Finding $\lim_{x\to 1 }\left( \frac{x-1-x\ln x }{x(x-1)^{2}}\right)$ without L'Hopital's rule I would like to calculate this limit but without using L'Hopital's rule
$$\lim_{x\to 1 }\left( \frac{x-1-x\ln(x) }{x(x-1)^{2}}\right)=-\dfrac{1}{2}$$
My thoughts:
Note that $\lim\limits_{x\to 1 }\dfrac{\ln(x)}{x-1}=1$ so I tried :
\begin{align*}
\lim_{x\to 1 }\left( \dfrac{x-1-x\ln(x) }{x(x-1)^{2}}\right)&=\lim_{x\to 1 }\left( \dfrac{1}{x(x-1)}-\dfrac{\ln(x)}{(x-1)^2}\right) \\
&=\lim_{x\to 1 }\left( \dfrac{1}{x(x-1)}\right)-\lim_{x\to 1 }\left(\dfrac{1}{(x-1)}\cdot\dfrac{\ln(x)}{(x-1)}\right) 
\end{align*}
but with no luck

Let us first make $x=y+1$. So, $$\dfrac{x-1-x\ln(x) }{x(x-1)^{2}}=\frac{y-(y+1) \log (y+1)}{y^2 (y+1)}$$ Now, consider Taylor expansion around $y=0$ $$\log(1+y)=y-\frac{y^2}{2}+\frac{y^3}{3}+O\left(y^4\right)$$ So, the numerator write $$y-(y+1) \log (y+1)=-\frac{y^2}{2}+\frac{y^3}{6}+O\left(y^4\right)$$
I am sure that you can take from here.
