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?
 A: It is much simpler to use Taylor polynomials at order $2$.
Set $x=1+h$. The 
\begin{align*}
\frac{1}{\ln(x)}  - \frac{1}{x - 1}&=\frac{x-1-\ln x}{(x - 1)\ln(x)}=\frac{h-\ln(1+h)}{h\ln(1+h)}=\frac{h-\bigl(h-\frac12h^2+o(h^2)\bigr)}{h\bigl(h+o(h\bigr)}\\
&=\frac{\frac12h^2+o(h^2)}{h^2+o(h^2)}=\frac{\frac12+o(1)}{1+o(1)}\to\frac12.
\end{align*}
A: Keep using L'hopital's rule if you have indeterminate forms $\left( \frac{0}{0} \right)$ 
$$\lim_{x\to1^+}\left( \frac{1}{\ln(x)}  - \frac{1}{x - 1} \right)=\lim_{x\to1^+}\left(\frac{(x-1)-\ln(x)}{\ln(x)(x-1)}\right)$$
Using L'hopital as we have indeterminate form
$$\lim_{x\to1^+}\left(\frac{(x-1)-\ln(x)}{\ln(x)(x-1)}\right)=\lim_{x\to1^+}\left(\frac{1-\frac1x}{\frac{(x-1)}{x}+\ln(x)}\right)$$
$$=\lim_{x\to1^+}\left(\frac{x-1}{x-1+x\ln(x)}\right)$$
Still indeterminate type so use L'hopital again
$$\lim_{x\to1^+}\left(\frac{x-1}{x-1+x\ln(x)}\right)=\lim_{x\to1^+}\left(\frac{1}{2+\ln(x)}\right)$$
$$=\color{blue}{\fbox{0.5}}$$
So altogether L'hopital was used twice.
A: We know $\lim_{x\to 1^+}((x-1)-\ln (x))=0$ and $\lim_{x\to 1^+}\ln(x)(x-1)=0$, because if $\lim_{x\to a}f(x)$ and $\lim_{x\to a}g(x)$ are finite, then $\lim_{x\to a}(f(x)+g(x))=\lim_{x\to a}f(x)+\lim_{x\to a}g(x)$ and $\lim_{x\to a}f(x)g(x)=\lim_{x\to a}f(x)\lim_{x\to a}g(x)$.
So we can apply L'Hopital's rule on 
$$\lim_{x\to 1^+}\frac{(x-1)-\ln(x)}{\ln(x)(x-1)}=\lim_{x\to 1^+}\frac{1-\frac{1}{x}}{\frac{x-1}{x}+\ln(x)},$$
which is again of the form $\frac{0}{0}$ for similar reasons, so we can apply L'Hopital's rule again:
$$=\lim_{x\to 1^+}\frac{\frac{1}{x^2}}{\frac{1}{x^2}+\frac{1}{x}}=\frac{\frac{1}{1^2}}{\frac{1}{1^2}+\frac{1}{1}}=\frac{1}{2}$$
A: 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
$$
\lim_{t\to0^+}\frac{t-\ln(1+t)}{t^2}=
\lim_{t\to0^+}\frac{1-\dfrac{1}{1+t}}{2t}=
\lim_{t\to0^+}\frac{t}{2t(1+t)}=\frac{1}{2}
$$
