I would like to show that $$\lim_{z\to \infty}\frac{2z^3-1}{z^2+1}= \infty,$$ using this

Definition: Let $f:D \to \mathbb C$ be a single-valued complex function, and suppose that $\infty$ is an accumulation point of $D$. The point $\infty$ is said to be the limit of $f(z)$ as $z$ approaches $\infty$ if for every $K>0$ there is an $M>0$ (depending of $K$) such that $$ |f(z)|>K \;\text{ whenever }\; |z|>M \;\text{ and }\; z\in D. $$

By the way, I know we can do this $$\lim_{z\rightarrow \infty}\frac{2z^3-1}{z^2+1}=\infty \quad \text{since}\quad \lim_{z\rightarrow 0}\frac{(1/z^2)+1}{(2/z^3)-1}=\frac{z+z^3}{2-z^3}=0.$$ But I would like to use the above definition. I am trying to understand how this definition works.

So I need to find $M>0$ such that if $|z|>M$ then $|f(z)|>K$ for every $K>0$.

This is my attempt so far: Considering $|z|=K>0$, then $$|2z^3-1|\geq |2|z|^3-1|=2K^3-1$$ and $$|z^2+1|\leq |z|^2+1=K^2+1$$ Thus $$\frac{|2z^3-1|}{|z^2+1|}\geq\frac{2K^3-1}{K^2+1}$$ However, I am not sure how to proceed from here. Can I take $M=\dfrac{2K^3-1}{K^2+1}?$ Or what am I missing? Is this a good approach?

Any help will be appreciated. Thanks.


1 Answer 1


A simpler approach might be to notice that $\left| \frac{2z^3 - 1}{z^2 + 1}\right| \geq \frac{2|z|^3 - 1}{|z|^2 + 1}$ (as you arleady did) and to use what you know about what it means to tend toward $+\infty$ in real analysis and relating it to your definition over $\mathbb C$ (notice that $|z| \in \mathbb R$). I think you are not meant to construct again the theory of limits here, but to use what you already know.

In your proof, you have taken an arbitrary $K$ and found $M$ such that if $|z| > K$, then $|f(z)| > M$. But you were supposed to take an arbitrary $K$ and find $M > 0$ such that if $|z| > M$, then $|f(z)| > K$. You have a mix-up between $M$ and $K$. And I think it would be quite tidious to find the $M$ that works for a given $K$.

  • 1
    $\begingroup$ In this particular case, if we assume $|z|\geq1$, then we can go slightly further: $\left| \frac{2z^3 - 1}{z^2 + 1}\right| \geq \frac{2|z|^3 - 1}{|z|^2 + 1}\geq\frac{|z|^3+0}{|z|^2+|z|^2}=\frac{|z|}{2}$. So, given $K>0$, the choice $M=1+2K$ works. $\endgroup$
    – peek-a-boo
    Dec 22, 2022 at 10:45
  • $\begingroup$ Oh, thanks for that. Sorry I am mixing-up $M$ and $K$. So in cases like this is better to use the method I mentioned above after the definition? Do you know an easier example where I can use the definition? Or can you point me to book/reference were I can find an example to apply the definition? Thanks in advance. $\endgroup$ Dec 22, 2022 at 10:48
  • $\begingroup$ I think the comment from peek-a-boo just before your comment gives a good way to use the definition with the Ms and the Ks in your case ! $\endgroup$
    – Zag
    Dec 22, 2022 at 10:51
  • $\begingroup$ I will analyse it. Thank you peek-a-book and Zag. $\endgroup$ Dec 22, 2022 at 10:54
  • $\begingroup$ @peek-a-boo So I can start with $|z|>M = 1+2K$, which implies $\dfrac{|z|-1}{2}>K$. Then I use the inequality you mentioned in the first comment to arrive at $|f(z)|>K$. Is this correct? $\endgroup$ Dec 22, 2022 at 11:36

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .