First, multiply by the conjugate of 1-x, which is \(\frac{1-x^2}{1+x}\).
Then, simplifying things, we get \(\lim\limits_{x\rightarrow1}(-\frac{x \ln x}{x-1})\).
Which then turns into this: \(-\lim _{x\to \:1}\left(\frac{\frac{1}{-1+x}\ln \left(x\right)}{\frac{1}{x}}\right) \\ \frac{\lim _{x\to \:1}\left(\frac{1}{-1+x}\ln \left(x\right)\right)}{\lim _{x\to \:1}\left(\frac{1}{x}\right)} \)
Using L'Hopital's rule and simplifying, we get that this equation is equal to 1.
We also need to find the value of \(\lim _{x\to \:1}\left(\frac{1}{x}\right)\) which is easily equal to 1.
Therefore your answer is \(-\frac{1}{1} = \boxed{-1}\).
