How would you show $\pi(x)\log(1-\frac{1}{x}) \sim \frac{1}{\log x}$?
Would you use $\lim_{x\to \infty}\frac{\pi(x)\log(1-\frac{1}{x})}{\frac{1}{\log x}} = 1$? and how would you show this?
Can you use the fact that $\log(1-\frac{1}{x}) = O(\frac{1}{x})$ and $\pi(x) \sim \frac{x}{\log x}$
$\log(1-1/x)=-1/x-O(1/x^2)$ and $\pi(x)=x/\log x+O(x/\log^2x)$ so their product is $-(1/x+O(1/x^2))(x/\log x+O(x/\log^2x))=-1/\log x-O(1/\log^2x).$