If `(x^(2)+x-2)/(x+3)le(f(x))/(x^(2))le(x^(2)+2x-1)/(x+3)` holds for a certain interval containing the value of `lim_(xto-1) f(x).`

Explanation

To solve the problem step by step, we will analyze the given inequalities and apply the limit to find \( \lim_{x \to -1} f(x) \).

### Step 1: Write down the inequalities
We are given the following inequalities:
\[ \frac{x^2 + x - 2}{x + 3} \leq \frac{f(x)}{x^2} \leq \frac{x^2 + 2x - 1}{x + 3} \]

### Step 2: Find the limits of the left and right expressions as \( x \to -1 \)
We will evaluate the limits of the left and right expressions as \( x \) approaches -1.

**Left Expression:**
\[ \lim_{x \to -1} \frac{x^2 + x - 2}{x + 3} \]
Substituting \( x = -1 \):
\[ = \frac{(-1)^2 + (-1) - 2}{-1 + 3} = \frac{1 - 1 - 2}{2} = \frac{-2}{2} = -1 \]

**Right Expression:**
\[ \lim_{x \to -1} \frac{x^2 + 2x - 1}{x + 3} \]
Substituting \( x = -1 \):
\[ = \frac{(-1)^2 + 2(-1) - 1}{-1 + 3} = \frac{1 - 2 - 1}{2} = \frac{-2}{2} = -1 \]

### Step 3: Apply the Squeeze Theorem
Now we have:
\[ \lim_{x \to -1} \frac{f(x)}{x^2} \text{ is squeezed between } -1 \text{ and } -1. \]
Thus, by the Squeeze Theorem:
\[ \lim_{x \to -1} \frac{f(x)}{x^2} = -1 \]

### Step 4: Solve for \( \lim_{x \to -1} f(x) \)
We know:
\[ \lim_{x \to -1} \frac{f(x)}{x^2} = -1 \]
This implies:
\[ \lim_{x \to -1} f(x) = -1 \cdot \lim_{x \to -1} x^2 \]
Calculating \( \lim_{x \to -1} x^2 \):
\[ \lim_{x \to -1} x^2 = (-1)^2 = 1 \]
Thus:
\[ \lim_{x \to -1} f(x) = -1 \cdot 1 = -1 \]

### Final Answer
\[ \lim_{x \to -1} f(x) = -1 \]