Having the sequence $(a_{n})_{n\geq1}$, $a_{n}=\int_{0}^{1} x^{n}(1-x)^{n}dx$, find $\lim\limits_{n{\rightarrow}\infty} \frac{a_{n+1}}{a_{n}}$ Having the sequence $(a_{n})_{n\geq1}$, $a_{n}=\int_{0}^{1} x^{n}(1-x)^{n}dx$, find $\lim\limits_{n{\rightarrow}\infty} \frac{a_{n+1}}{a_{n}}$. Is this limit equal to $\lim\limits_{n{\rightarrow}\infty}(a_{n})^{\frac{1}{2}}$? I am unsure if I can apply l'Hospital to this limit, as the integral is a real value. I tried applying integration by parts but I ended nowhere.
Later edit: I am so sorry, it is $(1-x)^{n}$, not $(1-x^{n})$, I did this mistake out of tiredness.

Starting with integration by parts we have
$$\begin{aligned} \displaystyle a_n \displaystyle & = \int_0^1 x^n(1-x)^n\,{dx} \\& = \frac{1}{n+1}\int_0^1 (x^{n+1})'(1-x)^n\,{dx} \\& = \frac{x^{n+1}}{n+1}(1-x)^n\,\bigg|_0^1 -\frac{1}{n+1}\int_0^1 x^{n+1}[(1-x)^n]'\,{dx}  \\& = \frac{n}{n+1}\int_0^1 x^{n+1}(1-x)^{n-1}\,{dx}  \end{aligned} $$
But also by symmetry this is the same as $\displaystyle a_n = \frac{n}{n+1}\int_0^1 x^{n-1}(1-x)^{n+1}\,{dx}$, therefore 
$$\begin{aligned} \displaystyle 2a_{n} & =   \frac{n}{n+1}\int_0^1 x^{n+1}(1-x)^{n-1}\,d{x}+ \frac{n}{n+1}\int_0^1 x^{n-1}(1-x)^{n+1}\,{dx} \\& = \frac{n}{n+1}\int_0^1 (2x^2-2x+1)x^{n-1}(1-x)^{n-1}\,{dx} \\& = -\frac{2n}{n+1} a_n+\frac{n}{n+1}a_{n-1}\end{aligned}$$
Therefore $\displaystyle a_{n} = \frac{na_{n-1}}{4n+2}$, hence  $\displaystyle a_{n+1} = \frac{(n+1)a_{n}}{4n+6}$ so $\displaystyle \frac{a_{n+1}}{a_n} = \frac{(n+1)}{4n+6} \to \frac{1}{4}. $ 
