Find $\lim_{(x,y) \to (0,0)} \frac{\sin{(x^3 + y^5)}}{x^2 + y^4}$. Prove your result. Find $\lim_{(x,y) \to (0,0)} \frac{\sin{(x^3 + y^5)}}{x^2 + y^4}$.  Prove your result.
I've attempted to apply the Squeeze Theorem as such:
$\frac{-1}{x^2 + y^4} \leq \frac{\sin{(x^3 + y^5)}}{x^2 + y^4} \leq  \frac{1}{x^2 + y^4}$.  Clearly, though, the leftmost and rightmost functions of this inequality tend to $\infty$  as $(x,y) \to 0$, so this result does not help.

$$
\left|\frac{\sin(x^{2+\color{red}{n}} + y^{4+\color{blue}{m}})}{x^2 + y^4}\right|
\leqslant\left|\frac{x^{2+\color{red}{n}} + y^{4+\color{blue}{m}}}{x^2 + y^4}\right|
\leqslant\frac{|x|^\color{red}{n}\cdot x^2+|y|^\color{blue}{m}\cdot y^4}{x^2+y^4}
\leqslant|x|^\color{red}{n}+|y|^\color{blue}{m}
$$
