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.
Of course, your explanation is the “correct” one - why it’s possible that x^0=1. Mine is the simple version that shows how logic checks out using algebraic rules.
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.
Simpler: x^1 = x, x^-1 = 1/x
x^1 * x^-1 = x^0 = x/x = 1.
Of course, your explanation is the “correct” one - why it’s possible that x^0=1. Mine is the simple version that shows how logic checks out using algebraic rules.
Of course you both assume x =/= 0 though.