Having the sequence $(a_{n})_{n\geq1}$, $a_{n}=\int_{0}^{1} x^{n}(1-x)^{n}dx$, find $\lim\limits_{n{\rightarrow}\infty} \frac{a_{n+1}}{a_{n}}$

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}. $

If the integrand is $x^n (1-x^n)$, you can integrate directly, as stated in the comments. But if it is $x^n (1-x)^n$, then this is the Beta Function, and we can rewrite it in terms of the gamma function as $$a_n = \frac{\Gamma (n+1) \Gamma(n+1)}{\Gamma(2n+2)}$$ And so $$a_{n}/a_{n+1} = \frac{\Gamma(n+1)^2}{\Gamma(n+2)^2} \cdot \frac{\Gamma(2n+4)}{\Gamma(2n+2)} = \frac{1}{(n+1)^2} \cdot \frac{(2n+3)(2n+2)}{1}$$ And the limit of this is simply 4.

This would not be equal to the limit of $a_n^{1/2}$, as that would just be $0$, the same limit as $a_n$.

You are considering $$ a_n = \int_0^1 x^n(1-x)^n\,{dx} $$ Let $x=\sin^2(t)$ to make $$a_n=2 \int_0^{\frac \pi 2}\sin ^{2 n+1}(t) \cos ^{2 n+1}(t)\,dt=B(n+1,n+1)=\frac{\sqrt{\pi } \,\, \Gamma (n+1)}{2^{2 n+1}\Gamma \left(n+\frac{3}{2}\right)}$$ from which follows NoName's result for ${a_{n + 1} \over a_{n}}$.

Interesting would be to take the logarithm of $a_n$, to use Stirling approximation for the gamma function and continuing with Taylor expansion to get $$\log(a_n)=-2 n \log (2)+\left(\frac{1}{2} \log \left(\frac{1}{n}\right)+\log \left(\frac{\sqrt{\pi }}{2}\right)\right)-\frac{3}{8 n}+O\left(\frac{1}{n^2}\right)$$ which makes $$a_n \sim 2^{-(2 n+1)} \sqrt{\frac{\pi}{n}}\, \left(1-\frac{3}{8 n}+\frac{25}{128 n^2}+O\left(\frac{1}{n^3}\right) \right)$$ which is quite accurate.