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. 
 A: 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 $y=0$.
