Proving that the limit at $(0,-1)$ does not exist

Question

I have the following function:

$$f(x,y) =\begin{cases} \frac {x^4 + (y+1)^3}{x^2 + (y+1)^2} &\text{ if }(x,y) \neq (0,-1)\\ 0 &\text{ if }(x,y) = (0,-1)\end{cases}$$

Prove that the limit of $f$ at $(0,-1)$ does not exists

I have been trying to prove this limit does not exist by composing $f(\alpha (t)) $ using different curves $\alpha (t)$ but I can't seem to find any that can prove it. Polynomials and circles don't seem to work. What else could I try?

Thanks in advance

Answer 1

If I'm not mistaken, the limit seems to exist:

$$ 0 \leq |f(x,y)| = \left|\dfrac{x^4}{x^2+(y+1)^2} + \dfrac{(y+1)^3}{x^2+(y+1)^2}\right| \leq x^2 + |(y+1)|$$

The right hand side goes to $0$ in the limit.

Answer 2

Note that wlog by a change of coordinates we can study $(x,y)\to(0,0)$ and by polar coordinates

$$\frac {x^4 +y^3}{x^2 + y^2}=r\cdot f(r,\theta)\to 0$$

since $f(r,\theta)=r\cos^4 \theta+\sin^3 \theta$ is bounded.