Question

Evaluate \(\lim _{x \rightarrow \infty} f(x)\) \(f(x)\) and \(\lim _{x \rightarrow-\infty} f(x)\) Then state the horizontal asymptote(s) of \(f\). Confirm your findings by plotting \(f\) $$f(x)=\frac{3 e^{x}+e^{-x}}{e^{x}+e^{-x}}$$

Short answer

Question: Determine the horizontal asymptote(s) of the function \(f(x) = \frac{3 e^{x} + e^{-x}}{e^{x} + e^{-x}}\) as \(x\) approaches positive and negative infinity. Answer: The horizontal asymptote of the function is \(y = 3\) as \(x\) approaches both positive and negative infinity.

Step-by-step solution

Simplify the function

To simplify the function \(f(x)\), we can divide numerator and denominator by a common factor, \(e^x\), to get: $$f(x) = \frac{3 e^{x} + e^{-x}}{e^{x} + e^{-x}} \cdot \frac{\frac{1}{e^x}}{\frac{1}{e^x}} = \frac{3 + e^{-2x}}{1 + e^{-2x}}$$

Compute the limit as x approaches positive infinity

As \(x \to \infty\), the exponential terms \(e^{-2x}\) will approach 0 since negative exponents lead to very small values as x increases. So we have: $$\lim_{x \to \infty} \frac{3 + e^{-2x}}{1 + e^{-2x}} = \frac{3 + 0}{1 + 0} = \frac{3}{1} = 3$$

Compute the limit as x approaches negative infinity

As \(x \to -\infty\), the exponential terms \(e^{-2x}\) will approach infinity since the exponent becomes positive and large. So we have: $$\lim_{x \to -\infty} \frac{3 + e^{-2x}}{1 + e^{-2x}} = \frac{3 + \infty}{1 + \infty} = \frac{\infty}{\infty}$$ In this case, as both numerator and denominator approach infinity, we can use L'Hopital's Rule to find the limit by taking the derivative of both numerator and denominator of \(f(x)\) and then compute the limit: $$\lim_{x \to -\infty} \frac{d}{dx}\left(\frac{3 + e^{-2x}}{1 + e^{-2x}}\right) = \lim_{x \to -\infty} \frac{-6e^{-2x}}{-2e^{-2x}}$$ After simplifying the expression, we get: $$\lim_{x \to -\infty} \frac{-6e^{-2x}}{-2e^{-2x}} = \frac{-6}{-2} = 3$$

Identify horizontal asymptotes

Based on the calculated limits, we can identify that there are two horizontal asymptotes at \(y=3\) as \(x\) approaches both positive and negative infinity.

Confirm findings by plotting the function

By plotting the function \(f(x) = \frac{3+e^{-2x}}{1+e^{-2x}}\), we can visually verify that the graph of the function converges to the horizontal lines \(y = 3\) as x approaches positive and negative infinity, confirming our earlier findings. In conclusion, the horizontal asymptote of the function is \(y = 3\) as \(x\) approaches both positive and negative infinity.

Key concepts

Limits at Infinity

Understanding limits at infinity is crucial in calculus, as it helps predict the behavior of functions as the input grows large or decreases without bound. Specifically, when we evaluate \(\lim_{x \to \infty} f(x)\) or \(\lim_{x \to -\infty} f(x)\), we’re seeking the value that a function approaches as \(x\) moves towards positive or negative infinity, respectively.

In our example with the function \(f(x) = \frac{3e^x + e^{-x}}{e^x + e^{-x}}\), direct substitution as \(x\) tends toward infinity isn’t feasible due to the indeterminate form of \(\frac{\infty}{\infty}\). However, by simplifying the function and then taking limits, we can determine that \(f(x)\) approaches 3 as \(x\) goes to both positive and negative infinity. This tells us the function has a horizontal asymptote at \(y=3\), implying that no matter how much \(x\) increases or decreases, the function’s value will get closer and closer to 3, but never actually reach it.

L'Hopital's Rule

L'Hopital's Rule is a powerful tool in calculus for evaluating limits that result in an indeterminate form, such as \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). It states that if the limit of \(\frac{f(x)}{g(x)}\) as \(x\) approaches a value results in one of these indeterminate forms, and if the derivatives of \(f\) and \(g\) exist near this point, the original limit will have the same value as the limit of the derivatives.

In the context of our problem, when calculating \(\lim_{x \to -\infty} \frac{3 + e^{-2x}}{1 + e^{-2x}}\), we end up with the indeterminate form \(\frac{\infty}{\infty}\). By applying L'Hopital's Rule and calculating the limit of the derivative of the numerator and denominator separately, we resolved the indeterminate form and found that the limit is 3. This process showcases L'Hopital's Rule as a crucial strategy for solving certain types of limit problems.

Exponential Functions

Exponential functions, such as \(e^x\), are vital in understanding growth processes and decay in real-world situations, from compound interest to population dynamics. These functions are characterized by a constant ratio of change as \(x\) varies—the rate of increase or decrease is proportional to the function's current value.

In our function \(f(x)\), both \(e^x\) and \(e^{-x}\) are exponentials with base \(e\), Euler's number. Here, \(e^x\) grows rapidly as \(x\) becomes large, while \(e^{-x}\) decreases rapidly. When \(x\) approaches negative infinity, \(e^{-x}\) behaves like an exponential function with positive growth, becoming infinitely large because the negative sign in the exponent flips. These behaviors of exponential functions are instrumental in simplifying expressions and finding limits at infinity.

Asymptotic Behavior

The asymptotic behavior of a function refers to its behavior as the input either gets very large or very small—how the function approaches a line, called an asymptote, without ever touching it. Horizontal asymptotes, like the one we’ve identified at \(y=3\) for \(f(x)\), show the value that a function will approach without ever crossing as \(x\) tends toward positive or negative infinity.

These horizontal lines can be seen in the graph of a function as a boundary line that the function curve approaches. Finding asymptotic behavior gives us a clear picture of the long-term behavior of functions and is especially helpful in fields like engineering, physics, and economics, where understanding end behavior of models is essential.