For $n=1,2,3,\cdots $, $f_n(x)=\frac{2nx^{n-1}}{1+x}$ for $x\in [0,1]$ then $\lim_{n\to \infty}\int_0^1f_n(x)=?$

Question

For $n=1,2,3,\cdots $, let $f_n(x)=\frac{2nx^{n-1}}{1+x}$ for $x\in [0,1]$ then $\lim_{n\to \infty}\int_0^1f_n(x)=?$

My attempt:

Since, $0\leq x\leq 1$, so $1\leq x+1 \leq 2 \implies \frac{1}{2} \leq \frac{1}{x+1} \leq 1$.

Now, $\frac{1}{2}2nx^{n-1}\leq f_n(x) \leq 2nx^{n-1}$

Integrating we get,

$1\leq \int_0^1f_n(x)\leq 2$

Taking $\lim_{n\to \infty}$ we get

$1\leq \lim_{n\to \infty} \int_0^1f_n(x)\leq 2$

But I am stuck with the actual value of the limit. Can anyone please help me? Thanks in advance.

Answer 1

We have $$1\leftarrow\frac{n}{n-1}=\int_0^1 nx^{n-2}dx=\int_0^1 \frac{2nx^{n-1}}{2x}dx\geq\int_0^1 f_n dx \geq \int_0^1 \frac{2nx^{n-1}}{2}dx=\int_0^1 nx^{n-1}dx=1$$ hence $$\lim_{n\to \infty} \int_0^1 f(x)dx =1$$