Limit of $f(x)=x^4$ as $x\to a$ by definition 
$\displaystyle\lim_{x\to a} x^4=a^4$

Note that $|x^4-a^4|=|x-a||x+a||x^2+a^2|$
If $|x-a|<1,$ then $|x+a|<1+2a.$
Then, $|x-a|<1$ implies $|x^4-a^4|<(1+2a)(|x^2+a^2|)$
Also, if $|x^2+a^2|<\frac{\varepsilon}{1+2|a|}$, then $|x^3-a^4|<\varepsilon$. Consequently, taking $\delta =min\left(min\left(1,1+2a\right),\frac {\varepsilon}{1+2|a|}\right)$ would work.
Is this correct? Is my choosing of $\delta$ ok?

Assume $a>0$.Taking $|x-a|<\delta$ gives $0<a-\delta < x< a+\delta$, when $\delta < a$. So $0<x+a=|x+a|<2a+\delta<3a$.
Analogically $0< x^2+a^2 =|x^2+a^2|<(a+\delta)^2+a^2<5a^2$. Now you need $|x^4-a^4|<\delta 15a^3<\varepsilon$ . Can you finish?
