Showing $\lim_{x\rightarrow0}\frac{1}{x^4}=\infty$ via $\epsilon$-$\delta$ 
Show that $\lim_{x\rightarrow0}\frac{1}{x^4}=\infty$.

Consider our preliminary work:
$$\begin{align}
|\frac{1}{x^4}-0|<\epsilon &\implies|x^4-0|<\frac{1}{\epsilon}  \tag1\\
&\implies|x^2-0||x^2+0|<\frac{1}{\epsilon} \tag2\\
&\implies|x-0||x+0||x^2+0|<\frac{1}{\epsilon} \tag3
\end{align}$$
Choose $\delta=1$, then
$$|x-0|<\delta=1\implies -1<x<1\Rightarrow|x-0|<1 \tag4$$
Thus,
$$|x-0||x+0||x^2+0|<\frac{1}{\epsilon}\implies|x-0|<\frac{1}{\epsilon} \tag5$$ Now, let $\epsilon>0$ and choose $\delta= \min\{1,\frac{1}{\epsilon}\}$.
What should I do next in writing this proof? I'm confused how to choose $\delta$ so that  $\frac{1}{\epsilon}$ becomes $\epsilon$. Also, do I need to pick multiple upper bounds since I broke my polynomial into three separate absolute values?

Let $M$ be given. Then, take $\delta = \dfrac{1}{\sqrt[4]{M}}$. Thus,
$$0< |x| < \delta \implies |x| < \dfrac{1}{\sqrt[4]{M}} \implies |x^4| < \dfrac{1}{M} \implies |\dfrac{1}{x^4}| > M $$
