# Evaluating a complex limit $\lim_{z\to 0}$ $\frac{x^2 y}{x^4+y^2}$ where $z=x+yi$

I am to determine whether $\lim_{z \to 0} \frac{x^2 y}{x^4+y^2}$ exists, where $z=x+yi$. My approach, as in multivariable calculus, is to look at $z$ approaching $0$ from different directions.

Along the the line $y=-x$ in the fourth quadrant, I have $$\lim_{z \to 0} \dfrac{x^2 y}{x^4+y^2} =\lim_{z\to 0} \dfrac{-(x^3)}{x^4+x^2}=-\infty.$$

Along the line $y=x$ in the first quadrant, I have

$$\lim_{z\to 0}\dfrac{x^2 y}{x^4+y^2}=\lim_{z\to 0} \dfrac{x^3}{x^4+x^2}=\infty.$$

Two questions. Is my approach correct? Second, since in the complex plane there is just a notion of $\infty$, and not $-\infty$, do these limits match up? In other words, do these two limits imply the limit does not exist or not?

• It looks just fine to me. – DonAntonio Feb 29 '16 at 17:06
• @joanpemo But does it imply the limit doesn't exist? – Sean I Feb 29 '16 at 17:09
• Indded so, in my opinion. You're identifying, correctly I think, $\;\Bbb C\cong\Bbb R^2\;$ ,so what we know from the real plane applies here as these two are identical topologically. – DonAntonio Feb 29 '16 at 17:10
• Yes: this implies that the limit does not exist. – Crostul Feb 29 '16 at 17:10
• Even simpler: use $y=kx^2$, so you have $\lim_{x\to0}\frac{kx^4}{x^4+k^2x^4}=\frac{k}{1+k^2}$ – egreg Feb 29 '16 at 17:11

## 1 Answer

You made a mistake with those two limits. They're both $0.$

I would look at the curves $y=0$ and $y=x^2,$ on which our function is, respectively, $0$ and $1/2.$

Added later: I see @egreg solved this in the comments.

• Thanks. I see my mistake now. – Sean I Feb 29 '16 at 20:56