Does $\lim_ {(x,y)\to (0,0 )} \frac{x^3+y^3}{x^2 + y^2}$ exist?

Question

My solution is the following:

approaching by the y-axis:

$\lim_ {(y)\to (0),(x=0)} =\lim_ {(y) \to (0)}=\frac{0+y^3}{0^2+y^2}=y=0$

approaching by $y=x$

$\lim_ {(y)\to (0),(y=x)} =\lim_ {y=x}=\frac{x^3+x^3}{x^2+x^2}=\frac{2x^3}{2x^2}=x=0$

So I think,that this limit exists. is it correct in this form?

Answer 1

HINT: $$ \frac{x^3+y^3}{x^2+y^2}=x\frac{x^2}{x^2+y^2}+y\frac{y^2}{x^2+y^2} $$

But your method doesn't answer the question.

Answer 2

Try coming at it from any other direction, e.g. y = mx.

Ced

Answer 3

Hint:

Use polar coordinates: $(x,y)\to (0,0)\iff r\to 0$. In polar coordinates, thexpression becomes $$\frac{x^3+y^3}{x^2+y^2}=r(\cos^3\theta+\sin^3\theta).$$

Answer 4

Since a limit is being taken as $(x,y) \to (0,0)$ we may assume that $(x,y) \neq (0,0)$ during the approach, and so write $x=r \cos t, y=r \sin t.$ [Here $r>0$ is unique, and $t$ is determined mod $2\pi.$] Then your function is $r \cdot (\cos ^3 t+ \sin ^3 t),$ whose magnitude is less than $2r$ and so goes to $0$ as $(x,y) \to (0,0)$ through values other than $(0,0).$