Solve multivariable limit $$\lim_{(x,y) \to (0,0)} \frac{x^3 + y^4}{x^2 + y^2}$$ 

I am almost sure it is equal to $0$ but I can't prove it.
Please give me some hint.
Use this inequality

$$0\le\left|\frac{x^3 + y^4}{x^2 + y^2}\right|\le\frac{|x|x^2 + |y|^2|y|^2}{x^2 + y^2}\le|x|+|y|^2\xrightarrow{(x,y)\to(0,0)}0$$
A more Rigerous solution one could show, which would be expected in Real Analysis. 

If the multivariate limit F(x,y) -> 0 for x and y approaching 0 ,then it must be true that {Xn} -> 0 and {Yn} -> 0 so that F({Xn}, {Yn}) -> 0 for all such sequences {Xn}, {Yn} that approach 0.  Examples of single variable sequences that approach 0 are {1/n}, {1/n^2}, etc. as n approaches infinity. 

In other words if there exist two sequences that both approach 0 as n goes to infinity such that F({Xn}, {Yn}) = 0, but then there exists other sequences (one is sufficient) such that F({Xn}, {Yn}) = L where L is not equal to 0.  Then F must be divergent, because it is impossible for a converging sequence to have more than one limit point.  

Let us have {Xn} = {1/n} and {Yn} = {1/n}, this implies

F({Xn}, {Yn}) =  F({1/n}, {1/n}) = lim n -> infinity ( (n+1)/2n^2 ) -> 0

So it was shown for two sequences, it would get very tiring to show all sequences approach 0.  The trick would be to assume that for two sequences approaching 0, that F({Xn}, {Yn}) -> L not equal to 0, and then show a contradiction arises if you do so.  By contradiction assume  there are two sequences that both approach 0 such that {Xn} -> 0 and {Ym} = {0} -> 0 for n and m approaching infinity.  (Yes {Ym} is constant), and 

F({Xn}, {Ym}) = L not equal to 0, which implies

lim n, m -> infinity ( ({Xn}^3 + 0) / ({Xn}^2 + 0) ) = lim n,m -> infinity ( {Xn} ) = L not equal to 0.  But we asserted that {Xn} -> 0, so a contradiction has arisen.  This must mean that F({Xn}, {Ym}) -> 0.

Q.E.D.