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.
A: 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)
A: $
\displaylines{
  \left\{ \begin{array}{l}
 t = 1 + u \\ 
 u \cong \ln t \\ 
 \end{array} \right. \cr 
   \Rightarrow  \cr 
  \mathop {\lim }\limits_{t \to 1} \left[ {\frac{{\left( {1 + t} \right)\ln t}}{{t\ln t - t + 1}}} \right] = \mathop {\lim }\limits_{u \to 2} \left[ {\frac{{\left( {2 + u} \right)u}}{{\left( {1 + u} \right)u - u}}} \right] \cr 
   = \mathop {\lim }\limits_{u \to 2} \left[ {\frac{{2u + u^2 }}{{u^2 }}} \right] = 2 \cr 
  \mathop {\lim }\limits_{t \to 1} \frac{{\ln t + t - 1}}{{t\ln t - t + 1}} =  - \mathop {\lim }\limits_{t \to 1} \left[ {\frac{{ - \ln t - t\ln t + t\ln t - t + 1}}{{t\ln t - t + 1}}} \right] \cr 
   =  - 1 + \mathop {\lim }\limits_{t \to 1} \left[ {\frac{{\left( {1 + t} \right)\ln t}}{{t\ln t - t + 1}}} \right] \cr 
   =  - 1 + 2 = 1 \cr 
  \mathop {\lim }\limits_{t \to 1} \frac{{\ln t + t - 1}}{{t\ln t - t + 1}} = 1 \cr}
$
othere way
\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}
A: $$
\displaylines{\mathop {\lim }\limits_{_{x \to 0} } \frac{{e^x  - x - 1}}{{x^2 }} = \frac{1}{2} \cdots \left( 1 \right) \cr}
$$
$$
\displaylines{ \mathop {\lim }\limits_{_{t \to 0} } \left( {\frac{{t^2 }}{{te^t  - e^t  + 1}}} \right) = 2 \cdots \left( 2 \right) \cr}
$$
$$\displaylines{  \mathop {\lim }\limits_{_{x \to 1} } \frac{{x - 1 - \ln \left( x \right)}}{{x\ln \left( x \right) - x + 1}} = \mathop {\lim }\limits_{_{t \to 0} } \frac{{e^t  - t - 1}}{{te^t  - e^t  + 1}} \cr 
   = \mathop {\lim }\limits_{_{t \to 0} } \left( {\frac{{e^t  - t - 1}}{{t^2 }} \times \frac{{t^2 }}{{te^t  - e^t  + 1}}} \right) \cr 
   = \mathop {\lim }\limits_{_{t \to 0} } \left( {\frac{{e^t  - t - 1}}{{t^2 }}} \right) \times \mathop {\lim }\limits_{_{t \to 0} } \left( {\frac{{t^2 }}{{te^t  - e^t  + 1}}} \right) \cr 
   = \frac{1}{2} \times 2 = 1 \cr}
$$
