Limit involving log

Question

• 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 $-\infty$.

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

Answer 1

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 $$\dfrac{x^2}{2}(\log(\log x)-\log(\log(x-1))-\dfrac{x}{2}-\dfrac{x}{\log x}=$$
$$=\dfrac{x^2}{2}\log(\log x)-\dfrac{x^2}{2}\log(\log(x-1))-\dfrac{x}{2}-\dfrac{x}{\log x} \sim$$
$$\sim \dfrac{x^2}{2}\log(\log x)-\dfrac{x^2}{2}\log(\log x)-\dfrac{x}{2}-\dfrac{x}{\log x}\underset{D.H.}{\sim}-\dfrac{x}{2}-\dfrac{x'}{(\log x)'}=$$
$$=-\dfrac{x}{2}-\dfrac{1}{\dfrac{1}{x}}=-\dfrac{3x}{2}\overset{x\to \infty}{\longrightarrow} -\infty.$$

Answer 2

The first thing I should work in order to get rid of is $$\log (\log (x))-\log (\log (x-1))=\log \left(\frac{\log (x)}{\log (x-1)}\right)$$ $$\frac{\log (x)}{\log (x-1)}=1+\frac{1}{x \log (x)}+\frac{\log (x)+2}{2 x^2 \log^2(x)}+O\left(\frac{1}{x^3}\right)$$ $$\log \left(\frac{\log (x)}{\log (x-1)}\right)=\frac{1}{x \log (x)}+\frac{\log (x)+1}{2 x^2 \log ^2(x)}+O\left(\frac{1}{x^3}\right)$$ All of that gives for the expression $$-\frac{x (\log (x)+1)}{2\log \left(x\right)}+\frac{1-3 \log (x)}{4 \log ^2(x)}+O\left(\frac{1}{x}\right)$$

For $x=10$, the above truncated expression gives $-7.45004$ to be compared to the exact value $-7.46945$.

For $x=100$, the above truncated expression gives $-61.0084$ to be compared to the exact value $-61.0093$.

We are quite far away from $\infty$ and it seems to be working quite well.