Evaluate `lim_(x to 3) [4x^(3) + 3x^(2) + 2x + 6]`

Answer

To evaluate the limit \( \lim_{x \to 3} [4x^3 + 3x^2 + 2x + 6] \), we can directly substitute \( x = 3 \) into the polynomial function since it is continuous everywhere.

### Step-by-Step Solution:

1. **Substitute \( x = 3 \)** into the expression:
\[ 4(3)^3 + 3(3)^2 + 2(3) + 6 \]

2. **Calculate \( 3^3 \)**:
\[ 3^3 = 27 \]
Therefore, \( 4(3^3) = 4 \times 27 = 108 \).

3. **Calculate \( 3^2 \)**:
\[ 3^2 = 9 \]
Therefore, \( 3(3^2) = 3 \times 9 = 27 \).

4. **Calculate \( 2(3) \)**:
\[ 2(3) = 6 \]

5. **Combine all the results**:
\[ 108 + 27 + 6 + 6 \]

6. **Add the values together**:
\[ 108 + 27 = 135 \]
\[ 135 + 6 = 141 \]
\[ 141 + 6 = 147 \]

Thus, the limit is:
\[ \lim_{x \to 3} [4x^3 + 3x^2 + 2x + 6] = 147 \]