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?

  • 1
    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
up vote 3 down vote accepted

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 $$\lim\limits_{(x,y)\to(0,0)}\frac{x^2y}{x^2+y}=\mbox{non existent}$$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.