How to calculate $\lim _{( x,y) \to (0,0) }\frac {x^2y^{3} }{\sin ( x^{8}+y^{4}) }$? How to calculate
$$\lim _{( x,y) \rightarrow (0,0)  }\dfrac {x^2y^{3} }{\sin ( x^{8}+y^{4}) }$$
Wolfram Alpha says it is equal to $0$, but if calculate by trying $y=x$ then I get the limit $\lim \limits_{ x \to 0  }\dfrac {x^5 }{\sin ( x^{8}+x^{4}) }$ which it says it's equal to $0$.
Now if I try $y=x^2$ then I get the limit  $\lim\limits_{ x \to 0  }\dfrac {x^8 }{\sin ( 2x^{8}) }$ which is equal to $\frac12$.
Which one is true, is there no limit or does limit equal to $0$?
 A: This is a good example for why one should not blindly trust Wolfram Alpha when it comes it multidimensional limits. Such limits are hard to do for a computer as the result can depend on the path taken towards the limit-point and there are infinitely many paths to approach any given point. When the limit does not exist the standard solution method is to find different paths that leads to different limits. What works in any given case if often based on experience, i.e. just 'seeing' what are some good paths to try. This type of approach is hard to implement as an algorithm on a computer. 
Wolfram Alpha gets this one wrong. To see this write the limit as
$$\lim_{(x,y)\to (0,0)} \frac{x^2y^3}{x^8+y^4} \cdot \frac{x^8+y^4}{\sin(x^8+y^4)} = \lim_{(x,y)\to (0,0)} \frac{x^2y^3}{x^8+y^4} \cdot \lim_{(x,y)\to (0,0)} \frac{x^8+y^4}{\sin(x^8+y^4)}$$
The latter limit is $1$ as $x^8+y^4\to 0$ as $(x,y)\to 0$ and $\lim\limits_{z\to 0}\frac{z}{\sin(z)} = 1$. The former limit does not exist as we can see by taking the limit along curves $x=k\sqrt{y}$ then
$$\lim_{(x,y)\to (0,0)} \left.\frac{x^2y^3}{x^8+y^4}\right|_{x=k\sqrt{y}} = \lim_{y\to 0} \frac{k^2y^4}{k^8y^4 + y^4} = \frac{k^2}{k^8 + 1}$$
and since the limit is path dependent (it depends on the value we take for $k$) it follows that $\lim\limits_{(x,y)\to (0,0)} \frac{x^2y^3}{x^8+y^4}$ does not exist.
