Find $\lim_{(x,y)\to(0,0)}{\frac{x^2y}{y+x^2}}$

Question

I need to find $$\displaystyle\lim_{(x,y)\to(0,0)}{\dfrac{x^2y}{y+x^2}}.$$

I tried with iterated limits, $y=mx,\;\;y=kx^2$ but everyone throws me that the limit is $0$.

I have been told that I have to see the contour lines, but I did not understand completely how to do it. For example

$C_0=\left\{(x,y)\in\mathbb R^2\mid \dfrac{x^2y}{x+y^2}=0\right\}=\{y=0\;\wedge\;(x,y)\neq (0,0)\}$, or

$C_1=\left\{(x,y)\in\mathbb R^2\mid \dfrac{x^2y}{x+y^2}=1\right\}=\left\lbrace y=\dfrac{x^2}{x^2-1}\;\wedge\;x\neq 1\;\wedge\;x\neq -1\right\rbrace$.

I think that in $ C_1 $ the point goes through the curve, and the curve exists. What does that tell me?

So how do I prove (or not) the existence of the limit?

Thanks!

Answer 1

If $y>0$ then $$\bigg|\frac{x^2y}{x^2+y} \bigg|\leq \bigg|\frac{x^2y}{y} \bigg|=x^2$$ Now checking $y<0$ one can see that for $y$ in the neighbourhood of $-x^2$ that the denominator goes to $0$ independent of the numerator.

If we write $y=-x^2+\epsilon$ then we get $$\lim_{(x,\epsilon)\to(0,0)}\frac{x^2(\epsilon -x^2)}{\epsilon}=\lim_{(x,\epsilon)\to(0,0)}x^2+\frac{-x^4}{\epsilon}=\lim_{(x,\epsilon)\to(0,0)}\frac{-x^4}{\epsilon}$$ So we can see that $\epsilon=kx^4$ gives a different value for the limit so it doesn't exist.

Answer 2

The limit does not exist.

In order for such limits to exist the denominator must have even exponents on $x$ and $y$. For example in your case let $$y=x^k-x^2$$ for any very large value of $k$. This will give (minus) infinity as a limit.

Answer 3

Note that

• for $x=y=t\to 0\implies{\dfrac{x^2y}{y+x^2}}=\frac{t^3}{t+t^2}=\frac{t^2}{1+t}\to 0$
• for $x=t \quad y=t^5-t^2 \quad t\to 0^+\implies{\dfrac{x^2y}{y+x^2}}=\frac{t^{7}-t^4}{t^5-t^2+t^2}=\frac{t^{7}-t^4}{t^5}=\frac{t^2-1}{t}\to -\infty$

thus the limit doesn't exist.