Assume that `lim_(thetararr-1) f(theta)` exists and `(theta^(2)+theta-2)/(theta+3)le(f(theta))/(theta^(2))le(theta^(2)+2theta-1)/(theta+3)` holds for certain interval containing the point `theta=-1 " then "lim_(thetararr-1) f(theta)`

Options

A. is equal to `f(-1)`
B. is equal to 1
C. is non-existent
D. is equal to `-1`

Explanation

To solve the problem, we need to analyze the given inequalities and apply the Squeeze Theorem (also known as the Sandwich Theorem) to find the limit of \( f(\theta) \) as \( \theta \) approaches \(-1\).

### Step-by-Step Solution:

1. **Understand the Given Inequalities:**
We have the inequalities:
\[ \frac{\theta^2 + \theta - 2}{\theta + 3} \leq \frac{f(\theta)}{\theta^2} \leq \frac{\theta^2 + 2\theta - 1}{\theta + 3} \]

2. **Substituting \(\theta = -1\):**
We will substitute \(\theta = -1\) into both sides of the inequality to find the bounds for \( f(-1) \).
- For the left side:
\[ \frac{(-1)^2 + (-1) - 2}{-1 + 3} = \frac{1 - 1 - 2}{2} = \frac{-2}{2} = -1 \]
- For the right side:
\[ \frac{(-1)^2 + 2(-1) - 1}{-1 + 3} = \frac{1 - 2 - 1}{2} = \frac{-2}{2} = -1 \]

3. **Establishing the Bounds:**
From the calculations, we have:
\[ -1 \leq \frac{f(-1)}{(-1)^2} \leq -1 \]
This simplifies to:
\[ -1 \leq f(-1) \leq -1 \]
Therefore, we can conclude:
\[ f(-1) = -1 \]

4. **Applying the Squeeze Theorem:**
Since we have established that:
\[ \lim_{\theta \to -1} \frac{\theta^2 + \theta - 2}{\theta + 3} = -1 \quad \text{and} \quad \lim_{\theta \to -1} \frac{\theta^2 + 2\theta - 1}{\theta + 3} = -1 \]
By the Squeeze Theorem, we can conclude that:
\[ \lim_{\theta \to -1} f(\theta) = f(-1) = -1 \]

### Final Answer:
Thus, we find that:
\[ \lim_{\theta \to -1} f(\theta) = -1 \]