- #1

duarthiago

- 11

- 0

## Homework Statement

Evaluate ##lim_{(x,y)\to(0,0)\frac{xy}{y-x^3}}##

## Homework Equations

## The Attempt at a Solution

I've tried move towards the origin along the path ##\gamma_{1} (t) = (t,2t^3)##, which leads to ##2t=0##. Then I've tried to use ##\gamma_{2} (t) = (0,t)## which leads to ##\frac{0}{t}##, then applying the L'Hospital's rule this limit is 0 too (can I always do it when I have a single-variable limit?).

I've suspected that the limit exists and its value is 0, then I've tried to use polar coordinates:

##lim_{(x,y)\to(0,0)\frac{xy}{y-x^3}}##

##lim_{r \to 0^{+}} \frac{r^2 cos(k)sin(k)}{r(sin(k)-r^2 cos^3 (k))}##

But I've got stuck there because I can't figure how to show that ##\frac{cos(k)sin(k)}{sin(k)-r^2 cos^3 (k)}## is bounded.

**A further question**: It is related to the question on L'Hospital's rule, above. In regards to the limit

##lim_{(x,y) \to (0,0)} \frac{xy(x-y)}{x^4 + y^4}##

The path ##\gamma_{A} (t) = (t,0)## leads to 0.

##\gamma_{B} (t) = (t,2t)## leads to ##\frac{-2t^3}{17t^4}=\frac{-2}{17t}##. Can I tend t to zero, conclude that the limit goes to the minus infinity and because of that say that the original limit doesn't exist? Maybe it seems a silly question, but until now I haven't seen something like this happening. It is always something bounded multiplied by zero.