`lim_(x->0^-)([x]+[x^2]+[x^3]++[x^(2n+1)]+n+1)/(1+[x^2]+|x|+2x), n in N` is equal to

Options

A. `n+1`
B. `n`
C. `1`
D. `0`

Explanation

To solve the limit

\[ \lim_{x \to 0^-} \frac{[x] + [x^2] + [x^3] + \ldots + [x^{2n+1}] + n + 1}{1 + [x^2] + |x| + 2x} \]

where \( n \in \mathbb{N} \), we will analyze the behavior of each term as \( x \) approaches \( 0 \) from the left.

### Step 1: Analyze the numerator

1. **Greatest Integer Function**: For \( x \to 0^- \), \( x \) is a negative number very close to \( 0 \). The greatest integer function \( [x] \) will be \( -1 \) because it rounds down to the nearest integer.

\[ [x] = -1 \]

2. **For \( x^2, x^3, \ldots, x^{2n+1} \)**: Since \( x^2 \) and all higher even powers will be positive and very small, their greatest integer will be \( 0 \):

\[ [x^2] = [x^3] = \ldots = [x^{2n}] = 0 \]

For odd powers (like \( x \) and \( x^3 \)), they will be negative and their greatest integer will also be \( -1 \):

\[ [x] = -1, \quad [x^3] = -1, \quad \ldots, \quad [x^{2n+1}] = -1 \]

3. **Counting Terms**: The total number of terms in the numerator is \( n + 1 \) (from \( [x], [x^3], \ldots, [x^{2n+1}] \)). The odd-indexed terms contribute \( -1 \) each, and there are \( n + 1 \) such terms.

Thus, the contribution from the odd powers is:

\[ -1 \times (n + 1) = -(n + 1) \]

4. **Final Numerator**: Therefore, the numerator becomes:

\[ [x] + [x^2] + [x^3] + \ldots + [x^{2n+1}] + n + 1 = -(n + 1) + (n + 1) = 0 \]

### Step 2: Analyze the denominator

1. **For \( [x^2] \)**: As established, \( [x^2] = 0 \).

2. **For \( |x| \)**: Since \( x \to 0^- \), \( |x| = -x \), which approaches \( 0 \) from the positive side.

3. **For \( 2x \)**: As \( x \to 0^- \), \( 2x \) approaches \( 0 \) from the negative side.

4. **Final Denominator**: Therefore, the denominator becomes:

\[ 1 + [x^2] + |x| + 2x = 1 + 0 + 0 + 0 = 1 \]

### Step 3: Combine the results

Now, substituting the results from the numerator and denominator into the limit:

\[ \lim_{x \to 0^-} \frac{0}{1} = 0 \]

### Conclusion

Thus, the limit is:

\[ \boxed{0} \]