$\lim_{n\to\infty}(\int_0^1 \frac{1}{1+x^n} dx)^n$ $$\lim_{n\to\infty}\left(\int_0^1 \frac{1}{1+x^n} dx\right)^n$$
I solved this problem as in the link:
Click here
I got the answer as $0$ but it says incorrect. I think the answer is wrong or there is some error with the question.
My try:
$$=\lim_{n\to\infty}\left(\int_0^1 \sum_{m=0}^{\infty} (-1)^mx^{mn}dx\right)^n$$
$$=\lim_{n\to\infty}\left( \sum_{m=0}^{\infty} \frac{(-1)^m}{1+mn}\right)^n$$
$$=\color{blue}{\lim_{n\to\infty}\left( \frac{\Phi(-1,1,\frac1n)}{n}\right)^n=0 ????}$$
Do you think the problem is incorrect or answer is incorrectly provided by the author ?
 A: You may just use

*

*$\lim_{t\to 0}(1-t)^{\frac 1t} = e^{-1}$
Now, write
\begin{eqnarray*}\left(\int_0^1 \frac{1}{1+x^n} dx\right)^n
& = & \left(1- \underbrace{\int_0^1 \frac{x^n}{1+x^n} dx}_{=:J_n}\right)^n
\end{eqnarray*}
Using integration by parts you get
$$J_n = \frac 1n\int_0^1 x\frac{nx^{n-1}}{1+x^n} dx = \frac 1n\left(\ln 2 - \underbrace{\int_0^1 \ln (1+x^n) dx}_{=:I_n}\right)$$
Now, DCT (or just squeezing) gives immediately
$$\lim_{n\to \infty }I_n = 0$$
All together
\begin{eqnarray*}\left(\int_0^1 \frac{1}{1+x^n} dx\right)^n
& = & \left(\left(1-\frac{\ln 2 - I_n}n\right)^{\frac{n}{\ln 2 - I_n}}\right)^{\ln 2 - I_n} \\
& \stackrel{n\to\infty}{\longrightarrow} & \left(e^{-1}\right)^{\ln2} = \frac 12
\end{eqnarray*}
A: Let $I_n$ be given by the integral $I_n=\int_0^1 \frac1{1+x^n}\,dx$.  Enforcing the substitution $x\mapsto x^{1/n}$ and reveals
$$\begin{align}
I_n&=\frac1n \int_0^1 \frac{x^{1/n}}{x(1+x)}\,dx\tag1
\end{align}$$

Using partial fraction expansion on the integrand in $(1)$, we obtain
$$\begin{align}
I_n&=\frac1n \int_0^1 x^{1/n-1}\,dx-\frac1n \int_0^1 \frac{x^{1/n}}{1+x}\,dx\\\\
&=1-\frac1n \int_0^1 \frac{x^{1/n}}{1+x}\,dx\tag2
\end{align}$$

Next, using $1+x\le e^x\le \frac1{1-x}$ for $x<1$ provides the estimates
$$e^{\frac1n\log(x)}-1\ge \frac{\log(x)}{n}$$
and for $x\in (0,1)$
$$\begin{align}
e^{\frac1n\log(x)}-1\le \frac{\frac{\log(x)}{n} }{1-\frac{\log(x)}{n}}\le \frac{\log(x)}{n}+\left(\frac{\log(x)}{n} \right)^2
\end{align}$$
Using these estimates in $(2)$, we can assert that
$$I_n=1-\frac{\log(2)}{n}+O\left(\frac1{n^2}\right)\tag3$$

Finally, raising both sides of $(3)$ to the $n$'th power and letting $n\to \infty$ yields the coveted limit
$$\begin{align}
\lim_{n\to \infty}I_n^n&=\lim_{n\to \infty}\left(1-\frac{\log(2)}n+O\left(\frac1{n^2}\right)\right)^n\\\\
&=\frac12
\end{align}$$
A: Picture of MMA input as requested:

A: Your answer cannot be correct.
Since, by your argument,  $$\int_{0}^{1} \frac{dx}{1+x^n}>1-\frac{1}{1+n}$$
A lower bound for your limit is:
$$\left(1-\frac1{n+1}\right)^n\to \frac1e$$
