Show $\lim_{(x, y) \to (0, 0)} \frac{x^2 + y^2}{y}$ does not exist

Question

I can see that the function $f(x, y) = \frac{x^2 + y^2}{y}$ is not defined wherever $y = 0$, however looking at the cone shaped graph of the function it looks like approaching (0,0) from any direction will give $f(x, y) = 0$.

But according to "everything" this limit does not exist, however I have (even after looking at a lot of similiar problems, however most of them with both $x$ and $y$ in the denominator) not found a way to show it yet. any hint? thanks in advance :)

Answer 1

$$\begin{align*}&(1)\;\;\;x=y\implies\lim_{x\to0}\frac{x^2+x^2}x=\lim_{x\to0}2x=0\\{}\\&(2)\;\;\;x=\sqrt y\implies\lim_{y\to0^+}\frac{y+y^2}y=\lim_{y\to0^)}(1+y)=1\end{align*}$$

Answer 2

Simply note that

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