Very strange limit involving factorial and golden ratio

Question

It's a complex limit I don't understand involving the factorial and golden ratio this is :

$$\lim_{x\to 0^-}\frac{\left(1-x!^{x^{x!}}-x^{x!^{x}}\right)}{x^{1.1}\ln\left(x\right)}=-\frac{i\left(-1+\sqrt{5}\right)+\sqrt{2\left(5+\sqrt{5}\right)}}{\sqrt{\left(1-\sqrt{5}\right)^{2}+2\left(5+\sqrt{5}\right)}}\infty$$

I find this result with Wolfram alpha and I'm really lost to understand it . What does it mean a complex infinite ?

Is this result purely symbolic ? How to show it ?

I wait for your explanation .

Answer 1

Diverging to infinity in the complex plane is similar to diverging to infinity on the real number line. In the reals, it is common to say that a function diverges to $+\infty$ or $-\infty$. The same holds for the complex plane, except here there are infinite directions that infinity lies in instead of just two. More formally, we say that a function $f(z)$ diverges to $e^{i \phi}\infty$ if $e^{-i\phi}f(z)$ diverges to $+\infty$.

If you wish to prove your specific example, you would need to show that

$$\lim_{x\to 0^{-}}\frac{i\left(-1+\sqrt{5}\right)-\sqrt{2\left(5+\sqrt{5}\right)}}{\sqrt{\left(1-\sqrt{5}\right)^{2}+2\left(5+\sqrt{5}\right)}}\cdot \frac{\left(1-x!^{x^{x!}}-x^{x!^{x}}\right)}{x^{1.1}\ln\left(x\right)}=+\infty$$

under the expanded definition of diverges to $+\infty$. This definition (for complex functions) is:

$$\lim_{x\to 0^-}f(x)=+\infty$$

if both

$$\lim_{x\to 0^-}|f(x)|=+\infty$$

$$\lim_{x\to 0^-}\text{Arg}(f(x))=0$$

Answer 2

Entering the expression (and please leave this as verbatim text, it isn't MathJax, it is the text entered into Wolfram Alpha!)

lim x->0- (1-(x!)^x^(x!) - x^(x!)^x)/(x^1.1ln(x))

in Wolfram Alpha, I get the result

$$(0.951057 - 0.309017i)\infty$$

which I suspect means the expression diverges to infinity in the direction of that unit complex number.

(A quick computation shows this is a numerical approximation of the expression on the right side of your result.)

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Comment: Given the "$-$" in front of your result, I suspect you have written the wrong sign in front of the real part within the numerator of the fraction (?)

Answer 3

The argument is explained by the denominator. Using the "principal value", $$ \lim_{x\to 0^-}\arg(x^{1.1}\ln(x)) = \frac{\pi}{10} $$ and $$ -\frac{i\left(-1+\sqrt{5}\right)+\sqrt{2\left(5+\sqrt{5}\right)}}{\sqrt{\left(1-\sqrt{5}\right)^{2}+2\left(5+\sqrt{5}\right)}} = -e^{i\pi/10} $$

Answer 4

Computing $$f(x)=\frac{1-(x!)^{x^{x!}}-x^{(x!)^x}}{x^{11/10} \log (x)}$$ to very high accuracy for $x=-10^{-k}$ and $200 \leq k \leq 600$

$$\log(\Re[f(x)])=a+b\,k$$

$$\begin{array}{l|lll} \text{} & \text{Estimate} & \text{Std Error} & \text{Confidence Interval} \\ \hline a & -5.766098 & 0.025488 & \{-5.817654,-5.714543\} \\ b & +0.227607 & 0.000061 & \{+0.227483,+0.227730\} \\ \end{array}$$ with $R^2=0.99999719$.

$$\frac{\Im[f(x)]}{\Re[f(x)]}=a +\frac b{1+c\,k}$$

$$\begin{array}{l|lll} \text{} & \text{Estimate} & \text{Std Error} & \text{Confidence Interval} \\ \hline a & -0.324920 & 1.7\times 10^{-10} & \{-0.324920,-0.324920\} \\ b & +3.402603 & 0.000094 & \{+3.402410,+3.402790\} \\ c & +2.255744 & 0.000063 & \{+2.255617,+2.255871\} \\ \end{array}$$ with $R^2>0.99999999$.

According to the $ISC$, $$a=-\tan \left(\frac{\pi }{10}\right)=\sqrt{1-\frac{2}{\sqrt{5}}}$$