If `3-x^(2)/(12) le f(x) le 3+x^(3)/9" for all "x ne 0`, then the value of `underset(h to 0)lim f(x)` is equal to

Options

A. `-1`
B. 1
C. 2
D. 0

Answer

To solve the problem, we need to find the limit of the function \( f(x) \) as \( x \) approaches 0, given the inequalities:

\[ 3 - \frac{x^2}{12} \leq f(x) \leq 3 + \frac{x^3}{9} \]

### Step 1: Identify the limits of the bounding functions

First, we need to evaluate the limits of the two bounding functions as \( x \) approaches 0.

1. **For the lower bound**:
\[ \lim_{x \to 0} \left( 3 - \frac{x^2}{12} \right) \]
Substituting \( x = 0 \):
\[ = 3 - \frac{0^2}{12} = 3 \]

2. **For the upper bound**:
\[ \lim_{x \to 0} \left( 3 + \frac{x^3}{9} \right) \]
Substituting \( x = 0 \):
\[ = 3 + \frac{0^3}{9} = 3 \]

### Step 2: Apply the Squeeze (Sandwich) Theorem

Since we have established that:
\[ 3 - \frac{x^2}{12} \leq f(x) \leq 3 + \frac{x^3}{9} \]
and both limits as \( x \) approaches 0 are equal to 3, we can apply the Squeeze Theorem.

According to the Squeeze Theorem:
\[ \lim_{x \to 0} f(x) = 3 \]

### Final Answer

Thus, the value of \( \lim_{h \to 0} f(x) \) is:
\[ \boxed{3} \]