# How can I find this limit involving thrice-iterated logarithm?

$$\lim_{x \to 0}\dfrac{\ln \ln \ln \left[x+(1+x)^{(1+x)^{1/x}/x}\right]+x\left[1-\dfrac{1}{e^{e+1}}\right]}{x^2}$$

How can I find the limit of this question? Any hint. Thank you so much.

• Maple outputs $$1/24\,{\frac {{{\rm e}^{-2}}{{\rm e}^{2}} \left( {{\rm e}^{{\rm e}}} \right) ^{2}+24\,{{\rm e}^{-2}}{{\rm e}^{2}}{{\rm e}^{{\rm e}}}+48\,{ {\rm e}^{-2}}{\rm e}{{\rm e}^{{\rm e}}}-12\,{{\rm e}^{-2}}{\rm e}-24\, {{\rm e}^{-2}}}{ \left( {{\rm e}^{{\rm e}}} \right) ^{2}}}.$$ – user64494 Sep 10 '14 at 5:51
• Dude, seriously, are you signaling for help to aliens from outer space ? :-) – Lucian Sep 10 '14 at 6:22
• Is this question from a textbook? – Paracosmiste Sep 10 '14 at 12:53
• it is not from textbook. – Simple Sep 10 '14 at 19:46
• It could be, then, as a somewhat harder than usual, but direct application of asymptotic computations with big $O$. – Jean-Claude Arbaut Sep 13 '14 at 5:46

You can find it with a lot of patience ! I give you what I did (hoping that a simpler solution will be provided) :

First, I look at the exponent and, going first to logarithms, obtain $$\frac{(x+1)^{\frac{1}{x}}}{x}=\frac{e}{x}-\frac{e}{2}+\frac{11 e x}{24}-\frac{7 e x^2}{16}+O\left(x^3\right)$$ Then $$(x+1)^{\frac{(x+1)^{\frac{1}{x}}}{x}}=e^e-e^{1+e} x+\frac{1}{24} e^{1+e} (25+12 e) x^2+O\left(x^3\right)$$ So$$A=x+(x+1)^{\frac{(x+1)^{\frac{1}{x}}}{x}}=e^e+\left(1-e^{1+e}\right) x+\frac{1}{24} e^{1+e} (25+12 e) x^2+O\left(x^3\right)$$ Now, let me play with the logarithms $$\log A=e+\left(e^{-e}-e\right) x+\left(\frac{25 e}{24}+e^{1-e}-\frac{e^{-2 e}}{2}\right) x^2+O\left(x^3\right)$$ $$\log\log A=1+\left(e^{-1-e}-1\right) x+\left(\frac{13}{24}-\frac{1}{2} e^{-2-2 e}-\frac{1}{2} e^{-1-2 e}+e^{-1-e}+e^{-e}\right) x^2+O\left(x^3\right)$$ $$\log\log\log A=\left(e^{-1-e}-1\right) x+\frac{1}{24} e^{-2-2 e} \left(-24-12 e+48 e^{1+e}+24 e^{2+e}+e^{2+2 e}\right) x^2+O\left(x^3\right)$$ where you can notice that the first term is $$-x\left[1-\frac{1}{e^{e+1}}\right]$$ So, the limit is $$\frac{1}{24} e^{-2-2 e} \left(-24-12 e+48 e^{1+e}+24 e^{2+e}+e^{2+2 e}\right)=\frac{1}{24}+\frac{1}{2} e^{-2 (1+e)} (2+e) \left(2 e^{1+e}-1\right)$$

All of the above used successively the development (Taylor series) of $\log(1+y)$ close to $y=0$.

• +1, and your note proves that this question was made up so that those trying the asymptotic approach (which is direct and "obvious", though quite harder than usual) will need to keep big $O$ up to $x^3$ in intermediate computations. It would be funny in "colle" :-) – Jean-Claude Arbaut Sep 13 '14 at 5:50
• @Jean-ClaudeArbaut. This is for sure ! Cheers – Claude Leibovici Sep 13 '14 at 5:55
• +1 for the really cool answer. My LHR approach is bit lengthy. – Paramanand Singh Sep 13 '14 at 6:46
• @ParamanandSingh. THanks, but I must confess that there is a trick ! I love Taylor !! I appreciated your answer which is more clever than mine. Cheers :-) – Claude Leibovici Sep 13 '14 at 6:48

Let's give it a try using elementary techniques. First we take care of $f(x) = (1 + x)^{(1 + x)^{1/x}/x}$. Clearly $$\log f(x) = (1 + x)^{1/x}\frac{\log(1 + x)}{x} \to e\tag{1}$$ so that $f(x) \to e^{e}$. We have to deal with the expression $$F(x) = \dfrac{\log \log \log(x + f(x)) + x(1 - e^{-e - 1})}{x^{2}}\tag{2}$$ It can be seen that both the numerator and denominator of $F(x)$ tend to $0$. I am somehow forced now to use L'Hospital Rule. This gives us another complicated expression $$G(x) = \frac{1}{2x}\left\{\left(1 - e^{-e - 1}\right) + \frac{1}{\log \log (x + f(x))}\cdot\frac{1}{\log(x + f(x))}\cdot\frac{1 + f'(x)}{x + f(x)}\right\}\tag{3}$$ The challenge now is to calculate $f'(x)$ and show that it tends to $-e^{e + 1}$ as $x \to 0$. This will ensure that we can apply LHR on $G(x)$ also. Assuming that we have done so we can see that the next application of LHR on $G(x)$ will give rise to the expression $$H(x) = \frac{1}{2}\left\{\frac{g(x)f''(x) - (1 + f'(x))g'(x)}{\{g(x)\}^{2}}\right\}\tag{4}$$ where $g(x) = (x + f(x))\cdot\log(x + f(x))\cdot\log\log(x + f(x))$.

