Applying L'Hopital on $\lim_{x\to1^+}\left( \frac{1}{\ln(x)} - \frac{1}{x - 1} \right)$ I am trying to apply L'Hopital's rule here:
$$\lim_{x\to1^+}\left( \frac{1}{\ln(x)}  - \frac{1}{x - 1} \right)$$
The indeterminate form here is $\infty - \infty$, so I need to somehow shape this limit to have a quotient instead. My attempt:
$$\frac{1}{\ln(x)}  - \frac{1}{x - 1} = \frac{\frac{(x-1)-\ln(x)}{\ln(x)}}{(x-1)}$$
This becomes
$$\frac{(x-1)-\ln(x)}{\ln(x)\cdot(x-1)}$$
That's
$$\frac{0^+-0^+}{0^+\cdot0^+}$$
Hmmmm... Does this become $\frac{0}{0}$? I am not sure.
To me, $0^+$ means "some very small number greater than $0$", but two $0^+$ don't necessarily refer to the same number... or do they? Because if they don't, then $0^+ - 0^+ \not= 0$.
The same goes for $0^+ \cdot 0^+$, they are numbers greater than $0$ so their product cannot be $0$.
What do you think? Or am I overthinking it?

If you make a single fraction, you get
$$
\frac{x-1-\ln x}{(x-1)\ln x}
$$
Both the numerator and the denominator are continuous at $1$, so this limit is of the form $\frac{0}{0}$. No need to consider signs, at this point.
You can avoid long applications of l'Hôpital with a simple observation:
$$
\lim_{x\to1^+}\frac{x-1-\ln x}{(x-1)\ln x}=
\lim_{t\to0^+}\frac{t-\ln(1+t)}{t\ln(1+t)}
$$
with the substitution $x-1=t$. Now you can rewrite the last limit as
$$
\lim_{t\to0^+}\frac{t-\ln(1+t)}{t^2}\frac{t}{\ln(1+t)}
$$
and you should know that the second fraction has limit $1$; so we have
$$
\lim_{t\to0^+}\frac{t-\ln(1+t)}{t^2}=
\lim_{t\to0^+}\frac{1-\dfrac{1}{1+t}}{2t}=
\lim_{t\to0^+}\frac{t}{2t(1+t)}=\frac{1}{2}
$$
