# Infinite limit at infinity complex variables using definition

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.

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 $$K$$.
• 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. Dec 22, 2022 at 10:45
• 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. Dec 22, 2022 at 10:48
• @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? Dec 22, 2022 at 11:36