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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?

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    $\begingroup$ It makes no sense to compute the log of 0, but you never get a zero in $\frac{x^3}{x^4+2}$, what you get is numbers that get closer and closer to zero, thus why the log gets closer and closer to $-\infty$ $\endgroup$
    – David
    Oct 20, 2021 at 21:21
  • $\begingroup$ @David I am agree with you, but I have not understood the reason of the result of my book because give $-\infty$. If you see my question I have written something that is similar to your comment. $\endgroup$
    – Sebastiano
    Oct 20, 2021 at 21:28
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    $\begingroup$ Both limits are $-\infty$. I see no issue with that... $\endgroup$
    – David
    Oct 20, 2021 at 21:53

3 Answers 3

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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.

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  • $\begingroup$ Thank you so much for your answer. I only agree with the answer if I would have explained before the limits of real functions of real variable first. But I did not. However, your answer confirms what I explained to my students about considering a $0^+$: then it makes sense to say that the limit is $-\infty$. Thanks again. (see my considerations on the question). $\endgroup$
    – Sebastiano
    Oct 20, 2021 at 21:53
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Hint: $$\log_2\left(\frac{n^3}{n^4+2}\right) = \log_2(n^3) - \log_2(n^4+2) \leq -\log_2(n)$$

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  • $\begingroup$ I positively accept an upper bond, and why it must be $\leq -\log_2(n)$. I can't ask students who are 17 or 18 to do creative upper bonds. $\endgroup$
    – Sebastiano
    Oct 20, 2021 at 19:56
  • $\begingroup$ @Sebastiano I suppose it depends on how comfortable they are with logarithms. It only requires basic facts and that it's increasing. $\endgroup$ Oct 20, 2021 at 20:01
  • $\begingroup$ For example I would never have thought of your solution. $\endgroup$
    – Sebastiano
    Oct 20, 2021 at 20:04
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$0<\frac {n^3}{n^4 + 2} < \frac {1}{n}$

$\log_2 \left( \frac {n^3}{n^4 + 2}\right) < \log_2 \left( \frac {1}{n}\right)$

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.

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  • $\begingroup$ I am agree with your answer. +1 :-) $\endgroup$
    – Sebastiano
    Oct 21, 2021 at 22:51

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