Limit as $x$ approaches $1$ from the right of $\frac{1}{\ln x}-\frac{1}{x-1}$ $$
\lim_{x\rightarrow 1^+}\;\frac{1}{\ln x}-\frac{1}{x-1}
$$
So I would just like to know how to begin to solve this limit, or what topic does this problem fall under so that I can search for examples online or in text. 
I know the answer is 1/2 and I have tried plugging in numbers  smaller or greater than one but as I assumed that was the wrong way to go. 
 A: No need to restrict to the right limit. By variable change $u=x-1$, this is 
$$
\lim_{u\rightarrow 0}\;\frac{1}{\ln (1+u)}-\frac{1}{u}=\lim_{u\rightarrow 0}\;\frac{u-\ln (1+u)}{u\ln(1+u)}.
$$
1) L'Hospital: if we differentiate once top and bottom, we get
$$
\frac{u}{(1+u)\ln(1+u)+u}.
$$
Once more, we find
$$
\frac{1}{\ln(1+u)+2}\longrightarrow \frac{1}{2}.
$$
So the initial limit is indeed $\frac{1}{2}$ by a double application of L'Hospital.
2) Taylor: we can use that $\ln(1+u)=u-\frac{u^2}{2}+O(u^3)$. Then
$$
\frac{u-\ln(1+u)}{u\ln(1+u)}=\frac{\frac{u^2}{2}+O(u^3)}{u^2+O(u^3)}=\frac{\frac{1}{2}+O(u)}{1+O(u)}\longrightarrow \frac{1}{2}.
$$
A: with attention to hopital rule we have:
$$
\lim_{x\to1^+}\frac{x-1-\ln x}{(x-1)\ln x}=
\lim_{x\to 1^+}\frac{1-\frac{1}{x}}{\ln x+\frac{x-1}{x}}=
\lim_{x\to 1^+}\frac{\frac{1}{x^2}}{\frac{1}{x}+\frac{1}{x^2}}=\frac{1}{2}
$$
A: It is generally easier to evaluate these problems if you can express the limit as one fraction. In this case, we have
$$
\frac{1}{\ln x} - \frac{1}{x-1} = \frac{x-1-\ln x}{\ln x(x-1)}.
$$
Applying L'Hopital's rule, we have
$$
\lim_{x\to1^+} \frac{1-\frac{1}{x}}{\frac{1}{x}(x-1) + \ln x}.
$$
We still have a limit of the form $\frac{0}{0}$, so again apply the rule again, and obtain
$$
\lim_{x\to1^+}\frac{\frac{1}{x^2}}{\frac{-1}{x^2}(x-1) + \frac{1}{x} + \frac{1}{x}} = \lim_{x\to1^+} \frac{x}{2x^2} = \frac{1}{2}.
$$
A: If you need some fun, you may solve like this: Since
$$ \frac{1}{\log x} + \frac{1}{1-x} = \int_{0}^{1} \frac{1-x^{t}}{1-x} \, dt, $$
we have
$$ \lim_{x\to 1^+} \left( \frac{1}{\log x} + \frac{1}{1-x} \right) = \int_{0}^{1} \lim_{x\to 1^+} \frac{1-x^{t}}{1-x} \, dt = \int_{0}^{1} t \, dt = \frac{1}{2}. $$
A: You can apply l'Hôpital's rule twice:
$$ \lim_{x\rightarrow 1^+}\frac{1}{\ln x}-\frac{1}{x-1} = \lim_{x\rightarrow 1^+}\frac{x-1-\ln x}{\ln x(x-1)} = \lim_{x\rightarrow 1^+}\frac{1-1/x}{1/x(x-1)+\ln x} = \lim_{x\rightarrow 1^+}\frac{x-1}{x-1+x\ln x} = \lim_{x\rightarrow 1^+}\frac{1}{1+\ln x +1} = \frac{1}{2}.$$
