Q:

limits of function without using L'Hopital's Rule $\mathop {\lim }\limits_{x \to 1} \frac{{x - 1 - \ln x}}{{x\ln x+ 1 - x}} = 1$

Good morning. 
I want to show that without L'Hopital's rule :
$\mathop {\lim }\limits_{x \to 1} \frac{{x - 1 - \ln x}}{{x\ln x + 1 - x}} = 1$
I did the steps
$
\begin{array}{l}
 \mathop {\lim }\limits_{x \to 1} \frac{{x - 1 + \ln \left( x \right)}}{{x\ln \left( x \right) - x + 1}} = \mathop {\lim }\limits_{y \to 0} \frac{{y + \ln \left( {y + 1} \right)}}{{\left( {y + 1} \right)\ln \left( {y + 1} \right) - y}} \\ 
 \ln \left( {y + 1} \right) = 1 - \frac{{y^2 }}{2} + o\left( {y^2 } \right);and\quad \mathop {\lim }\limits_{y \to 0} o\left( {y^2 } \right) = 0 \\ 
  \Rightarrow \left( {y + 1} \right)\ln \left( {y + 1} \right) = 1 + y - \frac{{y^2 }}{2} + o\left( {y^2 } \right) \\ 
 \mathop {\lim }\limits_{y \to 0} \frac{{y + \ln \left( {y + 1} \right)}}{{\left( {y + 1} \right)\ln \left( {y + 1} \right) - y}} = \frac{{1 + y - \frac{{y^2 }}{2}}}{{1 + y - \frac{{y^2 }}{2}}} = 1 \\ 
 \end{array}
$
help me what you please

A:

Let $y= \dfrac{x-1-\ln x}{x\ln x - 1+ x} \to yx\ln x - y + xy = x - 1 - \ln x \to (xy + 1)\ln x = (y-1)(1-x) \to (xy+1)\cdot \dfrac{\ln x}{x-1} = 1 - y$.
Now using a well-known fact that: $\dfrac{\ln x}{x-1} \to 1$ when $x \to 1$. Taking limit as $x \to 1$ both sides of the above equation we have: 
$y + 1 = 1 - y \to y = 0$.
Note: The answer obtained by L'hopital rule is $y = 0$ , not $1$ as claimed.

