Complex infinity and $\frac{0}{0}$ indeterminants

Question

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?

Answer 1

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.

Answer 2

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.