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

Question

Find the limit of $y+\frac{x^2}{y}$ when $(x,y) \to 0$.

I did polar coordinates, and got $\frac{r}{\sin(\theta)} \to0$ when $r \to 0$ since $-1 \le \sin(\theta)\le1$.

@Nameless that's right. You got me. – jacob Apr 20 '14 at 07:50 — jacob, Apr 20 '14 at 07:50

score 3 · Answer 1 · answered Apr 20 '14 at 07:30

3

The limit does not exist. $(x,y) = (t,t^2)$ and $(x,y) = (t,t)$ then the first case gives $1$ while the second gives $0$.

answered Apr 20 '14 at 07:30

DeepSea

77,651

score 1 · Answer 2 · answered Apr 20 '14 at 07:31

Take $x=\sqrt{a}\sqrt{y}$, for $a>0$ and $y>0$, then calculates the limite, you will find $$\lim_{x,y\to 0} y+\frac{x^2}{y} = \lim_{y\to 0^+} y+\frac{\sqrt{a}^2\sqrt{y}^2}{y} = \lim_{y\to 0^+} y+\frac{a|y|}{y} = a$$

Taking some $b>0, b\neq a$, and using the same argument, you can conclude that this limit also converges to $ b$. Therefore, you can find different values if you calculates from different paths. So, this limits does not exists.

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

2 Answers2