Q14 / 116 · NTA Abhyas (JEE Main) 2020 · Mathematics · Definite Integration

If $I_{n} = \int_{0}^{2} \frac{2 d x}{(1 - x^{n})}$ , then the value of $\lim_{n \to \infty} I_{n}$ is equal to

A
$1$
Correct answer
B
$2$
C
$\frac{1}{2}$
D
$\frac{1}{3}$

Solution

Ι_{n} = \int_{0}^{2} \frac{2 d x}{1 - x^{n}}

\underset{n \to \infty}{l i m} Ι_{n} = \underset{n \to \infty}{l i m} \int_{0}^{1} \frac{2 dx}{1 - x^{n}} + \underset{n \to \infty}{l i m} \int_{1}^{2} \frac{2 dx}{1 - x^{n}}

\lim_{n \to \infty} I_{n} = \underset{n \to \infty}{l i m} 2 \int_{0}^{1} \frac{d x}{1 - 0} + 0

= 2 (1 - 0) = 2

Hence, the correct answer is option: 'B'.

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