A) 1/2
B) 0
C) 1
D) 2
Correct Answer: A
Solution :
\[\underset{n\to \infty }{\mathop{\text{Lim}}}\,\int\limits_{{}}^{1}{\,\frac{n\cdot {{x}^{n-1}}}{1+x}}\,dx\] \[=\underset{n\to \infty }{\mathop{\lim }}\,\left[ \int\limits_{0}^{1}{\frac{1}{\underbrace{1+x}_{\text{I}}}\cdot \underbrace{n\cdot {{x}^{n-1}}}_{\text{II}}dx} \right]\] \[=\left. \frac{1}{1+x}\cdot {{x}^{n}} \right|_{0}^{1}+\int\limits_{0}^{1}{\,\frac{{{x}^{n}}}{{{(1+x)}^{2}}}dx}\] \[=\frac{1}{2}+0\] (as\[n\to \infty \])
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