To determine the asymptotes for each function in Exercise 59, let's analyze each given function individually:

a) $y = \frac{e^x}{x + 1}$

1. Vertical Asymptote: Vertical asymptotes occur where the denominator equals zero if the numerator does not cancel it out. Here, the denominator $x + 1 = 0 \Rightarrow x = -1$. Thus, there is a vertical asymptote at $x = -1$.
2. Horizontal Asymptote: As $x \to \infty$, $e^x$ grows much faster than $x+1$, so $y \to \infty$. Similarly, as $x \to -\infty$, $e^x \to 0$, making $y \to 0$. Thus, there is a horizontal asymptote at $y = 0$.

b) $y = x \, \text{arccot} \frac{2}{x}$

1. Vertical Asymptote: For this function, there is no value of $x$ that makes the argument inside arccot undefined, and thus no vertical asymptote.
2. Behavior at Infinity: As $x \to \infty$, $\frac{2}{x} \to 0$ and $\text{arccot}(0) = \frac{\pi}{2}$, so $y \approx x \cdot \frac{\pi}{2} \to \infty$. Similarly, as $x \to -\infty$, $\frac{2}{x} \to 0$ and $\text{arccot}(0) = \frac{\pi}{2}$, so $y \to -\infty$. There is no horizontal asymptote, but the behavior indicates that $y$ grows linearly as $x \to \infty$ or $x \to -\infty$.

c) $y = (x + 2) e^{1/x}$

1. Vertical Asymptote: No values of $x$ make this function undefined, so there is no vertical asymptote.
2. Horizontal Asymptote: As $x \to \infty$, $\frac{1}{x} \to 0$, so $e^{1/x} \to e^0 = 1$, and $y \approx x + 2$, which grows without bound. Similarly, as $x \to -\infty$, $e^{1/x} \to 1$, and $y \approx x + 2$. Thus, there are no horizontal asymptotes.

d) $y = \sqrt[3]{x^3 + x}$

1. Vertical Asymptote: This function is defined for all $x$, so there is no vertical asymptote.
2. Behavior at Infinity: As $x \to \infty$, $x^3 + x \approx x^3$, so $\sqrt[3]{x^3 + x} \approx x$. Thus, $y$ grows without bound as $x \to \infty$ and $x \to -\infty$, indicating no horizontal asymptote.

e) $y = e^x \ln x$

1. Vertical Asymptote: This function is defined for $x > 0$ only (since $\ln x$ is undefined for $x \leq 0$), but it does not exhibit a vertical asymptote; rather, it has a domain restriction at $x = 0$.
2. Behavior at Infinity: As $x \to \infty$, $e^x \ln x$ grows exponentially due to $e^x$, so there is no horizontal asymptote.

In summary:

• a) Vertical at $x = -1$, horizontal at $y = 0$
• b) No asymptotes
• c) No asymptotes
• d) No asymptotes
• e) No asymptotes, domain $x > 0$

Would you like further details on any specific function's asymptotic behavior?

Follow-up Questions:

1. What are asymptotes and why are they important in graph analysis?
2. How do you determine vertical asymptotes for more complex functions?
3. What role does exponential growth play in horizontal asymptotes?
4. Can logarithmic functions produce horizontal asymptotes, and under what conditions?
5. How would the asymptotes change if we modified each function slightly?

Tip:

When examining asymptotes, always start by identifying any domain restrictions for vertical asymptotes and evaluating limits at infinity for horizontal ones.