What is $\lim_{(x,y)\to(0,0)} \dfrac{x^3 y^4}{x^6+y^8}$? Why is this limit $0$. It seems to me that  $y^4/(x^6+y^8)$ is a non limited function and so is $x^3/(x^6+y^8)$. 
$$\lim_{(x,y)\to(0,0)} \dfrac{x^3 y^4}{x^6+y^8}$$

Note that $\displaystyle \frac{x^3}{x^6+y^8}\cdot \frac{y^4}{x^6+y^8}\neq \frac{x^3y^4}{x^6+y^8}.$  
What I would like to hint is: Compare the limits along the paths $x=y$ and $x^3=y^4$. That should do the trick.
