TITLE: Limit as $x$ approaches $1$ from the right of $\frac{1}{\ln x}-\frac{1}{x-1}$
QUESTION [6 upvotes]: $$
\lim_{x\rightarrow 1^+}\;\frac{1}{\ln x}-\frac{1}{x-1}
$$
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.

REPLY [2 votes]: 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
$$ \lim_{x\to 1^+} \left( \frac{1}{\log x} + \frac{1}{1-x} \right) = \int_{0}^{1} \lim_{x\to 1^+} \frac{1-x^{t}}{1-x} \, dt = \int_{0}^{1} t \, dt = \frac{1}{2}. $$