this post was submitted on 11 Oct 2023
986 points (96.3% liked)
Memes
45728 readers
1334 users here now
Rules:
- Be civil and nice.
- Try not to excessively repost, as a rule of thumb, wait at least 2 months to do it if you have to.
founded 5 years ago
MODERATORS
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.
load more comments
(2 replies)