Finding $\lim_{x\to 1 }\left( \frac{x-1-x\ln x }{x(x-1)^{2}}\right)$ without L'Hopital's rule

Question

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

Answer 1

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.

Answer 2

We can proceed as follows \begin{align} L &= \lim_{x \to 1}\frac{x - 1 - x\log x}{x(x - 1)^{2}}\notag\\ &= \lim_{x \to 1}\frac{x - 1 - x\log x}{1\cdot(x - 1)^{2}}\notag\\ &= \lim_{t \to 0}\frac{t - (1 + t)\log (1 + t)}{t^{2}}\notag\\ &= \lim_{t \to 0}\frac{t - \log (1 + t) - t\log(1 + t)}{t^{2}}\notag\\ &= \lim_{t \to 0}\frac{t - \log (1 + t)}{t^{2}} - \frac{\log(1 + t)}{t}\notag\\ &= \lim_{t \to 0}\dfrac{t - \left(t - \dfrac{t^{2}}{2} + o(t^{2})\right)}{t^{2}} - 1\notag\\ &= \lim_{t \to 0}\left(\frac{1}{2} + o(1)\right) - 1\notag\\ &= \frac{1}{2} - 1 = -\frac{1}{2}\notag \end{align}