It's been a while, but I think I remember this one. Lim 1/n =0 as n approaches infinity. Let x^0 be undefined. For any e>0 there exists an n such that |x^(1/n) -1| < e. If you desire x^(1/n) to be continuous at 0, you define x^0 as 1.

E2a: since x^(1/n)>1, you can drop the abs bars. I think you can get an inequality to pick n using logs.

Where are my programmer buddies? 🤘

I hate that I get this