Compute the limit $\lim_{n\to\infty} I_n(a)$ where $ I_n(a) :=\int_0^a \frac{x^n}{x^n+1}\,\mathrm{d}x, n\in N$. For $a>0$ we define
$$\space I_n(a)=\int_0^a\frac{x^n}{x^n+1}\,\mathrm{d}x , n\in N.$$


*

*Prove that $0\le I_n(1) \le \frac{1}{n+1}$

*Compute  $\lim_{n\to\infty} I_n(a)$
My attempt:


*

*I regard $I_n(1)=\int_0^1\frac{x^n}{x^n+1}$. If $x\in (0,1)$ then $x^n\in(0,1)$ and $x^n+1\in(1,2)$.
$$x^n>0 \Rightarrow x^n+1>1 \Rightarrow 1>\frac{1}{1+x^n }\Rightarrow x^n>\frac{x^n}{x^n+1}\Rightarrow \int_0^1\frac{x^n }{x^n+1}dx<\int_o^1 x^n \mathrm{d}x\\ \Rightarrow \int_0^1\frac{x^n }{x^n+1}dx<\frac{1}{n+1} \\ 0\le\frac{x^n}{x^n+1} \\ \text{In concusion } 0\le I_n(1) \le \frac{1}{n+1}.$$

*first case $a\in(0,1) \Rightarrow \lim_{n\to\infty} I_n(a) =0$. $I_n(a)\le\frac{1}{n+1})\text{case 2 . }a\in(1,\infty) \Rightarrow$ ???????
I don't believe the limit is $\infty$ because $\frac{x^n }{x^n+1}\le 1$.
I would appreciate some hints.

Lets consider the interval $(1, a)$. We have
$|\frac{x^n}{1 + x^n} - 1| = |\frac{1}{1+x^n}|$
By Bernoulli's Inequality, $x^n = (1 + (x - 1))^n \geq 1 + n(x - 1)$. This implies that
$|\frac{x^n}{1 + x^n} - 1| \leq \frac{1}{2 + n(x-1)}$
Now, fix $\epsilon > 0$ small enough and choose $n$ big enough such that $\frac{1}{2 + n(x-1)} < \frac{\epsilon}{2(a - 1)}$ for every $ x \in (1 + \frac{\epsilon}{2}, a)$. We then have
$\int_1^a |\frac{x^n}{1 + x^n} - 1| dx = \int_1^a |\frac{1}{1+x^n}| dx = $
$\int_{1}^{1 + \frac{\epsilon}{2}} |\frac{1}{1+x^n}| dx + \int_{1 +\frac{\epsilon}{2}}^{a} |\frac{1}{1+x^n}| dx \leq $
$\int_{1}^{1 + \frac{\epsilon}{2}} 1 dx + \int_{1 +\frac{\epsilon}{2}}^{a} \frac{1}{2 + n(x - 1)} dx < \epsilon $
Now separate the original integral in $(0, 1)$ and $(1, a)$, and conclude.
