https://file.aiquickdraw.com/mathbot/file/1727384364020n7vsc3mr.jpg

Let's go through each question in the image and solve them one by one.

### **Question 6:**
Evaluate the limit:

\[
\lim_{x \to \infty} \frac{6x + 4}{11x^2 - 2x + 5}
\]

#### Solution:
As \( x \to \infty \), the highest powers of \( x \) in the numerator and denominator dominate the behavior of the function. The numerator is linear (\(6x + 4\)), and the denominator is quadratic (\(11x^2 - 2x + 5\)).

1. The highest degree term in the numerator is \(6x\), and in the denominator, it's \(11x^2\).
2. As \(x \to \infty\), we can approximate the limit by comparing the dominant terms:

\[
\lim_{x \to \infty} \frac{6x}{11x^2} = \lim_{x \to \infty} \frac{6}{11x} = 0
\]

Thus, the limit is:
\[
\boxed{0}
\]

---

### **Question 7:**
Evaluate the limit:

\[
\lim_{x \to 2} \frac{x^2 - 9x + 18}{x^2 + 4x - 12}
\]

#### Solution:
First, factor both the numerator and the denominator:

- Numerator: \(x^2 - 9x + 18 = (x - 6)(x - 3)\)
- Denominator: \(x^2 + 4x - 12 = (x - 2)(x + 6)\)

Thus, the expression becomes:

\[
\frac{(x - 6)(x - 3)}{(x - 2)(x + 6)}
\]

Now, substitute \(x = 2\) directly into the simplified expression:

\[
\frac{(2 - 6)(2 - 3)}{(2 - 2)(2 + 6)} = \frac{(-4)(-1)}{(0)(8)}
\]

Since the denominator becomes zero when \(x = 2\), the limit does not exist in a typical sense. However, the behavior near \(x = 2\) may indicate a vertical asymptote. Thus, the limit is:

\[
\boxed{\infty} \quad (\text{or "oo" as requested})
\]

---

### **Question 8:**
If \( h(x) = 2 - 2x^3 \), find \( h'(2) \).

#### Solution:
1. Differentiate \( h(x) = 2 - 2x^3 \) using basic derivative rules:

\[
h'(x) = \frac{d}{dx} \left(2 - 2x^3\right) = -6x^2
\]

2. Now substitute \(x = 2\):

\[
h'(2) = -6(2^2) = -6(4) = -24
\]

So, \(h'(2) = \boxed{-24}\).

---

Next, use this to find the equation of the tangent line to \( y = 2 - 2x^3 \) at the point \( (2, -14) \).

The equation of the tangent line is given by:
\[
y - y_1 = m(x - x_1)
\]
where \(m = h'(2) = -24\) is the slope, and the point is \( (x_1, y_1) = (2, -14) \).

Substitute the values:
\[
y + 14 = -24(x - 2)
\]
Simplify:
\[
y = -24x + 48 - 14
\]
\[
y = -24x + 34
\]

Thus, the equation of the tangent line is:
\[
y = \boxed{-24x + 34}
\]

And the slope \(m\) is \(-24\), and the intercept \(b\) is \(34\).

---

### **Summary:**
- **Question 6:** \(\boxed{0}\)
- **Question 7:** \(\boxed{\infty}\)
- **Question 8:** \(h'(2) = \boxed{-24}, y = \boxed{-24x + 34}\)

---

Do you want details on any step or have additional questions? Here are 5 related questions you can explore:

1. How do you handle limits at infinity for rational functions?
2. What is L'Hopital's Rule and when is it applied in limits?
3. How do you determine vertical asymptotes from a rational function?
4. How do you differentiate higher powers of polynomials?
5. What are the applications of tangent lines in calculus?

**Tip:** When calculating limits of rational functions, always focus on the highest degree terms in the numerator and denominator when \(x\) approaches infinity.

Math problems from a homework assignment involving limits and differentiation.

Explore detailed solutions to advanced calculus problems, including limits at infinity and point-specific evaluations, as well as differentiation of polynomial functions. This guide is perfect for students in advanced high school or early college calculus.

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