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The value of $$\displaystyle \lim_{n \to \infty} \left[ (n+1) \int_0^1 x^n \ln(1+x) dx \right]$$ is

- $0$
- $\ln 2$
- $\ln 3$
- $\infty$

0 votes

**Answer: $B$ **

$$\underset{n \to \infty}{\lim} \int^1_0x^n\ln(1+x)dx$$

can be simplified to:

$$\ln(2)-\underset{n \to \infty}{\lim}\int^1_0\frac{x^{n+1}}{1 + x}dx$$

$$\because \bigg |\int^1_0\frac{x^{n+1}}{1+x}dx\bigg | \le \int^1_0 \big |x^{n+1} \big|dx$$

Now, here the bound $1+x$ is always greater than equal to $1$

$\Rightarrow \frac{1}{1+x} \le1$

$$\therefore \underset{n \to \infty}{\lim}\int^1_0 x^n \ln(1+x)dx = \ln(2)$$

Thus, $B$ be is the correct option.

0 votes

Let $u= x^{n+1}$. Then the integral becomes

$$\lim_{n\to \infty} \int_0^1 \log\left(1+u^{\frac{1}{n+1}}\right)\:du = \int_0^1\log(1+1)\:du = \log 2$$

by dominated convergence.

for dominated convergence. u can check : https://www.math3ma.com/blog/dominated-convergence-theorem