If `I_(n)=int_(0)^(2)(2dx)/((1-x^(n)))`, then the value of `lim_(nrarroo)I_(n)` is equal to

Options

A. 1
B. 2
C. `(1)/(2)`
D. `(1)/(3)`

Answer

To solve the problem, we need to evaluate the limit of the integral

\[ I_n = \int_{0}^{2} \frac{2 \, dx}{1 - x^n} \]

as \( n \) approaches infinity.

### Step-by-step Solution:

1. **Split the Integral**:
We can split the integral into two parts:
\[ I_n = \int_{0}^{1} \frac{2 \, dx}{1 - x^n} + \int_{1}^{2} \frac{2 \, dx}{1 - x^n} \]

2. **Evaluate the First Integral**:
For the first part, as \( n \to \infty \), \( x^n \) approaches 0 for \( x \) in the interval \( [0, 1) \). Therefore:
\[ \lim_{n \to \infty} \frac{2}{1 - x^n} = 2 \]
Thus,
\[ \lim_{n \to \infty} \int_{0}^{1} \frac{2 \, dx}{1 - x^n} = \int_{0}^{1} 2 \, dx = 2 \cdot [x]_{0}^{1} = 2 \cdot (1 - 0) = 2 \]

3. **Evaluate the Second Integral**:
For the second part, \( x \) is in the interval \( [1, 2] \). Here, as \( n \to \infty \), \( x^n \) approaches infinity for \( x > 1 \). Therefore:
\[ \lim_{n \to \infty} \frac{2}{1 - x^n} \to 0 \]
Thus,
\[ \lim_{n \to \infty} \int_{1}^{2} \frac{2 \, dx}{1 - x^n} = 0 \]

4. **Combine the Results**:
Now, we can combine the results from both integrals:
\[ \lim_{n \to \infty} I_n = \lim_{n \to \infty} \left( \int_{0}^{1} \frac{2 \, dx}{1 - x^n} + \int_{1}^{2} \frac{2 \, dx}{1 - x^n} \right) = 2 + 0 = 2 \]

5. **Final Result**:
Therefore, the value of \( \lim_{n \to \infty} I_n \) is:
\[ \boxed{2} \]