How prove $\lim_{x\to 0,y\to 0}\frac{2y^2x}{y^4+x^3}=0$

Question

prove or disprove $$\lim_{x\to 0,y\to 0}\dfrac{2y^2x}{y^4+x^3}=0？$$

consider

$$x^3+y^4>x^4+y^4,(x,y)\to (0,0)$$ so $$0\le |\dfrac{2y^2x}{x^3+y^4}|\le\dfrac{2y^2x}{x^4+y^4}|\le |\dfrac{2y^2x}{2x^2y^2}|=|\dfrac{1}{x}|$$

then I can't

Answer 1

$$\lim_{ (x,y) \to (0,0)}\dfrac{2y^2x}{y^4+x^3}$$

Consider the approach along $y = kx$

$$\lim_{x \to 0}\dfrac{2k^2x^3}{k^4x^4+x^3}$$ $$\lim_{x \to 0}\dfrac{2k^2}{k^4x+1}$$ $$2k^2$$

So you can see that $\lim_{ (x,y) \to (0,0)}$ depends on the direction of approach $k$ and therefore doesn't exist. (If it didn't depend on $k$, you would need to check a more general $y = f(x)$ to see if the limit exists.)

As for $\lim_{ x \to 0}\left(\lim_{y \to 0} \left( \dots \right)\right)$, see Mark Fischler's answer.

Answer 2

The limit isn't $0$. Consider the limit along the line $y=x$. The limit along this line is $2$.

Answer 3

And along the line $y=x^{\frac{2}{3}}$ the limit is infinite. Perhaps you mean $$ \lim_{x\rightarrow 0}\left( \lim_{y\rightarrow 0}\frac{2y^2x}{y^4+x^3} \right) $$ which is $$ \lim_{x\rightarrow 0}\left(0\right) = 0 $$