<s>[INST] 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
$$\space I_n(a)=\int_0^a\frac{x^n}{x^n+1}\,\mathrm{d}x , n\in N.$$


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

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


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*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.
 [/INST] Note that we have 
$$\begin{align}
\int_0^a \frac{x^n}{1+x^n}\,dx&=\int_0^1 \frac{x^n}{1+x^n}\,dx+\int_1^a \frac{x^n}{1+x^n}\,dx\\\\
&=\int_0^1 \frac{x^n}{1+x^n}\,dx+(a-1)-\int_1^a \frac{1}{1+x^n}\,dx
\end{align}$$
For $x\in [0,1]$, $0\le \frac{x^n}{1+x^n}\le x^n$ and for $x\in[1,a]$, $\frac{1}{1+x^n}\le \frac1{x^n}$.  Therefore, 
$$\left|\int_0^1 \frac{x^n}{1+x^n}\,dx\right|\le \frac1{n+1}$$
and
$$\left|\int_1^a \frac{1}{1+x^n}\,dx\right|\le \frac{1-a^{1-n}}{n-1}$$
Can you finish now?
</s>