search
Log In
0 votes
124 views

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

  1. $0$
  2. $\ln 2$
  3. $\ln 3$
  4. $\infty$
in Calculus
recategorized by
124 views

2 Answers

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

Related questions

0 votes
0 answers
1
80 views
If $f$ is continuous in $[0,1]$ then $\displaystyle \lim_ {n \to \infty} \sum_{j=0}^{[n/2]} \frac{1}{n} f \left(\frac{j}{n} \right)$ (where $[y]$ is the largest integer less than or equal to $y$) does not exist exists and is equal to $\frac{1}{2} \int_0^1 f(x) dx$ exists and is equal to $ \int_0^1 f(x) dx$ exists and is equal to $\int_0^{1/2} f(x) dx$
asked Sep 23, 2019 in Calculus Arjun 80 views
0 votes
1 answer
2
167 views
Consider the function $f(x) = \begin{cases} \int_0^x \{5+ \mid 1-y \mid \} dy & \text{ if } x>2 \\ 5x+2 & \text{ if } x \leq 2 \end{cases}$ Then $f$ is not continuous at $x=2$ $f$ is continuous and differentiable everywhere $f$ is continuous everywhere but not differentiable at $x=1$ $f$ is continuous everywhere but not differentiable at $x=2$
asked Sep 23, 2019 in Calculus Arjun 167 views
0 votes
0 answers
3
106 views
Given that $\int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt{\pi}$, the value of $ \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-(x^2+xy+y^2)} dxdy$ is $\sqrt{\pi/3}$ $\pi/\sqrt{3}$ $\sqrt{2 \pi/3}$ $2 \pi / \sqrt{3}$
asked Sep 23, 2019 in Calculus Arjun 106 views
0 votes
0 answers
4
124 views
Let $0 < \alpha < \beta < 1$. Then $ \Sigma_{k=1}^{\infty} \int_{1/(k+\beta)}^{1/(k+\alpha)} \frac{1}{1+x} dx$ is equal to $\log_e \frac{\beta}{\alpha}$ $\log_e \frac{1+ \beta}{1 + \alpha}$ $\log_e \frac{1+\alpha }{1+ \beta}$ $\infty$
asked Sep 23, 2019 in Calculus Arjun 124 views
...