axis. They're typically given by \(x = c\), where \(c\) is a value that causes the denominator to become zero.

In the case of our function, \(f(x) = \frac{2 e^{x} + 10 e^{-x}}{e^{x} + e^{-x}}\), the denominator \(e^{x} + e^{-x}\) does not equal zero for any real value of \(x\). This is because the exponential function \(e^{x}\) is always positive, and thus their sum remains positive as well. Therefore, there are no vertical asymptotes in this function.

Function Analysis

Function analysis involves studying the characteristics and properties of functions to understand their behavior. When analyzing \(f(x) = \frac{2 e^{x} + 10 e^{-x}}{e^{x} + e^{-x}}\), it's crucial to evaluate different limits and see how the function acts under various conditions.

• At \(x = 0\): \(f(x)\) simplifies to \( \frac{12}{2} = 6\), meaning the function reaches an actual point at this x-value.
• As \(x \rightarrow \infty\): The function approaches 2, indicating a flattening near this value as \(x\) grows large.
• As \(x \rightarrow -\infty\): The function gets closer to 10, another sign of behavior where it flattens as \(x\) decreases.

Studying these limits helps highlight the overall behavior and stabilizing points where the function stabilizes.

Exponential Functions

Exponential functions have the form \(f(x) = a \, e^{bx}\), where \(e\) is the base, which is approximately equal to 2.718. These functions model exponential growth or decay and have unique properties affecting how they look on a graph. In the function from our exercise, \(e^{x}\) grows extremely large as \(x \rightarrow \infty\), while \(e^{-x}\) shrinks towards zero. Conversely, as \(x \rightarrow -\infty\), \(e^{x}\) approaches zero, while \(e^{-x}\) expands infinitely.

Understanding the basic properties of exponential functions can help predict how they behave:

• They never touch the x-axis, as their values are always positive.
• They grow or decay at rates proportional to their size, leading to a consistent increase or decrease.

In our given function, the balance between \(e^{x}\) and \(e^{-x}\) is what shapes the overall behavior and limits we've analyzed.