# 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 $$x\to +\infty$$.

$$\color{green}{\lim_{x\to 0^+}\log_2\left(\frac{x^3}{x^4+2}\right)=-\infty}$$

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?

• 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 $-\infty$ Oct 20, 2021 at 21:21
• @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. Oct 20, 2021 at 21:28
• Both limits are $-\infty$. I see no issue with that... Oct 20, 2021 at 21:53

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.

• 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). Oct 20, 2021 at 21:53

Hint: $$\log_2\left(\frac{n^3}{n^4+2}\right) = \log_2(n^3) - \log_2(n^4+2) \leq -\log_2(n)$$

• 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. Oct 20, 2021 at 19:56
• @Sebastiano I suppose it depends on how comfortable they are with logarithms. It only requires basic facts and that it's increasing. Oct 20, 2021 at 20:01
• For example I would never have thought of your solution. Oct 20, 2021 at 20:04

$$0<\frac {n^3}{n^4 + 2} < \frac {1}{n}$$

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

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.

• I am agree with your answer. +1 :-) Oct 21, 2021 at 22:51