I need to prove that
$$\lim_{x\to a} x^4 = a^4 $$
for $a$ an arbitrary value so is it alright when I am into the proof allowing me to find what each is strictly less than! and then allowing me to write that delta is equivalent to $$\min\left(\frac{\varepsilon}{2},1\right)$$
First if $a=0$ one only needs that $|x|<\delta$ implies $|x^4|<\varepsilon$ achieved by $\delta=\varepsilon^{\frac14}.$
Now suppose $a>0$ and assume that $\delta<1$ (which will be put with another inequality to get the final $\delta$). If we factor $|x^4-a^4|$ we have one factor as $|x-a|$ and the other is $$|x^3+x^2a+xa^2+a^3|,\tag{1}$$ which needs bounding. From $|x-a|<1$ we have $a-1<x<a+1$ and since in this case $a>0$ we know $|x|<a+1.$ Now using the triangle inequality on $(1)$ and properties of absolute value, we get that the factor in $(1)$ is bounded above by $4\cdot (a+1)^3.$ [Here have also used $a<a+1$.]
So if we now choose $$\delta = \min [ 1, \frac{\varepsilon}{4\cdot (a+1)^3}],$$ we can show from the above by tracing the steps backward that from $|x-a|<\delta$ it follows that $|x^4-a^4|<\varepsilon.$
There is also the case $a<0$ to treat, for which the inequality $a-1<x<a+1$ from above leads to the different looking upper bound $|a-1|$ for $|x|$, but otherwise this case should run similarly.
This is equivalent to showing that $f(x) = x^4$ is continuous at the point $a$. It is easy to show that $g(x) = x$ is continuous, just let $\delta = \epsilon$. Now just use the fact (or prove it for yourself) that a product of continuous functions is continuous and use induction.
