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 1: Split the Integral
We can split the integral into two parts:

\[ I_n = \int_0^1 \frac{2 \, dx}{1 - x^n} + \int_1^2 \frac{2 \, dx}{1 - x^n} \]

### Step 2: Evaluate the First Integral
For the first integral, as \( n \to \infty \), \( x^n \) approaches 0 for \( x \in [0, 1) \). Thus, we have:

\[ \lim_{n \to \infty} \int_0^1 \frac{2 \, dx}{1 - x^n} = \int_0^1 \frac{2 \, dx}{1 - 0} = \int_0^1 2 \, dx \]

Calculating this integral:

\[ \int_0^1 2 \, dx = 2 \left[ x \right]_0^1 = 2(1 - 0) = 2 \]

### Step 3: Evaluate the Second Integral
For the second integral, \( x \) is in the range \( [1, 2] \). As \( n \to \infty \), \( x^n \) approaches infinity for \( x > 1 \), hence \( 1 - x^n \) approaches negative infinity. Therefore, the integrand approaches 0:

\[ \lim_{n \to \infty} \int_1^2 \frac{2 \, dx}{1 - x^n} = 0 \]

### Step 4: Combine the Results
Now, combining both parts, we have:

\[ \lim_{n \to \infty} I_n = 2 + 0 = 2 \]

### Final Result
Thus, the value of

\[ \lim_{n \to \infty} I_n = 2 \]