The problem says: Find a $\delta$ such that $|f(x)-l| < \epsilon$ for all x satisfying $0 < |x-a| < \delta$ when $f(x) = x^4; l = a^4$.
What I did so far was $|x^4-a^4| < \epsilon$ so $|x^2+a^2||x+a||x-a| < \epsilon$ which means $|x-a|< {\epsilon\over{|x+a||x^2+a^2|}}$. I could make $|x-a|<1$ for convenience and say that $|x|-|a| \le |x-a| \lt 1 \rightarrow |x| < |a| + 1$.
Using $|x-a|< {\epsilon\over{|x+a||x^2+a^2|}}$ with $|x| < |a| + 1$, $|x-a| < {\epsilon\over{|(|a| + 1)+a||(|a| + 1)^2+a^2|}}$ so I can say $\delta = min(1,{\epsilon\over{|(|a| + 1)+a||(|a| + 1)^2+a^2|}})$.
Is this correct? I feel like I did something wrong with the inequalities when combining $|x-a|< {\epsilon\over{|x+a||x^2+a^2|}}$ with $|x| < |a| + 1$. Please correct me if I'm wrong and give me hints because if this doesn't work, I'm lost.