Limit involving log 
*

*I was asking why this limit is $-\infty$ and not $+\infty\ ?$.

*I thought it was $+\infty$, but Wolfram Alpha says $-\infty$.

*My guess is that $\log\left(\log\left(x\right)\right) -\log\left(\log\left(x - 1\right)\right)$ goes to zero faster than $-0.5x$ goes to $-\infty$.

$$
\lim_{x \to \infty}\left\{\left(\frac{x^{2}}{2} - x\right) \left[\log\left(\log\left(x\right)\right) -
\log\left(\log\left(x - 1\right)\right)\right] -
\frac{x}{2} -\frac{x}{\log\left(x\right)}\right\}
$$

We'll study the expression for $x\to\infty$:
$\Big(\dfrac{x^2}{2}-x\Big)(\log(\log x)-\log(\log(x-1))-\dfrac{x}{2}-\dfrac{x}{\log x}$, which is asymptotic to
$$\dfrac{x^2}{2}(\log(\log x)-\log(\log(x-1))-\dfrac{x}{2}-\dfrac{x}{\log x}=$$
$$=\dfrac{x^2}{2}\log(\log x)-\dfrac{x^2}{2}\log(\log(x-1))-\dfrac{x}{2}-\dfrac{x}{\log x} \sim$$
$$\sim \dfrac{x^2}{2}\log(\log x)-\dfrac{x^2}{2}\log(\log x)-\dfrac{x}{2}-\dfrac{x}{\log x}\underset{D.H.}{\sim}-\dfrac{x}{2}-\dfrac{x'}{(\log x)'}=$$
$$=-\dfrac{x}{2}-\dfrac{1}{\dfrac{1}{x}}=-\dfrac{3x}{2}\overset{x\to \infty}{\longrightarrow} -\infty.$$
