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,
$$0< |x| < \delta \implies |x| < \dfrac{1}{\sqrt[4]{M}} \implies |x^4| < \dfrac{1}{M} \implies |\dfrac{1}{x^4}| > M $$