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.

Short answer

Question: Using the analysis provided, determine the horizontal and vertical asymptotes of the function \(f(x) = \frac{2 e^{x}+10 e^{-x}}{e^{x}+e^{-x}}\). Answer: The horizontal asymptotes of the function are \(y = 10\) and \(y = 2\). There are no vertical asymptotes for this function.

Step-by-step solution

Find the limit as x approaches 0

To find the limit of f(x) as x approaches 0, we simply plug in x = 0 into the function f(x): \(\lim_{x\rightarrow 0} f(x)=\lim_{x\rightarrow 0} \frac{2 e^{x}+10 e^{-x}}{e^{x}+e^{-x}}\) By plugging x = 0, we have \(\lim_{x\rightarrow 0} f(x)= \frac{2 e^{0}+10 e^{0}}{e^{0}+e^{0}}\) which simplifies to \(\lim_{x\rightarrow 0} f(x)= \frac{2+10}{1+1} = \frac{12}{2} =6\)

Find the limit as x approaches negative infinity

To find the limit of f(x) as x approaches negative infinity, we rewrite the function using a substitution of variables. Let \(u = -x\), thus as x approaches negative infinity, u approaches positive infinity: \(\lim_{x\rightarrow -\infty} f(x)=\lim_{u\rightarrow \infty} \frac{2 e^{-u}+10 e^{u}}{e^{-u}+e^{u}}\) Now, divide both the numerator and the denominator by \(e^u\), we get \(\lim_{u\rightarrow \infty} \frac{2 e^{-u}+10 e^{u}}{e^{-u}+e^{u}} = \lim_{u\rightarrow \infty} \frac{2 e^{-2u}+10 }{e^{-2u}+1}\) As \(u \rightarrow \infty\), \(e^{-2u} \rightarrow 0\), thus: \(\lim_{x\rightarrow -\infty} f(x) = \frac{0+10}{0+1} =10\)

Find the limit as x approaches positive infinity

Similarly, as x approaches positive infinity, we can rewrite the function as follows: \(\lim_{x\rightarrow \infty} f(x) = \lim_{x\rightarrow \infty} \frac{2 e^{x}+10 e^{-x}}{e^{x}+e^{-x}}\) Divide both the numerator and the denominator by \(e^x\), we have \(\lim_{x\rightarrow \infty} \frac{2 e^{x}+10 e^{-x}}{e^{x}+e^{-x}} = \lim_{x\rightarrow \infty} \frac{2 +10 e^{-2x}}{1+e^{-2x}}\) As \(x \rightarrow \infty\), \(e^{-2x} \rightarrow 0\), thus: \(\lim_{x\rightarrow \infty} f(x) = \frac{2+0}{1+0} = 2\)

Determine the horizontal and vertical asymptotes

Based on the limit analysis in Steps 1-3, we can determine the horizontal and vertical asymptotes of f(x): Horizontal Asymptotes: As \(x \rightarrow -\infty\), \(f(x) \rightarrow 10\), so the horizontal asymptote is \(y = 10\). As \(x \rightarrow \infty\), \(f(x) \rightarrow 2\), so the horizontal asymptote is \(y = 2\). There are no vertical asymptotes for this function as the denominator never becomes 0.

Plot the function f(x) to verify the results

By plotting the function, we can verify the horizontal asymptotes at \(y = 10\) and \(y = 2\), and no vertical asymptotes. The limit as x approaches 0 is indeed 6. The plot will help confirm that our analysis is consistent with the graphical behavior of the function f(x).

Key concepts

Horizontal Asymptotes

Horizontal asymptotes in a function describe the line that a graph approaches as the input reaches positive or negative infinity. They give us insights into the behavior of a function over a long range. In the function given, \( f(x)=\frac{2 e^{x}+10 e^{-x}}{e^{x}+e^{-x}} \), analyzing horizontal asymptotes helps us understand the behavior of the function as \( x \) becomes very large or very small.
To determine horizontal asymptotes, calculate the limits at infinity. As \( x \rightarrow \infty \) and \( x \rightarrow -\infty \), we find the function approaches two different constant values:

• As \( x \rightarrow \infty \), \( f(x) \) approaches 2. This suggests a horizontal asymptote at \( y = 2 \).
• As \( x \rightarrow -\infty \), \( f(x) \) approaches 10. This suggests a horizontal asymptote at \( y = 10 \).

These asymptotes indicate that the curve of the function will level off around these horizontal lines as it moves towards positive and negative infinity.

Vertical Asymptotes

Vertical asymptotes occur in a function where the function value grows without bound as it approaches a certain point \( x = a \). This usually happens when division by zero is possible in the function expression. For the given function, vertical asymptotes are not present, and here's why.
Examine the denominator \( e^{x} + e^{-x} \). For a vertical asymptote to occur, the denominator would need to equal zero, causing the function to become undefined. However, because \( e^{x} \) and \( e^{-x} \) are always positive, the denominator \( e^{x} + e^{-x} \) can never be zero.
Concluding that there are no vertical asymptotes in this function agrees with the function's smooth behavior along its entire domain. Unlike rational functions, exponential functions like this typically don't have vertical asymptotes unless specifically manipulated for that purpose.

Exponential Functions

Exponential functions are a class of functions defined by an initial quantity growing or decreasing at a constant proportional rate. Generally, they take the form \( f(x) = a \cdot b^{x} \). In our exercise, the function includes terms like \( e^{x} \) and \( e^{-x} \), where \( e \) (approximately 2.718) is the base of natural logarithms.
Key characteristics of exponential functions include:

• Rapid growth or decay: These functions enhance or diminish quickly. For \( e^{x} \), the rate of growth is constant and fairly rapid as \( x \) increases.
• Asymptotic behavior: As seen in the exercise, asymptotes often appear due to how exponentially changing values catch up or fall away when extrema are reached.

Understanding how exponential elements in a function behave helps predict long-term trends, control systems, and model natural processes like population growth or radioactive decay.