Sufficient bound to conclude limit has certain value. $\lim {\left( {\int_0^1 {{{dx} \over {1 + {x^n}}}} } \right)^n}=\frac 1 2 $ I am trying to show that
$$\lim {\left( {\int\limits_0^1 {{{dx} \over {1 + {x^n}}}} } \right)^n}=\frac 1 2 $$
Now, this can be done as follows. Using $x\mapsto x^{-1}$ we get that
$$\int\limits_0^1 {{{dx} \over {1 + {x^n}}}}  =\int\limits_1^\infty {{{nx^{n-1}} \over {1 + {x^n}}}} \frac{dx}{nx}$$
Upon integrating by parts we get
$$\int\limits_0^1 {{{dx} \over {1 + {x^n}}}}  =  - {{\log 2} \over n} + \int\limits_1^\infty  {{{\log \left( {1 + {x^n}} \right)} \over {n{x^2}}}dx} $$
or 
$$\int\limits_0^1 {{{dx} \over {1 + {x^n}}}}  = 1 - {{\log 2} \over n} + \int\limits_1^\infty  {{{\log \left( {1 + {x^n}} \right) - n} \over {n{x^2}}}dx} $$ since $\int_1^\infty x^{-2}dx=1$.
Using $x\mapsto x^{-1}$ once again on the RHS, we get
$$\int\limits_0^1 {{{dx} \over {1 + {x^n}}}}  = 1 - {{\log 2} \over n} + \int\limits_0^1 {{{\log \left( {1 + {x^n}} \right) - n\log x - n} \over n}dx} $$
But since $\int_0^1 \log x=-1=\int_0^1 dx$
$$\int\limits_0^1 {{{dx} \over {1 + {x^n}}}}  = 1 - {{\log 2} \over n} + {1 \over n}\int\limits_0^1 {\log \left( {1 + {x^n}} \right)dx} $$
Now, since for $x\geq 0$, $\log(1+x)\leq x$, we get $$\int\limits_0^1 {{{dx} \over {1 + {x^n}}}}  \le 1 - {{\log 2} \over n} + {1 \over n}\int\limits_0^1 {{x^n}dx}  = 1 - {{\log 2} \over n} + {1 \over {n\left( {n + 1} \right)}}$$ which means 
$$\int\limits_0^1 {{{dx} \over {1 + {x^n}}}}  = 1 - {{\log 2} \over n} + O\left( {{1 \over {{n^2}}}} \right)$$
Is this enough to conclude that 
$${\left( {\int\limits_0^1 {{{dx} \over {1 + {x^n}}}} } \right)^n} = {e^{ - \log 2}} = {1 \over 2}\text{ ? }$$
If so, how? 
 A: Your question can be recast as: if
$$
a_n=1+\frac{a}{n}+O\Bigl(\frac{1}{n^2}\Bigr),
$$
is it true that
$$
\lim_{n\to\infty}a_n^n=e^a?
$$
The answer is yes. Since $\log(1+x)=x+O(x^2)$ as $x\to0$, we have
$$
\log a_n=\frac{a}{n}+O\Bigl(\frac{1}{n^2}\Bigr).
$$
Then
$$
\lim_{n\to\infty}a_n^n=\lim_{n\to\infty}e^{n\log a_n}=e^a.
$$
A: $$
\begin{align}
\hspace{-2cm}\lim_{n\to\infty}\left(\int_0^1\frac{\mathrm{d}x}{1+x^n}\right)^n
&=\lim_{n\to\infty}\left(1-\int_0^1\frac{x^n\mathrm{d}x}{1+x^n}\right)^n\\
&=\lim_{n\to\infty}\left(1-\frac1n\int_0^1\frac{x\mathrm{d}x^n}{1+x^n}\right)^n\\
&=\lim_{n\to\infty}\left(1-\left[\frac1nx\log(1+x^n)\right]_0^1+\frac1n\int_0^1\log(1+x^n)\mathrm{d}x\right)^n\\
&=\lim_{n\to\infty}\left(1-\frac1n\log(2)+\frac1n\int_0^1\log(1+x^n)\mathrm{d}x\right)^n\\
&=\lim_{n\to\infty}\left(1-\frac1n\log(2)+\frac1n\int_0^1O(x^n)\,\mathrm{d}x\right)^n\tag{$\log(1+x^n)\le x^n$}\\
&=\lim_{n\to\infty}\left(1-\frac1n\log(2)+O\left(\frac1{n^2}\right)\right)^n\\
&=\lim_{n\to\infty}\left(1-\frac1n\left(\log(2)+O\left(\frac1n\right)\right)\right)^n\\
&=e^{-\log(2)}\\
\end{align}
$$
Note that $\lim\limits_{n\to\infty}f_n(x_n)=\lim\limits_{n\to\infty}f_n\left(\lim\limits_{n\to\infty}x_n\right)$ when $\{f_n\}$ are equicontinuous at $\lim\limits_{n\to\infty}x_n$.
In this case
$$
f_n(x)=\left(1-\frac xn\right)^n
$$
and
$$
x_n=\log(2)+O\left(\frac1n\right)
$$
A: $$\lim\limits_{n\rightarrow\infty}\left(\int^{1}_{0} \frac{dx}{1+x^{n}}\right)^{n} = e^{\lim\limits_{n\rightarrow\infty}\left(\int^{1}_{0} \frac{dx}{1+x^{n}}-1\right)n}$$
But $$\lim\limits_{n\rightarrow\infty}\left(\int^{1}_{0} \frac{dx}{1+x^{n}}-1\right)n = -\lim\limits_{n\rightarrow\infty}\left(\int^{1}_{0}x[\ln(1+x^{n})]'dx\right)=$$
$$ -\lim\limits_{n\rightarrow\infty}x\ln(1+x^{n})|^{1}_{0} + \lim\limits_{n\rightarrow\infty}\int^{1}_{0}\ln(1+x^{n})dx=-\ln 2$$ 
because $\ln(1+x^{n}) \leq x^{n}$ it involves $$\lim\limits_{n\rightarrow\infty}\int^{1}_{0}\ln(1+x^{n})dx = 0$$
The conclusion is immediate.
