If `f(x)=(3x^2+a x+a+1)/(x^2+x-2),` then which of the following can be correct (a) `("lim")_(xvec 1)f(x) for a=-2` (b) `("lim")_(xvec-2)f(x) for a=13` (c) `("lim")_(xvec 1)f(x)=4/3` (d) `("lim")_(xvec-2)f(x)=-1/3`

Options

A. `"ln "a_(1)`
B. `e^(a_(n))`
C. a_(1)`
D. `a_(n)`

Explanation

To solve the problem, we need to evaluate the limit of the function \( f(x) = \frac{3x^2 + ax + (a + 1)}{x^2 + x - 2} \) for different values of \( a \) and at specific points.

### Step-by-Step Solution

1. **Identify the function and the limits to evaluate:**
The function is given as:
\[ f(x) = \frac{3x^2 + ax + (a + 1)}{x^2 + x - 2} \]
We need to evaluate the limits for the following conditions:
- (a) \( \lim_{x \to 1} f(x) \) for \( a = -2 \)
- (b) \( \lim_{x \to -2} f(x) \) for \( a = 13 \)
- (c) \( \lim_{x \to 1} f(x) = \frac{4}{3} \)
- (d) \( \lim_{x \to -2} f(x) = -\frac{1}{3} \)

2. **Evaluate limit (a): \( \lim_{x \to 1} f(x) \) for \( a = -2 \)**
- Substitute \( x = 1 \):
\[ f(1) = \frac{3(1)^2 + (-2)(1) + (-2 + 1)}{(1)^2 + (1) - 2} = \frac{3 - 2 - 1}{1 + 1 - 2} = \frac{0}{0} \]
- Since we have a \( 0/0 \) form, we apply L'Hôpital's Rule:
- Differentiate the numerator and denominator:
\[ \text{Numerator: } \frac{d}{dx}(3x^2 - 2x - 1) = 6x - 2 \]
\[ \text{Denominator: } \frac{d}{dx}(x^2 + x - 2) = 2x + 1 \]
- Now, applying L'Hôpital's Rule:
\[ \lim_{x \to 1} f(x) = \lim_{x \to 1} \frac{6x - 2}{2x + 1} = \frac{6(1) - 2}{2(1) + 1} = \frac{4}{3} \]
- Since \( a = -2 \) gives us \( \frac{4}{3} \), option (a) is correct.

3. **Evaluate limit (b): \( \lim_{x \to -2} f(x) \) for \( a = 13 \)**
- Substitute \( x = -2 \):
\[ f(-2) = \frac{3(-2)^2 + 13(-2) + (13 + 1)}{(-2)^2 + (-2) - 2} = \frac{12 - 26 + 14}{4 - 2 - 2} = \frac{0}{0} \]
- Again, we apply L'Hôpital's Rule:
- Differentiate the numerator and denominator:
\[ \text{Numerator: } \frac{d}{dx}(3x^2 + 13x + 14) = 6x + 13 \]
\[ \text{Denominator: } \frac{d}{dx}(x^2 + x - 2) = 2x + 1 \]
- Now, applying L'Hôpital's Rule:
\[ \lim_{x \to -2} f(x) = \lim_{x \to -2} \frac{6x + 13}{2x + 1} = \frac{6(-2) + 13}{2(-2) + 1} = \frac{-12 + 13}{-4 + 1} = \frac{1}{-3} = -\frac{1}{3} \]
- Since \( a = 13 \) gives us \( -\frac{1}{3} \), option (b) is correct.

4. **Evaluate limit (c): \( \lim_{x \to 1} f(x) = \frac{4}{3} \)**
- From the previous calculation, we already found that:
\[ \lim_{x \to 1} f(x) = \frac{4}{3} \]
- Therefore, option (c) is correct.

5. **Evaluate limit (d): \( \lim_{x \to -2} f(x) = -\frac{1}{3} \)**
- From the previous calculation, we found that:
\[ \lim_{x \to -2} f(x) = -\frac{1}{3} \]
- Therefore, option (d) is also correct.

### Conclusion
All options (a), (b), (c), and (d) are correct.