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Let's analyze and solve the questions presented step by step.

### Given Function
The function \( f(x) \) is defined as:
\[
f(x) = \frac{3 + 3x - 5x^2}{1 - 16e^{-5x}}
\]

Our goal is to understand its behavior as \( x \to \pm \infty \), as well as the gaps in its domain.

### **a) First compute \(\lim_{x \to \infty} f(x)\).**

To compute the limit as \( x \to \infty \):
- As \( x \to \infty \), \( e^{-5x} \to 0 \), so the denominator approaches \( 1 \).
- The numerator is dominated by the highest power term, \( -5x^2 \).

Thus:
\[
\lim_{x \to \infty} f(x) = \frac{-5x^2}{1} = -\infty
\]

### **b) Compute the limit as \( x \to -\infty \)**

For \( x \to -\infty \), \( e^{-5x} \to \infty \), so the dominant term in the denominator is \( -16e^{-5x} \), and the numerator is still dominated by \( -5x^2 \).

Hence:
\[
f(x) \approx \frac{-5x^2}{-16e^{-5x}} = \frac{5x^2}{16e^{-5x}}
\]
This expression tends to 0 as \( x \to -\infty \).

Thus:
\[
\lim_{x \to -\infty} f(x) = 0
\]

### **c) Rewrite \( f(x) \) as a fraction \( \frac{P(x)}{Q(x)} \)**
The original function is already in the form \( \frac{P(x)}{Q(x)} \), where
\[
P(x) = 3 + 3x - 5x^2 \quad \text{and} \quad Q(x) = 1 - 16e^{-5x}.
\]

### **d) Using the fraction form from (c), compute \( \lim_{x \to -\infty} f(x) \) again.**
As calculated previously:
\[
\lim_{x \to -\infty} f(x) = 0.
\]

### **e) Find the point(s) \( x = a \) where \( f(x) \) is undefined.**

The function is undefined where \( Q(x) = 0 \):
\[
1 - 16e^{-5x} = 0 \implies e^{-5x} = \frac{1}{16}.
\]
Taking the natural logarithm:
\[
-5x = \ln(16) \implies x = -\frac{\ln(16)}{5} = -\frac{4\ln(2)}{5}.
\]

### **f) Analyzing the behavior at the point \( x = a \)**
The point found is \( x = -\frac{4\ln(2)}{5} \).

For each value \( a \):
#### **Left-hand and right-hand limits as \( x \to a \):**
Let \( a = -\frac{4\ln(2)}{5} \).

As \( x \to a^- \), \( Q(x) \to 0^- \) and \( P(x) \to \text{finite nonzero} \); hence, \( f(x) \to +\infty \).

As \( x \to a^+ \), \( Q(x) \to 0^+ \) and \( P(x) \to \text{finite nonzero} \); hence, \( f(x) \to -\infty \).

### **Answers summary:**
a) \( \lim_{x \to \infty} f(x) = -\infty \).

b) \( \lim_{x \to -\infty} f(x) = 0 \).

c) \( P(x) = 3 + 3x - 5x^2 \), \( Q(x) = 1 - 16e^{-5x} \).

d) \( \lim_{x \to -\infty} f(x) = 0 \).

e) The point where \( f(x) \) is undefined: \( x = -\frac{4 \ln(2)}{5} \).

f) Limits at \( x = -\frac{4\ln(2)}{5} \):
- \( \lim_{x \to a^-} f(x) = +\infty \)
- \( \lim_{x \to a^+} f(x) = -\infty \)

Would you like more details or have any further questions?

### **5 Relative Questions:**
1. What happens if we change the exponent in \( e^{-5x} \) to another value?
2. How would the results change if the numerator was a different degree polynomial?
3. How do we analyze limits involving functions with exponentials and polynomials generally?
4. What if the sign of the exponential term in the denominator were reversed?
5. How do asymptotes manifest in rational functions with exponential components?

### **Tip:**
When dealing with limits at infinity involving exponentials, remember that exponentials grow much faster than polynomial terms, which often dictates the behavior.

Consider the function f(x) = (3 + 3x - 5x^2) / (1 - 16e^{-5x}). The task is to analyze the behavior of f(x) as x tends to ±∞ and to find points where the function is undefined.

This detailed solution explores the behavior of the function f(x) = (3 + 3x - 5x^2) / (1 - 16e^{-5x}) as x approaches infinity and negative infinity, finds undefined points, and calculates limits using techniques like L'Hopital's Rule. Ideal for Calculus II students.

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