Note that $g(x) \to e^{e + 1}$ so I hope that we don't need further application of LHR on $H(x)$ and the final limit would be $$\dfrac{A}{2e^{2e + 2}}\tag{5}$$ where $$A = \lim_{x \to 0}g(x)f''(x) - (1 + f'(x))g'(x)\tag{6}$$ We can now see that the real challenge is to evaluate $f'(x), f''(x), g'(x)$ and it would require a reasonable amount of calculation. I will have to leave my keyboard and go for a pen-paper calculation to handle this. Will post the calculation if I succeed.

The easiest part is $g'(x)$ given by \begin{aligned}g'(x) &= (1 + f'(x))\log(x + f(x))\log\log(x + f(x))\\ &\,\,\,\,+\,\, (1 + f'(x))\log\log(x + f(x))\\ &\,\,\,\,+\,\, (1 + f'(x))\\ &= ( 1 + f'(x))\{1 + \log\log(x + f(x)) + \log(x + f(x))\log\log(x + f(x))\}\end{aligned} Based on the assumption made in bold we can see that $(1 + f'(x))g'(x)$ tends to $$(1 - e^{e + 1})^{2}(e + 2)\tag{7}$$ From $(1)$ we can see that $\log f(x) = p(x)\log p(x)$ where $p(x) = (1 + x)^{1/x}$. Hence \begin{align}\frac{f'(x)}{f(x)} &= p'(x)(1 + \log p(x))\notag\\ &= p(x)(1 + \log p(x))(\log p(x))'\notag\\ &= p(x)(1 + \log p(x))\left(\frac{\log (1 + x)}{x}\right)'\notag\\ &= p(x)(1 + \log p(x))\left(\frac{x - (1 + x)\log(1 + x)}{x^{2}(1 + x)}\right)\tag{8}\end{align} Now we can see that $p(x) \to e$ so that $f'(x)/f(x)$ tends to $(2e)(-1/2) = -e$. Since $f(x) \to e^{e}$ we see that $f'(x) \to -e^{e + 1}$ and our assumption (mentioned in bold earlier) is verified (which at the same time verifies $(7)$).

Now the hardest part is to differentiate $(8)$ and obtain $f''(x)$. We have \begin{aligned}f''(x) &= f'(x)p(x)\{1 + \log p(x)\}\left(\frac{x - (1 + x)\log(1 + x)}{x^{2}(1 + x)}\right)\\ &\,\,+\,\,f(x)\{1 + \log p(x)\}p(x)\left(\frac{x - (1 + x)\log(1 + x)}{x^{2}(1 + x)}\right)^{2}\\ &\,\,+\,\,f(x)p(x)\left(\frac{x - (1 + x)\log(1 + x)}{x^{2}(1 + x)}\right)^{2}\\ &\,\,+\,\,f(x)\{1 + \log p(x)\}p(x)\left(\frac{x - (1 + x)\log(1 + x)}{x^{2}(1 + x)}\right)'\end{aligned} Hence we can see that $f''(x)$ tends to $$e^{e + 2} + \frac{e^{e + 1}}{2} + \frac{e^{e + 1}}{4} + \frac{4e^{e + 1}}{3} = \frac{12e^{e + 2} + 25e^{e + 1}}{12}$$ The desired limit $A$ is thus \begin{aligned}A &= \frac{12e^{2e + 3} + 25e^{2e + 2}}{12} - (e + 2)(1 - e^{e + 1})^{2}\\ &= \frac{12e^{2e + 3} + 25e^{2e + 2} - 12(e + 2)(1 - 2e^{e + 1} + e^{2e + 2})}{12}\\ &= \frac{12e^{2e + 3} + 25e^{2e + 2} - 12(e - 2e^{e + 2} + e^{2e + 3} + 2 - 4e^{e + 1} + 2e^{2e + 2})}{12}\\ &= \frac{e^{2e + 2} + 24e^{e + 2} + 48e^{e + 1} - 12e - 24}{12}\end{aligned} The final limit is $A/2e^{2e + 2}$ so that the final limit is given by $$\frac{e^{2e + 2} + 24e^{e + 2} + 48e^{e + 1} - 12e - 24}{24e^{2e + 2}} = \frac{1}{24} + \frac{2e^{e + 2} + 4e^{e + 1} - e - 2}{2e^{2e + 2}}$$ Note: The limit of $$a(x) = \frac{x - (1 + x)\log(1 + x)}{x^{2}(1 + x)} = \dfrac{\dfrac{x}{1 + x} - \log(1 + x)}{x^{2}} = -\frac{1}{2} + \frac{2}{3}x + \cdots$$ and its derivative were calculated using the Taylor expression given above leading to $a(x) \to -1/2$ and $a'(x) \to 2/3$ as $x \to 0$.