\(\displaystyle\lim_{\underset{>}{x\to 0}}\sqrt[x]{\frac{1+x}{1-x}}=\lim_{\underset{>}{x\to 0}}\left(\frac{1+x}{1-x}\right)^\frac1x =\frac{ \displaystyle\lim_{\underset{>}{x\to 0}}\:(1+x)^\frac1x}{\displaystyle\lim_{\underset{>}{x\to 0}}\:(1-x)^\frac1x}=\frac{e}{e^{-1}}=e^2\)