Use definition to verify $$\lim_{\vec{x}\to \vec{0}} \frac{7x_1^2x_2^4-9x_1^3x_3^3}{(x_1^2+2x_2^2+4x_3^2)^2}=0$$
My Attempts:
Let $\epsilon\gt0$ be given, we need to find $\delta$ such that if $0\lt \sqrt{x_1^2+x_2^2+x_3^2}\lt \delta$ then $| \frac{7x_1^2x_2^4-9x_1^3x_3^3}{(x_1^2+2x_2^2+4x_3^2)^2}-0|\lt \epsilon$. Note that if $0\lt \sqrt{x_1^2+x_2^2+x_3^2}\lt \delta$ , then $|x_1|\lt\delta,|x_2|\lt\delta,|x_3|\lt\delta$, take $\delta=\min\{\frac{\epsilon}{16},1\}$, then $$|\frac{7x_1^2x_2^4-9x_1^3x_3^3}{(x_1^2+2x_2^2+4x_3^2)^2}|\lt |7x_1^2|\cdot|\frac{x_2^4}{(x_1^2+2x_2^2+4x_3^2)^2}|+|9x_1^3|\cdot|\frac{x_3^3}{(x_1^2+2x_2^2+4x_3^2)^2}|\lt7\delta^2+9\delta^3\lt16\delta\lt \epsilon $$
So $$\lim_{\vec{x}\to \vec{0}} \frac{7x_1^2x_2^4-9x_1^3x_3^3}{(x_1^2+2x_2^2+4x_3^2)^2}=0$$
Have I make a rigorous proof?