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

$$
\displaylines{\mathop {\lim }\limits_{_{x \to 0} } \frac{{e^x  - x - 1}}{{x^2 }} = \frac{1}{2} \cdots \left( 1 \right) \cr}
$$
$$
\displaylines{ \mathop {\lim }\limits_{_{t \to 0} } \left( {\frac{{t^2 }}{{te^t  - e^t  + 1}}} \right) = 2 \cdots \left( 2 \right) \cr}
$$
$$\displaylines{  \mathop {\lim }\limits_{_{x \to 1} } \frac{{x - 1 - \ln \left( x \right)}}{{x\ln \left( x \right) - x + 1}} = \mathop {\lim }\limits_{_{t \to 0} } \frac{{e^t  - t - 1}}{{te^t  - e^t  + 1}} \cr 
   = \mathop {\lim }\limits_{_{t \to 0} } \left( {\frac{{e^t  - t - 1}}{{t^2 }} \times \frac{{t^2 }}{{te^t  - e^t  + 1}}} \right) \cr 
   = \mathop {\lim }\limits_{_{t \to 0} } \left( {\frac{{e^t  - t - 1}}{{t^2 }}} \right) \times \mathop {\lim }\limits_{_{t \to 0} } \left( {\frac{{t^2 }}{{te^t  - e^t  + 1}}} \right) \cr 
   = \frac{1}{2} \times 2 = 1 \cr}
$$
