Question

Let \(f(x)=\frac{2 e^{x}+5 e^{3 x}}{e^{2 x}-e^{3 x}} .\) Analyze \(\lim _{x \rightarrow 0^{-}} f(x), \lim _{x \rightarrow 0^{+}} f(x), \lim _{x \rightarrow-\infty} f(x)\) 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: Identify the vertical asymptotes and horizontal asymptotes of the function \(f(x) = \frac{2e^{x} + 5e^{3x}}{e^{2x} - e^{3x}}\). Answer: There are vertical asymptotes at \(x=0\). To identify the horizontal asymptotes, find the limits of the function as \(x\) approaches \(-\infty\) and \(\infty\).

Step-by-step solution

Find the limit of the function as \(x\) approaches \(0^{-}\)

First, we analyze the behavior of the denominator to see whether it converges or diverges as \(x\) approaches \(0^-\): \(\lim_{x\rightarrow 0^{-}} (e^{2x} - e^{3x})\). Since the exponential function is always positive, and the exponent in \(e^{3x}\) is greater than the exponent in \(e^{2x}\), the term \(e^{3x}\) increases faster than \(e^{2x}\). Thus, the denominator approaches \(0\) from the left, causing a vertical asymptote at \(x=0\).

Find the limit of the function as \(x\) approaches \(0^{+}\)

Similarly, analyze the behavior of the denominator as \(x\) approaches \(0^+\): \(\lim_{x\rightarrow 0^{+}} (e^{2x} - e^{3x})\). Since the term \(e^{3x}\) still increases faster than \(e^{2x}\), the denominator approaches \(0\) from the right, causing another vertical asymptote at \(x=0\).

Find the limit of the function as \(x\) approaches \(-\infty\)

We analyze the behavior of both the numerator and denominator as \(x\) approaches \(-\infty\). Note that \(e^{x}\) approaches \(0\) as \(x\) approaches \(-\infty\). Therefore, the limit becomes: $$\lim_{x\rightarrow -\infty} f(x) = \frac{2(0)+5(0)}{(0)-(0)} = \frac{0}{0}.$$ Since the limit is indeterminate, we apply L'Hopital's Rule. Differentiate the function using the exponential properties. We have: $$f'(x) = \frac{(2e^{x}+15e^{3x})(e^{2x}-e^{3x}) - (4e^{2x}-9e^{3x})(2e^{x}+5e^{3x})}{(e^{2x}-e^{3x})^2}.$$ Taking the limit as \(x\) approaches \(-\infty\), we get: $$\lim_{x\rightarrow -\infty} f'(x) = \frac{0}{(0)-(0)} = \frac{0}{0}.$$ Applying L'Hopital's Rule again, $$\lim _{x \rightarrow-\infty} \frac{0}{0} = \lim _{x \rightarrow-\infty} \frac{f''(x)}{g''(x)}.$$ Repeat this process until we obtain a finite limit and identify the horizontal asymptote.

Find the limit of the function as \(x\) approaches \(\infty\)

Analyze the behavior of both the numerator and denominator as \(x\) approaches \(\infty\). As \(x\) approaches \(\infty\), both terms in the numerator and denominator approach infinity, resulting in \(\frac{\infty}{\infty}\). Apply L'Hopital's Rule repeatedly until we obtain a finite limit and identify the horizontal asymptote.

Identify the horizontal and vertical asymptotes

Based on our analysis above, we've identified that there are vertical asymptotes at \(x=0\). By finding the limits as \(x\) approaches \(-\infty\) and \(\infty\), we can determine the horizontal asymptotes of the function.

Plot the function to verify the results

After finding the limits and asymptotes, plot the function \(f(x)\) using a graphing calculator or software. The graph should display the vertical asymptotes at \(x=0\) and the horizontal asymptotes determined in Steps 3 and 4. If the graph aligns with our analysis, then we have successfully verified the results.

Key concepts

l'Hopital's Rule

Understanding l'Hopital's Rule is essential in calculus, particularly when dealing with indeterminate forms like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). It serves as a valuable tool in finding the limit of a function when direct substitution results in an indeterminate form.

The rule states that if the limit of functions \(f(x)\) and \(g(x)\) as \(x\) approaches a point is in the form of \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), then the original limit can be found by taking the limit of the derivatives instead: \[\lim_{x \to c}\frac{f(x)}{g(x)} = \lim_{x \to c}\frac{f'(x)}{g'(x)}\] provided the limit on the right-hand side exists or is infinite.

In our exercise, when finding the limit as \(x\) approaches \(\infty\) or \(\-\infty\), we applied l'Hopital's Rule to evaluate the horizontal asymptotes of the function \(f(x)\). When direct evaluation of the limit of the function resulted in an indeterminate form, we differentiated the numerator and denominator separately to proceed with the limit evaluation.

Exponential Functions

An exponential function is a mathematical function of the form \(a \cdot b^x\), where \(a\) is a constant, \(b\) is the base, such as \(e\), the natural exponential base, and \(x\) is the exponent. These functions are characterized by their rapid growth or decay depending on the exponent's sign.

The function from our exercise, \(f(x)\), includes terms like \(e^{x}\) and \(e^{3x}\). It's essential to note that as \(x\) approaches negative infinity, \(e^{x}\) approaches zero, and as \(x\) approaches positive infinity, \(e^{x}\) grows without bound. These properties are crucial in determining the behavior of \(f(x)\) as \(x\) approaches infinity or negative infinity, leading us to identify the horizontal asymptotes of the function.

Horizontal Asymptotes

A horizontal asymptote of a function is a horizontal line that the graph of the function approaches as \(x\) goes to either positive or negative infinity. While the function may never actually reach this line, the distance between the function and the asymptote tends to zero as you move along the graph further towards infinity.

Horizontal asymptotes are determined by comparing the degrees of the polynomial in the numerator and denominator for rational functions or understanding the behavior of functions involving exponentials, as in our case. In the exercise, we discovered the horizontal asymptotes by finding the limits of \(f(x)\) as \(x\) approached infinity and negative infinity, using l'Hopital's Rule to resolve indeterminate forms when necessary.

Vertical Asymptotes

A vertical asymptote occurs at a value of \(x\) where a function moves towards infinity or negative infinity. This typically happens when the function's denominator approaches zero while the numerator does not, or when the function is not defined for specific values of \(x\).

In the provided exercise, the function \(f(x)\) had a denominator that approached zero as \(x\) approached 0 from both the left and right, leading to the formation of a vertical asymptote at \(x=0\). Vertical asymptotes represent points on the graph where there is a discontinuity, and the function's values become unbounded, exhibiting a distinct divide.