Question

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

Short answer

Answer: The function has horizontal asymptotes at \(y=10\) as \(x\rightarrow -\infty\), and \(y=2\) as \(x\rightarrow \infty\). It has no vertical asymptotes. The value of the function at \(x=0\) is \(f(0)=6\).

Step-by-step solution

Find the limit of f(x) as x approaches 0

To find \(\lim_{x\rightarrow 0} f(x)\), we simply substitute x=0 into the function: $$\lim _{x \rightarrow 0} f(x) = \frac{2 e^{0}+10 e^{-0}}{e^{0}+e^{-0}} = \frac{2\cdot 1 + 10\cdot 1}{1+1} = \frac{12}{2} = 6$$ So, \(\lim_{x\rightarrow 0} f(x) = 6\).

Find the limit of f(x) as x approaches negative infinity

To find \(\lim_{x \rightarrow -\infty} f(x)\), notice that as \(x\) becomes very large in the negative direction, the term \(e^x\) approaches 0 while the term \(e^{-x}\) becomes very large. Therefore, we can rewrite the function as: $$f(x) = \frac{2 e^{x}+10 e^{-x}}{e^{x}+e^{-x}} = \frac{2 \frac{1}{e^{-x}}+10 e^{-x}}{\frac{1}{e^{-x}}+e^{-x}}$$ Since \(e^{-x}\) becomes very large when \(x\) approaches negative infinity, the function simplifies to: $$f(x) \approx \frac{10 e^{-x}}{e^{-x}}$$ Now the limit becomes: $$\lim _{x \rightarrow -\infty} f(x) = \lim _{x \rightarrow -\infty} \frac{10 e^{-x}}{e^{-x}} = 10$$ So, \(\lim_{x\rightarrow -\infty} f(x)=10\).

Find the limit of f(x) as x approaches positive infinity

To find \(\lim_{x \rightarrow \infty} f(x)\), notice that as \(x\) becomes very large in the positive direction, the term \(e^x\) becomes very large while the term \(e^{-x}\) approaches 0. Therefore, we can rewrite the function as: $$f(x) = \frac{2 e^{x}+10 e^{-x}}{e^{x}+e^{-x}} = \frac{2 e^{x}}{e^{x}}$$ Now the limit becomes: $$\lim _{x \rightarrow \infty} f(x) = \lim _{x \rightarrow \infty} \frac{2 e^{x}}{e^{x}} = 2$$ So, \(\lim_{x\rightarrow \infty} f(x) = 2\).

Find the horizontal and vertical asymptotes

Based on the limits found in Steps 2 and 3, we can now determine the horizontal asymptotes: - As \(x\) approaches negative infinity, the function approaches the horizontal line \(y=10\). - As \(x\) approaches positive infinity, the function approaches the horizontal line \(y=2\). To find the vertical asymptote, we first analyze the denominator of \(f(x)\): \(e^x + e^{-x}\). There are no values of \(x\) that make the denominator equal to 0, so there are no vertical asymptotes.

Plot the function to verify the results

To verify the results, plot the function \(f(x)=\frac{2 e^{x}+10 e^{-x}}{e^{x}+e^{-x}}\). The plot should show the following characteristics: - The function approaches the horizontal line \(y=10\) as \(x\rightarrow -\infty\) - The function approaches the horizontal line \(y=2\) as \(x\rightarrow \infty\) - The function has no vertical asymptotes - The value of the function at \(x=0\) is \(f(0)=6\)

Key concepts

Infinity

When we talk about infinity in the context of limits, we refer to what happens to the value of a function as the input grows without bound in the positive or negative direction. In this exercise, evaluating the function as \(x\) approaches infinity (positive or negative) provides insight into the behavior of the function over large values of \(x\). The goal is to determine what \(f(x)\) tends toward as we let \(x\) become extremely large or extremely small.

• As \(x\to -\infty\), the function simplifies significantly because certain terms dominate over others. In this exercise, as \(x\) becomes large negatively, \(e^{x}\) approaches zero making \(e^{-x}\) very large. This alters \(f(x)\) to approach 10.
• As \(x\to \infty\), \(e^{-x}\) vanishes, and \(e^{x}\) terms dominate, resulting in \(f(x)\) approaching 2.

Understanding these limits helps to paint a big picture of the function's behavior over vast stretches of the \(x\)-axis.

Asymptotes

Asymptotes are lines that a graph of a function approaches but never actually reaches. They serve as invisible boundaries that mark the behavior of the function.In this context, horizontal asymptotes are particularly relevant. They describe the limit of \(f(x)\) as \(x\) approaches positive or negative infinity.

• The horizontal asymptote as \(x\to -\infty\) is \(y=10\). This means the function gets closer and closer to this line as \(x\) moves infinitely left.
• The horizontal asymptote as \(x\to \infty\) is \(y=2\). Similarly, as \(x\) moves infinitely to the right, the function approaches this value.

Vertical asymptotes occur where a function becomes undefined, typically when the denominator of a rational function equals zero. For this exercise, the function \(f(x)\) has no vertical asymptotes because its denominator, \(e^{x} + e^{-x}\), is never zero. These asymptotes provide a framework to anticipate the behavior of the function across its domain.

Exponential Function

An exponential function is a mathematical function of the form \(f(x) = a \cdot e^{bx}\), where \(e\) is Euler's number, approximately equal to 2.718. This type of function is crucial in modeling growth, decay, and compound interest, among many other applications.In this exercise, the function \(f(x)\) involves two exponential terms, \(e^{x}\) and \(e^{-x}\). Understanding their behavior individually helps to solve limits and find asymptotes.

• As \(x\to \infty\), \(e^{x}\) grows very large while \(e^{-x}\) shrinks toward zero. This dramatically shifts the behavior of the function, usually dominated by the larger exponential term.
• Similarly, as \(x\to -\infty\), \(e^{-x}\) becomes large and \(e^{x}\) approaches zero. This results in a different limiting behavior set by the dominating term.

Familiarity with these shifts allows us to simplify expressions intelligently and predict the long-term behavior of exponential functions. This knowledge is directly applicable to determining limits and asymptotes, tuning us into the nuanced dynamics at play within such mathematical expressions.