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

by use of Taylor series when ${x \to 1}$ you'll get $\ln x=(x-1)-\frac{{(x - 1)^2}}{2x^2}$
$\mathop {\lim }\limits_{x \to 1} \frac{{x - 1 - (x - 1)+\frac{{(x - 1)^2}}{2x^2}}}{{{x(x - 1)}-\frac{{(x - 1)^2}}{2x} + 1 - x}}$
after factoring $(x-1)$ you'll have
$\mathop {\lim }\limits_{x \to 1} \frac{{\frac{{(x - 1)}}{2x^2}}}{x-\frac{{(x - 1)}}{2x}-1}$
factor another $(x-1)$
$\mathop {\lim }\limits_{x \to 1} \frac{{\frac{1}{2x^2}}}{1-\frac{1}{2x}}=1$
here's the graph to ensure.
https://www.google.com/webhp?sourceid=chrome-instant&ion=1&espv=2&ie=UTF-8#q=y%3D(x-1-ln(x))%2F(x*ln(x)%2B1-x)
