How do you show that $\lim_{(x,y)\to(0,0)} \frac{xy}{x^2+y}$ doesn't exist?

Question

I have to prove that this limit doesn't exist.

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

I tried this parametrization: $\begin{cases} x = t \\ y = mt^\alpha\end{cases}$

obtaining as result that the previous limit in this specific case would be equivalent to

$$\lim_{t\to0} \frac{mt}{t^{2-\alpha}+m}$$

which would be null for each value of $\alpha,m$.

Using a polar coordinate system doesn't seem effective too.

How do I prove that this doesn't exist?

Answer 1

Let $y=-x^2+x^4$. Then,

$$\frac{xy}{x^2+y}=-x^{-1}+x$$

What happens now?

Answer 2

As an alternative to the solution by Mark Viola, note that

• for $x=t \quad y=t \quad t\to 0 \implies \frac{xy}{x^2+y}=\frac{t^2}{t^2+t}=\frac{t}{t+1}\to0$

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

Answer 3

If $y=\frac {x^2}{x-1} $, the limit is

$$\lim_{(x,y)\to (0,0)}\frac {xy}{xy}=1$$

but if $x=\sqrt {y} $, it gives $$\lim_{(x,y)\to (0,0^+)}\frac {\sqrt {y}}{2}=0.$$ So, the limit cannot exist.

Answer 4

Let $$m=\frac{xy}{x^2+y}$$ Then $$m(x^2+y)=xy$$ so $$y=\frac{mx^2}{x-m}$$

For any $m$ we have an equation of a curve, which meets $(0,0)$ and on which the expression $\frac{xy}{x^2+y}$ is constant, equal $m$. This way there are infinitely many paths leading to the origin, each of which results in a different limit $m$.

Hence the limit does not exist.