find $ \lim\limits_{n\to\infty}\int_{0}^{1}1/(1+x^{n})\,dx $ First of all, English is not my native language, but Chinses is. I tried to spilt the integration interval into 2 pieces: $ [0, 1-1/n] $ and $ [1-1/n, 1] $. In both intervals I use the mean value theorem:
    $$
 \int_{0}^{1-1/n}\frac{1}{1+x^{n}}\,dx=\frac{1}{1+\xi_{n}^{n}}\left( 1-\frac{1}{n} \right), \qquad \text{and} \qquad \int_{1-1/n}^{1}\frac{1}{1+x^{n}}\,dx=\frac{1}{1+\eta_{n}^{n}}\frac{1}{n},
 $$
    where $ \xi_{n}\in(0, 1-1/n), \eta_{n}\in(1-1/n, 1) $.I found that the latter formula has a limit of $ 0 $ when $ n\to\infty $. However I can't handle the previous formula. Does anyone has some thoughts? 
 A: $\int_0^{1}\frac  1 {1+x^{n}}dx =1-\int_0^{1}\frac  {x^{n}} {1+x^{n}}dx$. Note that $0 \leq \frac {x^{n}} {1+x^{n}} \leq x^{n}$ and $\int_0^{1} x^{n}dx=\frac 1 {n+1} \to 0$. Put these together to see that the required limit is $1$. 
A: Consider limit 
$$\lim_{n\to\infty}\int_{0}^{1}\frac{x^n}{1+x^{n}}\,dx=0.$$
then your limit 
$$\lim_{n\to\infty}\int_{0}^{1}\frac{1}{1+x^{n}}\,dx=1.$$
Hint：
$$0\leq\frac{x^n}{1+x^{n}}\leq x^n\implies
\lim_{n\to\infty}\int_{0}^{1}\frac{x^n}{1+x^{n}}\,dx=0.$$
A: Consider $\int_0^{1-n^{-1/2}}\frac1{1+x^n}\,dx$ and $\int_{1-n^{-1/2}}^1\frac1{1+x^n}\,dx$ instead. The second integral is still bounded above by $n^{-1/2}$.
For the first integral, $$1\ge \int_0^{1-n^{-1/2}}\frac1{1+x^n}\,dx=\frac{1}{1+\xi_n^{n}}(1-n^{-1/2})\ge \frac1{1+(1-n^{-1/2})^n}(1-n^{-1/2})$$
However, $\left(1-n^{-1/2}\right)^n=\left(\left(1-n^{-1/2}\right)^{n^{1/2}}\right)^{n^{1/2}}\stackrel{n\to\infty}{\longrightarrow} \left[\left(e^{-1}\right)^{\infty}\right]=0$, therefore $$\liminf_{n\to\infty}\int_0^{1-n^{-1/2}}\frac1{1+x^n}\,dx\ge \frac1{1+0}(1-0)=1$$
