How to solve $\lim_{x\to1}\left[\frac{x}{x-1}-\frac{1}{\ln(x)}\right]$ without using L'Hospital's rule? $$\lim_{x\to1}\left[\frac{x}{x-1}-\frac{1}{\ln(x)}\right]$$
How can I solve this without using the L'Hopital's rule? Any tips or hints would be greatly appreciated. I tried using the substitution $x=e^y$, but to no avail.

One may recall that, as $u \to0$, by the Taylor expansion,
$$
\frac1{1+u}=1-u+u^2+O(u^3)\tag1
$$ 
$$
\ln(1+u)=u-\frac12 u^2+O(u^3).\tag2
$$ Set $u:=x-1$ in your initial expression, we have $u \to 0$ as $x \to 1$, and we get
$$
\frac{x}{x-1}-\frac{1}{\ln(x)}=\frac{1+u}u-\frac{1}{\ln(1+u)}=1+\frac1u-\frac{1}{\ln(1+u)}. \tag3
$$ Then, using $(2)$ and $(1)$,
$$
\begin{align}
\frac{1}{\ln(1+u)}&=\frac{1}{u-\frac12 u^2+O(u^3)}\\
&=\frac1u\times\frac{1}{1-u/2+O(u^2)}\\
&=\frac1u\times\left( 1+\frac{u}2+O(u^2)\right)\\
&=\frac1u+\frac12+O(u) \tag4
\end{align}
$$ From $(4)$, as $u \to0$,  you deduce
$$
\frac1u-\frac{1}{\ln(1+u)} \to \color{red}{-}\frac12
$$ inserting in $(3)$ gives

$$\lim_{x\to1}\left[\frac{x}{x-1}-\frac{1}{\ln(x)}\right]=\frac12.$$

