I have been trying to determine the following two limits, Wolfram Alpha computes (1) to be equal to something it refers to as complex infinity, and (2) to be indeterminate. So I also would like to know the difference between "Complex Infinity" and an indeterminate, as well as the step by step working out for demonstrating these results.

$$\lim _{x\rightarrow 3/2}\Biggl(\frac{\lfloor\ln(x^{3})\rfloor}{\lfloor\ln(x)\rfloor}\Biggr)\quad\quad\quad\quad\quad\quad\quad\quad (1)$$

$$\lim _{x\rightarrow 3/2}\Biggl(\frac{\lfloor\ln(x^{2})\rfloor}{\lfloor\ln(x)\rfloor}\Biggr)\quad\quad\quad\quad\quad\quad\quad\quad(2)$$

As far as I can tell, (1) is an indeterminate limit of the form $\frac{1}{0}$ and (2) is an indeterminate of the form $\frac{0}{0}$, on what basis can we say that one is different to the other, or more so, what is the argument for there being a significant difference we should account for in all such cases that evaluate to the two differing "Categories" of indeterminate forms?

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    $\begingroup$ I would say "both of them undefined". $\endgroup$ – ℋolo Jul 17 '18 at 17:01
  • $\begingroup$ Agreed. But I just found it curious that their software has been coded to return a distinctly different answer so I guess that's the point of my post $\endgroup$ – Adam L Jul 17 '18 at 17:02
  • $\begingroup$ complex infinity is not widely used, but wolfram alpha use it when we have $|x|=\infty$ and $\arg(x)=$unknown/undefined. Undefined is when the value can be one of few possible values without way to determine which one it is. see also math.stackexchange.com/questions/1294852/… $\endgroup$ – ℋolo Jul 17 '18 at 17:05
  • $\begingroup$ Sure the definition they provided for it was orientated around the programming of wolfram alpha, but I just felt as if such a definition was one with a substantial load in terms mathematically fundamental meaning, so it naturally raised an eye brow $\endgroup$ – Adam L Jul 17 '18 at 17:08
  • $\begingroup$ There are 13 Indeterminate forms, starting with 0/0. $\endgroup$ – Ed Pegg Jul 17 '18 at 17:19

The difference is in the limit of the top.

Note that $$\ln(3/2)=0.4054...$$ so its integer part is $0$

Similarly $$\ln(9/4)=0.810934054...$$ so its integer part is $0$ as well.

On the other hand $$\ln(27/8)=1.21639...$$ so iteger part is $1$

That is the first limit is of $$ 0/0 $$ form while the second is of $$1/0$$ form.

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    $\begingroup$ The first limit is not $0/0$: it's not defined in a punctured neighborhood of $3/2$, so it makes no sense at all. $\endgroup$ – egreg Jul 17 '18 at 17:37
  • $\begingroup$ @egreg you are correct in that the quotient is not defined . I was talking abot the form not the limit. $\endgroup$ – Mohammad Riazi-Kermani Jul 17 '18 at 17:45
  • $\begingroup$ Sure ok thankyou sir, I suppose I need more elaboration in that I was aware of the difference in the numerators of the forms, more as to how this should affect the way in which we contemplate the functions and the arguments for which they evaluate to these differing forms $\endgroup$ – Adam L Jul 17 '18 at 17:48
  • $\begingroup$ @Adam Thanks for your attention . $\endgroup$ – Mohammad Riazi-Kermani Jul 17 '18 at 17:50

Suppose a function $f$ is defined on a set $D$. In order for $$ \lim_{x\to a}f(x) $$ to at least make sense, $a$ should be an accumulation point (aka limit point) of $D\setminus\{a\}$. The two expressions you have are defined where $\ln(x)$ exists and $$ \lfloor\ln(x)\rfloor\ne0 $$ Now $\lfloor\ln(x)\rfloor=0$ if and only if $0\le\ln(x)<1$, which is the same as $1\le x<e$. Thus the functions are defined over $(0,1)\cup(e\infty)$.

As you see, $3/2$ is not an accumulation point of the domain of the functions, so neither limit makes sense.

Wolframalpha does strange things for assigning a meaning to undefined expressions: don't take its answers as revealed truth.


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