Easy Limits

$\displaystyle \begin{aligned} \log\left(\lim_{n\to\infty}\prod_{r=0}^{n-1}\left[\dfrac{n-r}{n+r}\right]^{\dfrac{1}{n}}\right) &= \lim_{n\to\infty}\dfrac{1}{n}\sum_{r=0}^{n-1} \log\left(\dfrac{1-\dfrac{r}{n}}{1+\dfrac{r}{n}}\right)\\ &= \int_0^1 \log\left(\dfrac{1-x}{1+x}\right)\; dx \\ &= -\log 4\end{aligned}$

This makes the answer $\boxed{0.25}$

Relevant wiki: Riemann Sums

Similar solution with @Aditya Narayan Sharma 's

$\begin{aligned} L & = \lim_{n \to \infty} \prod_{r=0}^{n-1} \left(\frac {n-r}{n+r} \right)^\frac 1n \\ & = \lim_{n \to \infty} \exp \left(\ln \left( \prod_{r=0}^{n-1} \left(\frac {n-r}{n+r} \right)^\frac 1n \right)\right) \\ & = \exp \left({\color{#3D99F6} \lim_{n \to \infty} \frac 1n \sum_{r=0}^{n-1} \ln \left( \frac {1-\frac rn}{1+\frac rn}\right)}\right) & \small \color{#3D99F6} \text{By Riemann sums: }\lim_{n \to \infty} \frac 1n \sum_{k=a}^b f \left(\frac kn\right) = \int_{\lim_{n \to \infty} \frac an}^{\lim_{n \to \infty} \frac bn} f(x) \ dx \\ & = \exp \left({\color{#3D99F6} \int_0^1 \ln \left( \frac {1-x}{1+x}\right) dx}\right) \\ & = \exp \left(\int_0^1 (\ln (1-x) - \ln(1+x)) \ dx \right) \\ & = \exp \left(-(1-x)\ln (1-x)- (1+x)\ln(1+x) \ \bigg|_0^1 \right) \\ & = e^{-2\ln 2} = e^{\ln \frac 14} = \frac 14 = \boxed{0.25} \end{aligned}$