To compute $\lim\limits_{n\to +\infty} \log_2\left(\frac{n^3}{n^4+2}\right)=-\infty, \lim\limits_{x\to+\infty}\log_2\left(\frac{x^3}{x^4+2}\right)$ On my textbook of Maths for students of my high school (there are really only two such limits) I have found, for example, a limit of a succession with this solution:
$$\color{magenta}{\lim_{n\to +\infty} \log_2\left(\frac{n^3}{n^4+2}\right)=-\infty} \tag 1$$
I am a bit perplexed about the result. In fact it is true that $\forall n\in \Bbb N\smallsetminus \{0\}$ we have:
$$\frac{n^3}{n^4+2}>0$$
and the domain of the logarithm is guaranteed ($n\ne 0$). Being
$$\lim_{n\to +\infty} \log_2\left(\frac{n^3}{n^4+2}\right)=\log_2\left(\lim_{n\to +\infty} \frac{n^3}{n^4+2}\right)$$
but $$\lim_{n\to +\infty} \frac{n^3}{n^4+2}=0$$
hence it has no sense to calculate $$\require{cancel} \color{red}{\cancel{\log_2 0}}$$
Considering the function
$$y=f(x)=\log_2\left(\frac{x^3}{x^4+2}\right) \tag 2$$
we know that the domain is $]0,+\infty[$. The allowable limits are for $x\to 0^+$ and $x\to +\infty$.
It is true, instead, that
$$\color{green}{\lim_{x\to 0^+}\log_2\left(\frac{x^3}{x^4+2}\right)=-\infty}$$
but if I plot the $(2)$ with Desmos, when I compute the $\lim\limits_{x\to +\infty}\log_2\left(\frac{x^3}{x^4+2}\right)$ I have precisely $-\infty$.

So why must the limit of $(1)$ be $-\infty$? What am I doing wrong and why? So it must be $\lim_n \log_c a_n=-\infty$ ($c>0 \wedge c\ne 1$), when $\{a_n\}\to 0$, i.e. infinitesimal?

 A: Hint: $$\log_2\left(\frac{n^3}{n^4+2}\right) = \log_2(n^3) - \log_2(n^4+2) \leq -\log_2(n)$$
A: The problem is your equation
$$
\lim_{n \to +\infty} \log_2\left(\frac{n^3}{n^4+2}\right) = \log_2\left(\lim_{n \to +\infty} \frac{n^3}{n^4+2}\right).
$$
This equation is incorrect because $\log_2(0)$ is undefined.  You are trying to apply the theorem that if $\lim_{n \to +\infty} f(n) = L$ and $g$ is continuous at $L$, then $\lim_{n \to +\infty} g(f(n)) = g(L)$.  In this case, $g$ is $\log_2$ and $L=0$, and $g$ is not continuous at $L$ because $g(L)$ is undefined.  So the theorem does not apply.
Here is a better way to think about it.  Let $u = n^3/(n^4 + 2)$.  Then since $\lim_{n \to +\infty} n^3/(n^4+2) = 0$, we can say that as $n \to +\infty$, $u \to 0$.  In fact, for every positive $n$, $u$ is positive, so we can say that as $n \to +\infty$, $u \to 0^+$.  Also, $\lim_{u \to 0^+} \log_2 u = -\infty$, so as $u \to 0^+$, $\log_2 u \to -\infty$.  Stringing together the statements "as $n \to +\infty$, $u \to 0^+$" and "as $u \to 0^+$, $\log_2 u \to -\infty$," you can conclude that as $n \to +\infty$, $\log_2(n^3/(n^4+2)) = \log_2 u \to -\infty$.
Note:  You have to be a little bit careful when stringing together such statements.  For details, see my book Calculus: A Rigorous First Course, Section 2.6.
A: $0<\frac {n^3}{n^4 + 2} < \frac {1}{n}$
$\log_2 \left( \frac {n^3}{n^4 + 2}\right) < \log_2 \left( \frac {1}{n}\right)$
For any positive number $N$ setting $n = 2^N \implies \log_2 \left(\frac {n^3}{n^4+2}\right) < -N$
Letting $N$ become arbitrarily large will cause $\log_2 \left(\frac {n^3}{n^4+2}\right)$ to become equally negative.
