Multivariable limit exists? Does the limit 
$$\lim_{(x,y)\rightarrow (0,0)} \frac{y^4}{x^\beta(x^2+y^4)}$$ exists for $\beta>0$? I don't think it exists but how do you prove it rigorously. Thanks 
 A: Approach the origin along the x-axis, where $y=0$. Then the limit along this direction is zero. Now approach the origin along the parabola $x=y^2$. This is equivalent to taking the limit $\lim_{x\to 0} \frac{x^2}{x^\beta(2x^2)}=\lim_{x\to 0} \frac{1}{2x^\beta}$. Since $\beta>0$, this limit will be infinite. Since the limit depends on the direction that you approach the origin, it does not exist. 
A: $$\lim\limits_{(x,y)\to (0,0)} \frac{y^4}{x^\beta\left(x^2+y^4\right)}$$
Using polar coordinates, we have
$$\lim\limits_{r\to 0^+} \frac{(r\sin\phi)^4}{(r\cos\phi)^\beta\left((r\cos\phi)^2+(r\sin\phi)^4\right)}$$
$$=\lim\limits_{r\to 0^+} \frac{r^4\sin^4\phi}{r^\beta\left(\cos^\beta\phi\right)\left(r^2\cos^2\phi+r^4\sin^4\phi\right)}$$
$$=\lim\limits_{r\to 0^+} \frac{r^4\sin^4\phi}{r^\beta r^2\left(\cos^\beta\phi\right)\left(\cos^2\phi+r^2\sin^4\phi\right)}$$
$$=\lim\limits_{r\to 0^+} \frac{r^2\sin^4\phi}{r^\beta\left(\cos^\beta\phi\right)\left(\cos^2\phi+r^2\sin^4\phi\right)}$$
$$=\lim\limits_{r\to 0^+} \frac{\sin^4\phi}{r^{\beta-2}\left(\cos^\beta\phi\right)\left(\cos^2\phi+r^2\sin^4\phi\right)}$$
Now let's attempt to find a bound that is independent of $\phi$
$$\left|\frac{\sin^4\phi}{\left(\cos^\beta\phi\right)\left(\cos^2\phi+r^2\sin^4\phi\right)}\right|\leq \left|\frac{1}{\left(\cos^\beta\phi\right)\left(\cos^2\phi+r^2\sin^4\phi\right)}\right| $$
The right hand side is unbounded, which implies that the limit is dependent on $\phi$. Therefore 
$$\lim\limits_{(x,y)\to (0,0)} \frac{y^4}{x^\beta\left(x^2+y^4\right)}\Rightarrow \mbox{does not exist}$$
