This is not the algebraic answer that I am sure you need, but....

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 \(-\frac{1}{1} = \boxed{-1}\).

I always forget about L'hopital's rule - thanks for you answer and thanks for reminding me.

Bur....

Why not just use l'hopital's rule to start with?

\(\displaystyle\lim_{x\rightarrow 1}\;\;\frac{xlnx}{1-x}\\ =\displaystyle\lim_{x\rightarrow 1}\;\;\frac{lnx+1}{-1}\\ =-1\)