How to develop a solution in limit without L'Hopital? 
Find $\displaystyle\lim_{x\to1}\left(\dfrac{1}{1-x}-\dfrac{1}{\ln x}\right).$

I don't know how to remove $\infty-\infty$ uncertainty in the question. Please explain it and how many different solution can we do in the question?    
 A: You could try to use Taylor Series expansions.
$\displaystyle\lim_{m\to0}\frac1m-\frac1{\ln(1-m)}, m=1-x$
$=\displaystyle\lim_{m\to0}\frac1m+\frac1{m+\frac{m^2}2+\frac{m^3}{3}...}=\displaystyle\lim_{m\to0}\frac1m(1+\frac1{1+\frac m2+\frac{m^2}3...})$ 
which diverges to $-\infty$ when $m\to0^-\implies x\to1^+$, and $+\infty$ when $m\to0^+\implies x\to1^-$.

A: Write ${1 \over 1-x} - {1 \over \log x} = {\log x +x-1\over (1-x) \log x}$.
The Taylor expansions are
$\log x +x-1 = 2(x-1)+\cdots$ and  $(1-x) \log x = -(x-1)^2+ \cdots $.
Hence we expect the behaviour for $x$ close to $1$ to be $\approx -{2 \over x-1}$.
A: I want to rewrite the limit slightly:
$$\lim_{x\to1} \left(\frac{1}{1-x} - \frac{1}{\ln(x)}\right) = -\lim_{x\to1} \left(\frac{1}{\ln(x)} + \frac{1}{x-1} \right). $$
Note that now our two summands have the same sign when $x<1$ and when $x>1$. The limit from the left goes to $(-\infty)+ (-\infty) = (-\infty)$ and the limit from the right goes to $\infty + \infty = \infty$. Thus, the limit does not exist.
No indefinite forms needed! 
A: We have that by $y=x-1 \to 0$
$$\lim_{x\to1} \left(\dfrac{1}{1-x}-\dfrac{1}{\ln x}\right) =\lim_{y\to0} \left(-\dfrac{1}{y}-\dfrac{1}{\ln (1+y)}\right)$$
and by standard limits since $\frac{\ln(1+y)}y\to 1$
$$-\dfrac{1}{y}-\dfrac{1}{\ln (1+y)}=-\dfrac{\ln(1+y)+y}{y\ln (1+y)}=-\dfrac{\frac{\ln(1+y)}y+1}{\ln (1+y)}\to \begin{cases}-\infty\quad y\to 0^+\\\\+\infty\quad y\to 0^- \end{cases}$$
A: I guess the first term should have been $\frac{1}{x-1}$ rather than $\frac{1}{1-x}$. In the former case the substitution $x=e^t$ reduces the problem to the evaluation of 
$$ \lim_{t\to 0}\left(\frac{1}{e^t-1}-\frac{1}{t}\right)=-\lim_{t \to 0}\frac{e^t-1-t}{t(e^t-1)}=-\lim_{t\to 0}\frac{\frac{t^2}{2}+O(t^3)}{t^2+O(t^3)}=\color{red}{-\frac{1}{2}}. $$
The Maclaurin series of $e^t$ is either a trivial consequence of the series definition of $e^t$ or a simple consequence of the dominated convergence theorem / integration by parts:
$$ e^t-1-t = t^2\int_{0}^{1}(1-x)e^{tx}\,dx \to t^2 \int_{0}^{1}(1-x)\,dx = \frac{t^2}{2}.$$
