# Multivariable function limit

How to approach this: $\lim\limits_{(x,y)\to(0,0)}\frac{x^2y}{x^2+y}$?

Been able to grind $\lim\limits_{(x,y)\to(0,0)}\frac{x^2y}{x^2+y^2}$, it is in(link) finnish, but formulas and idea should be selfevident. However $\dot +y$ instead of $\dots+y^2$ in divisor confuses me.

Or am I thinking in completely wrong direction?

• What happens to the function when $(x,y)\to(0,0)$ in such a way that it is very close to the parabola $y=-x^2$? Compare to what happens when the approach is along the curve $y=x^2$. Depending on the details of the definition of the limit we may already tumble because the function is not defined in any punctured neighborhood of the origin. – Jyrki Lahtonen Sep 12 '15 at 18:45
• Aha, wolframalpha.com/input/?i=limit+x^2y%2F%28x^2%2By%29+%28x%2Cy%29-%3E%280%2C0%29 this is what caused confusion then? (Wolframalpha implies L=0) Tricky question.. if $y\neq-x^2$.. Stop - does it mean, that if there is in a undefined path in neighborhood of origin, than there is no limit for origin? – Timo Junolainen Sep 12 '15 at 19:03
• Yes, completely wrong direction. The limit in the example exists and that in your question does not :) – Miguel Sep 12 '15 at 19:03
• @Timo You could exclude the points $y=x^2$, even then the limit does not exist. Consider trajectories $x= \sqrt{y^2-y}$ – Miguel Sep 12 '15 at 19:17

## 3 Answers

First of all, you need to consider your space restricted to $\mathbb{R}^2 - \{(x,y):x^2+y=0\}$. Otherwise you cannot form a neighbourhood of the origin where the function is defined.

Even then, consider the trajectories $x=0$ and $x^2=y^2-y$: $$\lim_{x=0,y\to 0} \frac{x^2 y}{x^2+y} = \lim_{y\to 0} \frac{0}{y}=0$$

$$\lim_{x^2=y^2-y,y\to 0} \frac{x^2 y}{x^2+y} = \lim_{y\to 0} \frac{y^3-y^2}{y^2}=-1$$

Since the limit differs along the two trajectories, then the limit does not exist.

The morale is: it is tricky and requires some experience to find a suitable trajectory, but at least you need to have the intuition that the limit does not exist. A useful observation is the presence of odd powers.

• I guess you want to remove the zero locus of $x^2+y=0$ from the reckoning (there is a typo on the first line). Other than that this looks good. – Jyrki Lahtonen Sep 13 '15 at 5:49

$f(x,y)=\frac{x^2y}{x^2+y}$

Suppose the limit exist and it is finite. Then, let's take $\epsilon > 0$. There is $\delta > 0$ so that $|\frac{x^2y}{x^2+y} - L|<\epsilon$ for all $x,y$ so that $x^2 + y^2 < \delta ^2$.

Let's consider $x_n= \frac{1}{\sqrt n}, y_n= -\frac{1}{n + 1}$. There is N so that $x_n^2 + y_n^2<\delta^2, \forall n > N$. We have $f(x_n,y_n)=-1$.

So $|-1 - L| < \epsilon$ for arbitrary $\epsilon$, therefore $L = -1$.

Now, consider $x_n= \frac{1}{\sqrt n}, y_n= -\frac{1}{n + 2}$. By following the same steps as above, we'll get $L = -\frac{1}{2}$.

It is proven that there is no finite limit. Similar for infinite limit

• I like it. I had never thought about taking sequences... – Miguel Sep 12 '15 at 19:36

Here's another approach $$\lim\limits_{(x,y)\to(0,0)}\frac{x^2y}{x^2+y}$$ Using polar coordinates, we have $$\lim\limits_{r\to 0^+}\frac{r^2\cos^2\phi\sin\phi}{r\cos^2\phi+\sin\phi}$$ Now lets attempt to find bounds that are independent of $\phi$ $$\frac{r^2\left|\cos^2\phi\sin\phi\right|}{\left|r\cos^2\phi+\sin\phi\right|}\leq\frac{r^2}{\left|r\cos^2\phi+\sin\phi\right|}$$ Since this limit is dependent on $\phi$, we can conclude that $$\lim\limits_{(x,y)\to(0,0)}\frac{x^2y}{x^2+y}=\mbox{non existent}$$