# 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 $$-\infty$$.

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

• Pls clarify! And ask a concise question! Indentation will improve readability a lot, and double double dollars like this \$\$ \log\log (e^e) \$\$ produce detached formulae: $$\log\log (e^e) =1$$ Jan 5, 2021 at 18:45
• Hint: by the mean value theorem, $\log(\log(x))-\log(\log(x-1))$ is about $\frac1{x\log x}$. Jan 5, 2021 at 21:19

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

• Thank you guys. But Vajra, I don't see the necessity to use L'hopital's Rule here. With your argument, if we had +x/log(x) instead of -x/log(x), then the limit would go to +infinity. Jan 6, 2021 at 7:21
• Infact I used It only to write in "symbols" the fact that $x/\log x\to \infty$. It was't necessary Jan 6, 2021 at 8:48

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.