I have an interesting method but not a normal ...

\(\displaystyle \frac{x}{x-1} - \frac{1}{\ln{x}} \)

\(\displaystyle = \frac{ (x-1) + 1}{x-1} - \frac{1}{\ln{x}} \)

\(\displaystyle = 1 + \left( \frac{1}{x-1} - \frac{1}{\ln{x}} \right )\)

I let \(\displaystyle L = \lim_{x\to 1} \left( \frac{1}{x-1} - \frac{1}{\ln{x}} \right ) \)

we know \(\displaystyle \frac{1}{t^2 - 1} = \frac{1}{2} \frac{1}{t-1} - \frac{1}{2} \frac{1}{t+1} \) and so

\(\displaystyle \frac{1}{x -1} = \frac{1}{2} \frac{1}{\sqrt{x} -1} - \frac{1}{2} \frac{1}{\sqrt{x}+1} \)

so we have

\(\displaystyle L = \lim_{x\to 1} \left(\frac{1}{2} \frac{1}{\sqrt{x} -1} - \frac{1}{2} \frac{1}{\sqrt{x}+1} - \frac{1}{\ln{x}} \right ) \)

also \(\displaystyle \ln{x} = 2\ln{\sqrt{x}}\)

\(\displaystyle L = \lim_{x\to 1} \left( \frac{1}{2} ( \frac{1}{\sqrt{x} -1} - \frac{1}{ \ln{\sqrt{x}}}) \right) - \frac{1}{2} \lim_{x\to 1} \frac{1}{\sqrt{x} + 1}\)

Let \(\displaystyle u = \sqrt{x} \)

\(\displaystyle L = \frac{L}{2} - \frac{1}{4} \)

\(\displaystyle L = -\frac{1}{2}\)

so the final answer is \(\displaystyle 1 - \frac{1}{2} = \frac{1}{2} \)

More surprisingly , if we continue to 'break down' the fraction \(\displaystyle \frac{1}{ \sqrt{x} - 1} = \frac{1}{2} \frac{1}{\sqrt[4]{x} -1} - \frac{1}{2} \frac{1}{\sqrt[4]{x}+1} \) and put \(\displaystyle \ln{x} = 4\ln{\sqrt[4]{x}}\) and so on for \(\displaystyle 8,16,32 ,...\) . All of the situations can reach the answer obtained !!

If there is no mistake in my method , it is interesting that the answer will not be affected for different choices of the base of the log. function !!

EDIT:

The answer will not be affected for different choices of the base of the log. function .

This statement is false because at first we don't know if the limit exists , we have \(\displaystyle L = \frac{L}{2} - \frac{1}{4} \) , but we are also forced to believe that \(\displaystyle \infty = \frac{ \infty}{2} - \frac{1}{4}\) . The limit only exists when the base is \(\displaystyle e \) .