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Let's solve these limits step by step.

We are given the following rational functions for the limits:

\[
f(x) = \frac{3x^2 + 15x + 18}{x^2 - 6x + 8}
\]

### Step 1: Factorize the numerator and denominator.

1. **Numerator: \( 3x^2 + 15x + 18 \)**  
   Factor out the common factor of 3:
   \[
   3(x^2 + 5x + 6)
   \]
   Then factor the quadratic \( x^2 + 5x + 6 \):
   \[
   3(x + 2)(x + 3)
   \]

2. **Denominator: \( x^2 - 6x + 8 \)**  
   Factor the quadratic:
   \[
   (x - 2)(x - 4)
   \]

Thus, the simplified form of the function is:
\[
f(x) = \frac{3(x + 2)(x + 3)}{(x - 2)(x - 4)}
\]

### Step 2: Evaluate each limit.

#### (a) \( \lim_{x \to 2} \frac{3x^2 + 15x + 18}{x^2 - 6x + 8} \)

Substitute \( x = 2 \) into the simplified form:
\[
f(2) = \frac{3(2 + 2)(2 + 3)}{(2 - 2)(2 - 4)}
\]
This results in:
\[
f(2) = \frac{3(4)(5)}{(0)(-2)}
\]
This gives a **division by zero**, so we need to apply **L'Hopital's Rule** or factor and cancel terms.

After factoring, we see that the term \( (x - 2) \) cancels out, giving us:
\[
\lim_{x \to 2} f(x) = \frac{3(x + 3)}{x - 4} = \frac{3(2 + 3)}{2 - 4} = \frac{3(5)}{-2} = -\frac{15}{2}
\]

Thus, the limit is:
\[
\boxed{-\frac{15}{2}}
\]

#### (b) \( \lim_{x \to 2^+} \frac{3x^2 + 15x + 18}{x^2 - 6x + 8} \)

Since the limit from the right-hand side of \( x = 2 \) is the same as from the left-hand side (there is no discontinuity other than the removable one), the result is the same:
\[
\boxed{-\frac{15}{2}}
\]

#### (c) \( \lim_{x \to 4} \frac{3x^2 + 15x + 18}{x^2 - 6x + 8} \)

Substitute \( x = 4 \) into the simplified form:
\[
f(4) = \frac{3(4 + 2)(4 + 3)}{(4 - 2)(4 - 4)}
\]
This gives:
\[
f(4) = \frac{3(6)(7)}{(2)(0)} = \frac{126}{0}
\]
Since this results in division by zero (with a non-zero numerator), the limit **does not exist**, as it approaches infinity or negative infinity.

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

#### (d) \( \lim_{x \to 4^-} \frac{3x^2 + 15x + 18}{x^2 - 6x + 8} \)

For \( x \to 4^- \), the same logic applies as for part (c). The function approaches infinity, so the limit does not exist (it tends to \( +\infty \) or \( -\infty \)).

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

### Summary of the Limits:
- (a) \( \boxed{-\frac{15}{2}} \)
- (b) \( \boxed{-\frac{15}{2}} \)
- (c) \( \boxed{\text{DNE}} \)
- (d) \( \boxed{\text{DNE}} \)

Would you like any further explanation or a deeper dive into one of these steps?

Here are 5 related questions you might find interesting:
1. What is L'Hopital's Rule and how is it applied in limit problems?
2. How do you factor quadratic expressions quickly?
3. What happens to the limit when both numerator and denominator approach 0?
4. How do you determine if a limit does not exist (DNE)?
5. How does the concept of removable discontinuity relate to limits?

**Tip**: Always factor the expression first when dealing with limits of rational functions—this helps simplify the evaluation.

Evaluate the following limits: (a) lim as x -> 2 of (3x^2 + 15x + 18) / (x^2 - 6x + 8); (b) lim as x -> 2^+ of (3x^2 + 15x + 18) / (x^2 - 6x + 8); (c) lim as x -> 4 of (3x^2 + 15x + 18) / (x^2 - 6x + 8); (d) lim as x -> 4^- of (3x^2 + 15x + 18) / (x^2 - 6x + 8)

Learn how to evaluate limits involving rational functions with techniques such as factoring and L'Hopital's Rule. This problem set covers step-by-step solutions for limits at specific points, including division by zero cases and removable discontinuities. Suitable for college-level calculus students.

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