Can be solved without L'Hopital? $\lim_{x\to 1} \frac{x-1-\ln x}{x\ln x+1-x}=?$

Question

I would appreciate if somebody could help me with the following problem:

Q: Can be solved without L'Hopital? $$\lim_{x\to 1} \frac{x-1-\ln x}{x\ln x+1-x}=?$$

Answer 1

Hint: Use $\ln(x) = \ln(1+(x-1)) = (x-1)-\dfrac{(x-1)^2}{2} + \dfrac{(x-1)^3}{3} - \cdots $, and $x\ln(x) = ((x-1)+1)\ln(x) = (x-1)\ln(x) + \ln(x)$.

Answer 2

Let $1-x=h\iff x=1-h$

So, we have $$\lim_{h\to0}\frac{h-\ln(1-h)}{(1-h)\ln(1-h)+h}$$

$$=\lim_{h\to0}\frac{1+\dfrac{\ln(1-h)}{-h}}{-(1-h)\cdot\dfrac{\ln(1-h)}{-h}+1}$$

use $\lim_{u\to0}\dfrac{\ln(1+u)}u=1$