Answer
$$\lim_{x\to1}\Bigg(\frac{x}{x-1}-\frac{1}{\ln x}\Bigg)=\frac{1}{2}$$
Work Step by Step
$$A=\lim_{x\to1}\Bigg(\frac{x}{x-1}-\frac{1}{\ln x}\Bigg)$$
$$A=\lim_{x\to1}\frac{x\ln x-x+1}{(x-1)\ln x}$$
As $x\to1$, $(x\ln x-x+1)\to(1\times\ln 1-1+1)=1\times0+0=0$ and $(x-1)\ln x\to(1-1)\ln 1=0\times0=0.$
This is an indeterminate form of $\frac{0}{0}$, so we can apply L'Hospital's Rules:
$$A=\lim_{x\to1}\frac{[x\ln x-x+1]'}{[(x-1)\ln x]'}$$
$$A=\lim_{x\to1}\frac{(1\times\ln x+x\times\frac{1}{x})-1+0}{(x-1)'\ln x+(x-1)(\ln x)'}$$
$$A=\lim_{x\to1}\frac{\ln x+1-1}{1\times\ln x+\frac{x-1}{x}}$$
$$A=\lim_{x\to1}\frac{\ln x}{\ln x+\frac{x-1}{x}}$$
$$A=\lim_{x\to1}\frac{\ln x}{\frac{x\ln x+x-1}{x}}$$
$$A=\lim_{x\to1}\frac{x\ln x}{x\ln x+x-1}$$
$\lim_{x\to1}(x\ln x+x-1)=1\times\ln 1+1-1=1\times0+0=0$ and $\lim_{x\to1}(x\ln x)=1\times\ln 1=1\times0=0.$
Another indeterminate form of $\frac{0}{0}$. L'Hospital's Rule would be used here:
$$A=\lim_{x\to1}\frac{(x\ln x)'}{(x\ln x+x-1)'}$$
$$A=\lim_{x\to1}\frac{1\times\ln x+x\times\frac{1}{x}}{1\times\ln x+x\times\frac{1}{x}+1-0}$$
$$A=\lim_{x\to1}\frac{\ln x+1}{\ln x+1+1}$$
$$A=\lim_{x\to1}\frac{\ln x+1}{\ln x+2}$$
$$A=\frac{\ln 1+1}{\ln 1+2}$$
$$A=\frac{0+1}{0+2}=\frac{1}{2}$$