Using the line $y=x$ and the parabola $y=x^{2}$ in the calculation of $\lim_{\left ( x,y \right )\rightarrow \left ( 0,0 \right )}\displaystyle\frac{x^4 \sin(x^2 + y^2)}{x^4 + y^2}$ it turns out that this limit is $0$. But to affirm that $\lim_{\left ( x,y \right )\rightarrow \left ( 0,0 \right )}\displaystyle\frac{x^4 \sin(x^2 + y^2)}{x^4 + y^2}=0$, it is necessary to use the definition of limit, this is $\left | \displaystyle\frac{x^4 \sin(x^2 + y^2)}{x^4 + y^2} \right |< \varepsilon $, whenever $0< \sqrt{x^{2}+y^{2}}< \delta $.
How to find the $\varepsilon =\varepsilon \left ( \delta \right )$ relationship?
Yes. We see that \begin{align*} \left|\dfrac{x^{4}}{x^{4}+y^{2}}\sin(x^{2}+y^{2})\right|&\leq\dfrac{x^{4}(x^{2}+y^{2})}{x^{4}+y^{2}}\\ &\leq x^{2}+y^{2}\\ &<\delta^{2}\\ &<\epsilon, \end{align*} by choosing $\delta=\sqrt{\epsilon}$.